Differential Equations

$ \begin{aligned} &\text { Equation of circle is }\\ &\begin{aligned} & (x-\alpha)^2+(y-\beta)^2=a^2 \\ & \Rightarrow \quad(x-\alpha)+2(y-\beta) \times y_1=0 \\ & \Rightarrow \quad x-\alpha+(y-\beta) \times y_1=0 \\ & \Rightarrow \quad(y-\beta)^2\left(1+y_1^2\right)=a^2 \\ & \Rightarrow \quad(y-\beta)^2=\frac{a^2}{1+y_1^2} \\ & \Rightarrow \quad y-\beta=\frac{a}{\sqrt{1+y_1^2}} \\ & \Rightarrow \quad y-\frac{a}{\sqrt{1+y_1^2}}=\beta \\ & \Rightarrow \quad y_1=\left(\frac{-a \times \frac{1}{2 \sqrt{1+y_1^2}} \times 2 y_1 y_2}{\left(1+y_1^2\right)}\right)=0 \\ & \Rightarrow \quad y_1+\frac{a y_1 y_2}{\left(1+y_1^2\right)^3}=0 \\ & \Rightarrow \quad \frac{a^2 y_1^2 y_2^2}{\left(1+y_1^2\right)^3}=y_1^2 \\ & \Rightarrow \quad a^2 y_2^2=\left(1+y_1^2\right)^3 \end{aligned} \end{aligned} $

$ \begin{aligned} & \text { } \begin{aligned} & \frac{d x}{d y}+x \cdot \frac{1}{1+y^2}=\frac{\tan ^{-1} y}{1+y^2} \\ & \quad \text { IF }=e^{\int \frac{d y}{1+y^2}} \\ &=e^{\tan ^{-1} y} \\ & x e^{\tan ^{-1} y}=\int \frac{e^{\tan ^{-1} y} \cdot \tan ^{-1} y}{1+y^2} d y \\ & \Rightarrow x e^{\tan ^{-1} y}=\int e^\theta \cdot \theta d \theta \quad[\text { put } y=\tan \theta] \\ & \Rightarrow x e^{\tan ^{-1} y}=\theta \int \theta d \theta-\int\left(\frac{d \theta}{d \theta} \int e^\theta d \theta\right) d \theta \\ & \Rightarrow \theta e^\theta-e^\theta+C \\ & \Rightarrow x e^{\tan ^{-1} y}=\tan ^{-1} y e^{\tan ^{-1} y}-e^{\tan ^{-1} y}+C \\ & \Rightarrow x=\tan ^{-1} y-1+c e^{-\tan ^{-1} y} \end{aligned} \end{aligned} $

So, $f(y)=\tan ^{-1} y-1$

$ \begin{aligned} & \text { } f^{\prime}(x)-\frac{4 \sin 2 x}{1+\cos ^2 x}-\frac{f(x) \sin 2 x}{1+\cos ^2 x}=0 \\ & f^{\prime}(x)-\frac{f(x) \sin 2 x}{1+\cos ^2 x}=\frac{4 \sin 2 x}{1+\cos ^2 x} \\ & \text { Let } y=f(x) \\ & \Rightarrow \quad \frac{d y}{d x}-\frac{y \sin 2 x}{1+\cos ^2 x}=\frac{4 \sin 2 x}{1+\cos ^2 x} \end{aligned} $

$ \begin{array}{ll} \Rightarrow & y \times \mathrm{IF}=\int(Q \times \mathrm{IF}) d x \\ \Rightarrow & \mathrm{IF}=e^{-\int \frac{\sin 2 x d x}{1+\cos ^2 x}} \\ \Rightarrow & \mathrm{IF}=1+\cos ^2 x \\ \Rightarrow & y \times\left(1+\cos ^2 x\right)=\int \frac{4 \sin 2 x}{1+\cos ^2 x} \times 1+\cos ^2 x d x \end{array} $

$ \begin{array}{cc} \Rightarrow & y\left(1+\cos ^2 x\right)=4 \int \sin 2 x d x \\ \Rightarrow & y\left(1+\cos ^2 x\right)=-2 \cos 2 x+C \\ & y(0)=0 \\ & 0(1+1)=-2+C \Rightarrow C=2 \\ \Rightarrow & y\left(1+\cos ^2 x\right)=2(1-\cos 2 x) \\ \Rightarrow & y(x)=\frac{4 \sin ^2 x}{1+\cos ^2 x}=f(x) \\ \Rightarrow & f\left(\frac{\pi}{3}\right)=\frac{4 \sin ^2 60^{\circ}}{1+\cos ^2 60^{\circ}} \\ & =\frac{4 \times \frac{3}{4}}{1+\frac{1}{4}} \\& = \frac{12}{5} \end{array} $

$ \begin{aligned} & \text { } \frac{x^2}{a^2}+\frac{y^2}{4}=1 \\ & \frac{2 x}{a^2}+\frac{2 y \cdot y^{\prime}}{4}=0 \end{aligned} $

(differentiating both sides)

$ \begin{aligned} & \frac{x}{a^2}=\frac{-y y^{\prime}}{4} \\ \Rightarrow \quad & \frac{1}{a^2}=-\frac{y y^{\prime}}{4 x} \end{aligned} $

So, differential equation of such ellipse.

$ \begin{array}{rlrl} & x^2\left(\frac{-y y^{\prime}}{4 x}\right)+\frac{y^2}{4}=1 \\ \Rightarrow & \frac{-x y y^{\prime}}{4}+\frac{y^2}{4}=1 \\ \Rightarrow & \frac{y^2-4}{4}=\frac{x y y^{\prime}}{4} \\ \Rightarrow & x y \frac{d y}{d x}=y^2-4 \end{array} $

$ \begin{aligned} &\text { We have, }\\ &\begin{aligned} & \frac{d y}{d x}+(\sec x \cdot \operatorname{cosec} x) y=\cos ^2 x \\ & \text { IF }=\mathrm{e}^{\int \sec x \operatorname{cosec} x d x} \end{aligned} \end{aligned} $

$ \begin{aligned} &\begin{aligned} & =e^{\int \frac{2}{\sin 2 x} d x}=e^{2 \int \operatorname{cosec} 2 x d x} \\ & =e^{2 \frac{\ln |\operatorname{cosec} 2 x-\cot 2 x|}{2}} \\ & =(\operatorname{cosec} 2 x-\cot 2 x) \\ & =\frac{1-\cos 2 x}{\sin 2 x}=\frac{2 \sin ^2 x}{2 \sin x \cos x} \\ & =\frac{\sin x}{\cos x} \end{aligned}\\ &\text { Solution }\\ &\begin{aligned} y(\mathrm{IF}) & =\int \cos ^2 x(\mathrm{IF}) d x+k \\ \Rightarrow y \tan x & =\int \sin x \cos x d x+k \\ \Rightarrow y \tan x & =\frac{\sin ^2 x}{2}+k \\ \Rightarrow 2 y \tan x & =\sin ^2 x+C \end{aligned} \end{aligned} $

$ \begin{aligned} & \text { } y=A e^x+B \sin x \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)\\ & \frac{d y}{d x}=A e^x+B \cos x \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)\\ & \frac{d^2 y}{d x^2}=A e^x-B \sin x \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(iii)\end{aligned} $

From Eqs. (i) and (iii),

$ \frac{1}{2}\left(y+\frac{d^2 y}{d x^2}\right)=A e^x $

From Eqs. (i) and (ii),

$ B=\left(y-\frac{d y}{d x}\right) \frac{1}{\sin x-\cos x} $

∴ From Eq. (iii),

$ \begin{array}{r} \Rightarrow \quad \frac{d^2 y}{d x^2}=\frac{1}{2}\left(y+\frac{d^2 y}{d x^2}\right)-\left(y-\frac{d y}{d x}\right) \frac{\sin x}{\sin x-\cos x} \end{array} $

$ \begin{aligned} & \Rightarrow \frac{1}{2} \frac{d^2 y}{d x^2}=\frac{y}{2}-\frac{\sin x}{\sin x-\cos x}\left(y-\frac{d y}{d x}\right) \\ & \begin{array}{r} \Rightarrow(\sin x-\cos x) \frac{d^2 y}{d x^2}=(\sin x-\cos x) y \\ \quad-2 \sin x \cdot y+2 \sin x \frac{d y}{d x} \end{array} \\ & \Rightarrow(\sin x-\cos x) \frac{d^2 y}{d x^2}-2 \sin x \frac{d y}{d x} +(2 \sin x-\sin x+\cos x) y=0 \\ & \Rightarrow(\sin x-\cos x) \frac{d^2 y}{d x^2}-2 \sin x \frac{d y}{d x} +(\sin x+\cos x) y=0 \\ & \begin{array}{r} \therefore f(x)+g(x)+h(x) \\ =\sin x-\cos x-2 \sin x+\sin x+\cos x \\ \end{array} \end{aligned} $

$ =0 $

Let centre $(h, k)$ lies on $y=2 x$

$ \begin{array}{ll} \Rightarrow & k=2 h \\ \therefore & \frac{(x-h)^2}{a^2}-\frac{(y-2 h)^2}{b^2}=1 \\ \Rightarrow & \frac{(x-h)^2}{a^2}-\frac{(y-2 h)^2}{2 a^2}=1 \\ & {\left[\because e=\sqrt{3} \Rightarrow b^2=2 a^2\right]} \end{array} $

On differentiating w.r.t. $x$, we get

$ \begin{aligned} & 4(x-h)-2(y-2 h) \frac{d y}{d x}=0 \\ \Rightarrow \quad & (y-2 h) y_1=2(x-h) \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)\end{aligned} $

Again, differentiating,

$ (y-2 h) y_2=2-y_1^2 $

From Eq. (i), $h=\frac{2 x-y y_1}{2\left(1-y_1\right)}$

∴ Differentiating equation is

$ \begin{aligned} & \left(y-2\left(\frac{2 x-y y_1}{2\left(1-y_1\right)}\right)\right) y_2=2-y_1^2 \\ \Rightarrow \quad & \left(y-y y_1-2 x+y y_1\right) y_2=2+y_1^3 -2 y_1-y_1^2 \\ \Rightarrow \quad & (y-2 x) y_2+y_1^2+2 y_1=y_1^3+2 \end{aligned} $

Given, $\left(x^3-y^3\right) d x=\left(x^2 y-x y^2\right) d y$

$ \begin{aligned} & \frac{d y}{d x}=\frac{x^3-y^3}{x^2 y-x y^2} \\ & \frac{d y}{d x}=\frac{x^2+x y+y^2}{x y} \end{aligned} $

Homogeneous differential equation

Put $\quad y=v x$

$ \begin{aligned} & \frac{d y}{d x}=v+x \frac{d v}{d x} \\ \Rightarrow & v+\frac{x d v}{d x}=\frac{1+v+v^2}{v} \\ \Rightarrow & x \frac{d v}{d x}=\frac{1+v}{v} \\ \Rightarrow & \frac{v}{v+1} d v=\frac{d x}{x} \\ \Rightarrow & \frac{v+1-1}{v+1} d v=\frac{d x}{x} \end{aligned} $

On taking integration both sides, we get

$ \begin{aligned} & \int \frac{v+1}{v+1} d v-\int \frac{1}{v+1} d v=\int \frac{d x}{x} \\ & v-\ln |v+1|=\ln x+\ln c \\ \Rightarrow & v=\ln |x(v+1) c| \\ \Rightarrow & \frac{y}{x}=\log [(y+x) c] \\ \Rightarrow & y=x \log (c|x+y|) \end{aligned} $

Given, $t^2 d x+\left(x^2-t x+t^2\right) d t=0$

$ \Rightarrow \quad \frac{d t}{d x}=-\frac{t^2}{x^2-t x+t^2} $

On putting $t=V x$

$ \begin{aligned} & \Rightarrow \quad \frac{d t}{d x}=V+x \frac{d V}{d x} \\ & \therefore \quad\left(V+x \frac{d V}{d x}\right)=-\frac{V^2 x^2}{x^2-V x^2+V^2 x^2} \\ & \Rightarrow \quad V+x \frac{d V}{d x}=-\frac{V^2}{1-V+V^2} \\ & \Rightarrow \quad x \frac{d V}{d x}=-\frac{V^2}{1-V+V^2}-V \\ & \Rightarrow \quad x \frac{d V}{d x}=\frac{-V^2-V+V^2-V^3}{1-V+V^2} \\ & \Rightarrow \quad x \frac{d V}{d x}=-\frac{V+V^3}{1-V+V^2} \\ & \Rightarrow \quad \frac{1-V+V^2}{V+V^3} d V=-\frac{d x}{x} \end{aligned} $

Which can be solve by variables separable method.

Equation of parabola with its axis parallel to the $Y$-axis.

$ \begin{aligned} & y=a x^2+b x+c, a \neq 0 \\ \Rightarrow \quad & \frac{d y}{d x}=2 a x+b \Rightarrow \frac{d^2 y}{d x^2}=2 a \\ \Rightarrow \quad & \frac{d^3 y}{d x^3}=0 \end{aligned} $

Given, $\cos x \frac{d y}{d x}=y \sin x-1$,

$ \begin{aligned} & x \neq(2 n+1) \frac{\pi}{2}, n \in Z \\ \Rightarrow \quad & \frac{d y}{d x}=\frac{y \sin x}{\cos x}-\frac{1}{\cos x} \\ & =y \tan x-\sec x \\ \Rightarrow \quad & \frac{d y}{d x}-y \tan x=-\sec x \end{aligned} $

This is in the form $\frac{d y}{d x}+P(x) y=Q(x)$,

where, $P(x)=-\tan x$ and $Q(x)=-\sec x$

$ \begin{aligned} \mathrm{IF} & =e^{\int P(x) d x}=e^{\int-\tan x d x} \\ & =e^{\ln |\cos x|}=|\cos x| \end{aligned} $

So, the general solution is

$ \begin{aligned} y \cdot \text { IF } & =\int A(x) \cdot \text { IF } d x+C \\ y \cdot \cos x & =\int-\sec x \cdot \cos x d x+C \\ & =-\int d x+C=-x+C \\ & =y=\frac{-x+C}{\cos x} \end{aligned} $

Given, $f(0)=1$

$ \Rightarrow \quad y=1 \text { and } x=0 $

So, $1=\frac{-0+C}{\cos (0)}$

$ \Rightarrow 1=C $

So, $y=\frac{-x+1}{\cos x}=(1-x) \sec x$

Given, differential equation

$ \begin{aligned} & 2 d x+d y=(6 x y+4 x-3 y) d x \\ & \Rightarrow \quad d y=(6 x y+4 x-3 y-2) d x \\ & \Rightarrow \quad \frac{d y}{d x}=6 x y+4 x-3 y-2 \\ & \Rightarrow \quad \frac{d y}{d x}=6 x y+4 x-2-3 y \\ & \Rightarrow \quad 2 x(3 y+2)-1(3 y+2) \\ & \Rightarrow \quad(3 y+2)(2 x-1) \\ & \Rightarrow \quad \frac{d y}{3 y+2}=d x(2 x-1) \end{aligned} $

Integrating to both sides w.r.t $x$, we get

$ \begin{aligned} & \int \frac{d y}{3 y+2}=\int(2 x-1) d x \\ & \Rightarrow \frac{1}{3} \log |3 y+2|=\frac{2 x^2}{2}-x+C_1 \\ & \Rightarrow \frac{1}{3} \log |3 y+2|=x^2-x+C_1 \\ & \Rightarrow \quad \log |3 y+2|=3 x^2-3 x+3 C_1 \\ & \quad=3 x^2-3 x+C \end{aligned} $

(where $C=3 C_1$ )

So, $\log |3 y+2|=3 x^2-3 x+C$

$ \frac{d y}{d x}=\frac{2 x-5 y+3}{5 x+2 y-3} $

Put, $ x=X+h $

and $ y=Y+K $

$ \frac{d y}{d x}=\frac{d Y}{d X}=\frac{2(X+h)-5(Y+K)+3}{5(X+h)+2(Y+K)-3} $

$ \frac{d Y}{d X}=\frac{2 X-5 Y+2 h-5 K+3}{5 X+2 Y+5 h+2 K-3} $

It is Homogeneous differential equation if

$ \begin{aligned} 2 h-5 K+3 & =0 \\\\ 5 h+2 K-3 & =0 \\\\ \frac{h}{15-6} & =\frac{-K}{-6-15}=\frac{1}{4+25} \\\\ \frac{h}{9} & =\frac{K}{21}=\frac{1}{29} \\\\ \Rightarrow \quad h & =\frac{9}{29} \text { and } K=\frac{21}{29} \\\\ \frac{d Y}{d X} & =\frac{2 X-5 Y}{5 X+2 Y} \end{aligned} $

Put $ Y=v X $

$ \begin{array}{c} v+X \frac{d v}{d X}=\frac{2-5 v}{5+2 v} \\\\ \frac{d Y}{d X}=v+X \frac{d v}{d X} \\\\ X \frac{d v}{d X}=\frac{2-5 v}{5+2 v}-v=\frac{2-5 v-5 v-2 v^{2}}{5+2 v} \\\\ X \frac{d v}{d X}=\frac{2-10 v-2 v^{2}}{2 v+5}=-2 \frac{\left(v^{2}+5 v-1\right)}{2 v+5} \\\\ \int \frac{2 v+5}{v^{2}+5 v-1} d v=\int-\frac{2 d x}{x} \\\\ \ln \left(v^{2}+5 v-1\right)=-2 \ln X+\ln C \\\\ \ln \left(v^{2}+5 v-1\right)+\ln X^{2}=\ln C \\\\ x^{2}\left(v^{2}+5 v-1\right)=C \end{array} $

$ \begin{array}{l} X^{2} \frac{\left(Y^{2}+5 X Y-X^{2}\right)}{X^{2}}=C \\\\ Y^{2}+5 X Y-X^{2}=C \\\\ \left(y-\frac{21}{29}\right)^{2}+5\left(x-\frac{9}{29}\right) \\\\ \left(y-\frac{21}{29}\right)-\left(x-\frac{9}{29}\right)^{2}=C \end{array} $

Since, curve passes through point $ (1,1) $

$ \left(1-\frac{21}{29}\right)^{2}+5\left(1-\frac{9}{29}\right) $

$ \left(1-\frac{21}{29}\right)-\left(1-\frac{9}{29}\right)^{2}=C $

$ \left(\frac{8}{29}\right)^{2}+5\left(\frac{20}{29}\right)\left(\frac{8}{29}\right)-\left(\frac{20}{29}\right)^{2}=C $

$ \begin{array}{l} \frac{1}{29^{2}}(64+800-400)=C \\\\ C=\frac{464}{29^2}=\frac{16}{29} \end{array} $

Equation of curve is

$ \begin{array}{l} \left(y-\frac{21}{29}\right)^{2}+5\left(x-\frac{9}{29}\right) \\\\ \left(y-\frac{21}{29}\right)-\left(x-\frac{9}{29}\right)^{2}=\frac{464}{29^{2}} \\\\ y^{2}+\frac{441}{29^{2}}-\frac{42}{29} y +5\left[x y-\frac{9}{29} y-\frac{21}{29} x+\frac{189}{29^{2}}\right] -\left[x^{2}+\frac{81}{29^{2}}-\frac{18 x}{29}\right]=\frac{464}{29^{2}} \end{array} $

$\begin{aligned} & \begin{array}{l}\Rightarrow y^2-x^2+5 x y+x \\\\ \begin{aligned} &\left(-\frac{105}{29}+\frac{18}{29}\right)+y\left(-\frac{42}{29}-\frac{45}{29}\right) +\frac{441+945-81}{29^2}=\frac{464}{29^2}\end{aligned} \\\\ \begin{array}{r}\Rightarrow \quad y^2-x^2+5 x y-\frac{87}{29} x-\frac{87}{29} y +\frac{1305}{29^2}=\frac{464}{29^2}\end{array} \\\\ \begin{array}{c}\Rightarrow y^2-x^2+5 x y-\frac{87}{29} x-\frac{87}{29} y+\frac{841}{29^2}=0\end{array} \\\\ \Rightarrow \quad y^2-x^2+5 x y-\frac{87}{29} x-\frac{87}{29} y+1=0 \\\\ \Rightarrow x^2-y^2-5 x y+\frac{87}{29} x+\frac{87}{29} y-1=0\end{array} \\\\ & x^2-5 x y-y^2+3 x+3 y-1=0\end{aligned}$

$ \left(6 x^{2}-2 x y-18 x+3 y\right) d x $ $ -\left(x^{2}-3 x\right) d y=0 $

$ M(x, y)=6 x^{2}-2 x y-18 x+3 y $

On differentiate $ M $ with respect to $ y $, whe

$ \begin{aligned} \frac{\partial M}{\partial y} & =\frac{d}{d y}\left[6 x^{2}-2 x y-18 x+3 y\right] \\\\ & =-2 x+3 \\\\ N(x, y) & =-\left(x^{2}-3 x\right) \end{aligned} $

On differentiate $ N $ with respect to $ x $, we get

$ \frac{\partial N}{\partial x}=\frac{d}{d x}\left[-\left(x^{2}-3 x\right)\right]=-(2 x-3) $

So, $ \frac{\partial M}{\partial y}=\frac{\partial N}{\partial x} $

The two sites have been shown to be equivalent the equation is an identity any the differential equation is event. Now, let $ \frac{\partial f}{\partial n}=M(x, y) $, then integrate with respect to $ x n $, heat $ y $ as constant as follows

$ \begin{array}{l} \int \frac{\partial f}{\partial x}=\int\left(6 x^{2}-2 x y-18 x+3 y\right) \partial x \\\\ f=2 x^{3}-\frac{y x^{2}}{2}-9 x^{2}+3 x y+H(y) \end{array} $

To solve for $ H(g) $, let $ \frac{\partial f}{\partial y}=N(x, y) $

Then, differentiate of with respect to $ y $, heat $ x $ as constant as follows.

$ \begin{aligned} \frac{\partial f}{\partial y}=0 & -x^{2}-0+3 x+\frac{\partial H(y)}{\partial y} \\\\ & =-x^{2}+3 x+\frac{\partial H(y)}{\partial y} \end{aligned} $

On using Eq. (ii), we get

$ \begin{aligned} -x^{2}+3 x+\frac{\partial H(y)}{\partial y} & =-\left(x^{2}-3 x\right) \\\\ -x^{2}+3 x+\frac{\partial H(y)}{\partial y} & =-x^{2}+3 x \\\\ \frac{\partial H(y)}{\partial y} & =0 \end{aligned} $

Now, integrate the above term as $ \int \partial H(y)=0 $

On substitute $ H(y)=0 $ in the Eq. (i), we get

$ \therefore $ We can equate of to $ C $, we get $ 2 x^{3}-x^{2} y-9 x^{2}+3 x y=C_{1} $

Hence, the general solution of the given differential equation is

$ 2 x^{3}-x^{2} y-9 x^{2}+3 x y+c=0 $

To determine the order and degree of the differential equation, let's start with the given expression:

$ \frac{dy}{dx} = \left(\frac{d^2y}{dx^2} + 2\right)^{\frac{1}{2}} + \frac{d^2y}{dx^2} + 5 $

Rearrange the terms to isolate the square root:

$ \left[\frac{dy}{dx} - \frac{d^2y}{dx^2} - 5\right] = \left(\frac{d^2y}{dx^2} + 2\right)^{\frac{1}{2}} $

Square both sides to eliminate the square root:

$ \left[\frac{dy}{dx} - \frac{d^2y}{dx^2} - 5\right]^2 = \frac{d^2y}{dx^2} + 2 $

Now, we have a polynomial in derivatives. The highest order derivative in this equation is $\frac{d^2y}{dx^2}$, indicating that the order of the differential equation is 2. Since the equation is a polynomial in terms of its highest derivative with the degree being that of the highest exponent applied to the derivative, and here the highest power is 2 (after squaring), the degree of the differential equation is also 2.

Therefore, the order and the degree of the differential equation are both 2.

Given the equation $ y = \sin x + A \cos x $, differentiate both sides with respect to $ x $:

Substitute $ A = \frac{y - \sin x}{\cos x} $ into the equation.

Differentiate:

$ \frac{d y}{d x} = \cos x - \left(\frac{y - \sin x}{\cos x}\right)\sin x $

Simplify:

$ \frac{d y}{d x} = \cos x - y \tan x + \frac{\sin^2 x}{\cos x} $

Further simplification leads to:

$ \frac{d y}{d x} + y \tan x = \frac{\cos^2 x + \sin^2 x}{\cos x} $

Since $\cos^2 x + \sin^2 x = 1$, rewrite it as:

$ \frac{d y}{d x} + y \tan x = \sec x $

Thus, the integrating factor is:

$ e^{\int \tan x \, dx} = e^{\log(\sec x)} = \sec x $

So, the integrating factor of the differential equation is $ \sec x $.

We have,

$ y=A e^{-x}+B \cos x $

On differentiating w.r.t. $ x $, we get

$ y^{\prime}=-A e^{-x}-B \sin x $

On differentiating w.r.t. $ x $, we get

$ y^{\prime \prime}=A e^{-x}-B \cos x $

On adding in Eqs (ii) and (iii), we get

$ \begin{array}{l} y^{\prime \prime}+y^{\prime}=-B(\cos x+\sin x) \\\\ B=\frac{-y^{\prime \prime}-y^{\prime}}{\cos x+\sin x}+x \end{array} $

On adding in Eqs, (i) and (iii), we get

$ \begin{array}{l} y^{\prime \prime}+y=2 A e^{-x} \\\\ e^{-x} A=\frac{y^{\prime \prime}+y}{2} \\\\ \Rightarrow y=\frac{y^{\prime \prime}+y}{2}+\frac{-y^{\prime \prime} \cos x-y^{\prime} \cos x}{\cos x+\sin x} \\\\ \Rightarrow 2 y(\cos x+\sin x)=y^{\prime \prime} \cos x+y^{\prime \prime} \sin x+ \\\\ y \cos x+y \sin x-2 y^{\prime \prime} \cos x-2 y^{\prime} \cos x \\\\ \Rightarrow 2 y(\cos x+\sin x)=y^{\prime \prime}(\sin x-\cos x)+y \\\\ \quad(\cos x+\sin x)-2 y^{\prime} \cos x \\\\ \Rightarrow y^{\prime \prime}(\sin x-\cos x)-y(\cos x+\sin x) \\\\ -2 y^{\prime} \cos x=0 \\\\ \text { Replace } y^{\prime \prime}=\frac{d^{2} y}{d x^{2}}, y^{\prime}=\frac{d y}{d x} \\\\ \because(\sin x-\cos x) \frac{d^{2} y}{d x^{2}}-2 \cos x \frac{d y}{d x}-y \\\\ (\cos x+\sin x)=0 \\\\ \Rightarrow(\cos x-\sin x) \frac{d^{2} y}{d x^{2}}+2 \cos x \frac{d y}{d x}+y \\\\ \quad(\cos x+\sin x)=0 \end{array} $

Given, $ \frac{d y}{d x}+\frac{\sin (2 x+y)}{\cos x}+2=0 $

Let $ 2 x+y=u $

$ \begin{array}{ll} & 2+\frac{d y}{d x}=\frac{d u}{d x} \Rightarrow \frac{d y}{d x}=\frac{d u}{d x}-2 \\\\ \Rightarrow & \frac{d u}{d x}-2+\frac{\sin (u)}{\cos x}+2=0 \\\\ \Rightarrow & \frac{d u}{d x}+\frac{\sin u}{\cos x}=0 \end{array} $

By variable seperable

$ \begin{array}{l} \Rightarrow \quad \frac{d u}{\sin u}=-\frac{d x}{\cos x} \\\\ \Rightarrow \quad \int \operatorname{cosec} u d u=-\int \sec x d x \\\\ \Rightarrow \log (\operatorname{cosec} u-\cot u) \\\\ \quad=-\log (\sec x+\tan x)+\log C \\\\ \Rightarrow \log (\operatorname{cosec} u-\cot u) \\\\ \quad+\log (\sec x+\tan x)=\log C \\\\ \Rightarrow(\operatorname{cosec} u-\cot u)(\sec x+\tan x)=C \\\\ \Rightarrow[\operatorname{cosec}(2 x+y)-\cot (2 x+y)] \\\\ \quad[\sec x+\tan x]=C, \end{array} $

$ \begin{aligned} &\text { Given, }\\\\ &\begin{aligned} & (9 x-3 y+5) d y=(3 x-y+1) d x \\\\ & \therefore \quad \frac{d y}{d x}=\frac{3 x-y+1}{9 x-3 y+5} \end{aligned} \end{aligned} $

$ \begin{aligned} &\text { Let } \quad 3 x-y=V\\\\ &\begin{array}{ll} \Rightarrow & 3-\frac{d y}{d x}=\frac{d V}{d x} \\\\ \therefore & 3-\frac{d V}{d x}=\frac{V+1}{3 V+5} \end{array}\\\\ &\begin{array}{lr} \Rightarrow & \frac{d V}{d x}=3-\frac{V+1}{3 V+5}=\frac{8 V+14}{3 V+5} \\\\ \Rightarrow & \left(\frac{3 V+5}{8 V+14}\right) d V=d x \\\\ \Rightarrow & \frac{3}{8}\left(\frac{8 V+\frac{40}{3}}{8 V+14}\right) d V=d x \\\\ \Rightarrow & \frac{3}{8}\left[\frac{8 V+14}{8 V+14}+\frac{\frac{40}{3}-14}{8 V+14}\right] d V=d x \\\\ \Rightarrow & {\left[\frac{3}{8}+\left(-\frac{1}{8}\right)\left(\frac{1}{4 V+7}\right)\right] d V=d x} \end{array} \end{aligned} $

On integrating both sides, we get

$ \begin{aligned} & \frac{3}{8} V-\frac{1}{8} \times \frac{\log |4 V+7|}{4}=x+C_1 \\\\ \Rightarrow \quad & 12(3 x-y)-\log |4(3 x-y)+7| \\\\ \Rightarrow \quad & =32 x+32 C_1 \\\\ \Rightarrow \quad & 4 x-12 y-\log |12 x-4 y+7|=C \end{aligned} $

Where $C=32 C_1$

Given,

$ \begin{aligned} & \frac{d y}{d x}=\frac{2 y^2+1}{2 y^3-4 x y+y} \\\\ \Rightarrow \quad & \frac{d x}{d y}=\frac{2 y^3-4 x y+y}{2 y^2+1} \\\\ \Rightarrow \quad & \frac{d x}{d y}+\left(\frac{4 y}{1+2 y^2}\right) x=\frac{2 y^3+y}{2 y^2+1} \end{aligned} $

This is a linear differential equation

$ \begin{aligned} \text { IF } & =\exp \left(\int \frac{4 y}{1+2 y^2} d y\right) \\\\ & =\exp \left(\log \left(1+2 y^2\right)\right)=1+2 y^2 \end{aligned} $

Now, general solution is given by

$ \begin{aligned} & x \times \mathrm{IF}=\int\left[\left(\frac{2 y^3+y}{2 y^2+1}\right) \times \mathrm{IF}\right] d y+C \\\\ \Rightarrow & x\left(1+2 y^2\right)=\int\left(2 y^3+y\right) d y+C \\\\ \Rightarrow & x+2 x y^2=2\left(\frac{y^4}{4}\right)+\frac{y^2}{2}+C \\\\ \Rightarrow & 4 x y^2+2 x=y^4+y^2+C \end{aligned} $

We have,

$ \begin{aligned} & \left(3 x^2-2 x y\right) d y+\left(y^2-2 x y\right) d x=0 \\\\ & \Rightarrow \frac{d y}{d x}=-\left(\frac{y^2-2 x y}{3 x^2-2 x y}\right) \ldots \text { (i) } \end{aligned} $

Let $y=v x \Rightarrow \frac{d y}{d x}=v+x \frac{d v}{d x}$

From Eq. (i), we get

$ \begin{aligned} v+x \frac{d v}{d x} & =-\left(\frac{v^2 x^2-2 v x^2}{3 x^2-2 v x^2}\right) \\\\ x \frac{d v}{d x} & =-\left(\frac{v^2-2 v}{3-2 v}\right)-v \end{aligned} $

$\begin{gathered}x \frac{d v}{d x}=\frac{-v^2+2 v-3 v+2 v^2}{3-2 v} \\\\ x \frac{d v}{d x}=\frac{v^2-v}{3-2 v} \Rightarrow \frac{2 v-3}{v^2-v} d v=-\frac{d x}{x} \\\\ \left.\frac{2 v-1}{v^2-v}-\frac{2}{\left(v-\frac{1}{2}\right)^2-\left(\frac{1}{2}\right)^2}\right) d v=-\frac{d x}{x} \\\\ \log \left(v^2-v\right)-\frac{2}{2\left(\frac{1}{2}\right)} \log \left(\frac{v-\frac{1}{2}-\frac{1}{2}}{v-\frac{1}{2}+\frac{1}{2}}\right)\end{gathered}$

$\begin{aligned} & =-\log x-\log c \\\\ & \log \left(v^2-v\right)-2 \log \left(\frac{v-1}{v}\right)=-(\log c+\log x) \\\\ & \log \left(v^2-v\right)-\log \left(\frac{v-1}{v}\right)^2=-\log c x \\\\ & \Rightarrow \log \left\{\left[v(v-1] \times \frac{v^2}{(v-1)^2}\right\}=-\log \alpha\right.\end{aligned}$

$ \begin{aligned} \frac{v^3}{v-1} & =\frac{1}{c x} \Rightarrow \frac{\frac{y^3}{x^3}}{\frac{y}{x}-1}=\frac{1}{c x} \\\\ \frac{y^3 \cdot x}{x^3(y-x)} & =\frac{1}{c x} \Rightarrow \frac{y^3}{y-x}=\frac{x}{c} \\\\ \Rightarrow \quad x y-x^2 & =c y^3 \end{aligned} $

Which is the general solution of given differential equation.

$ \begin{aligned} & \text { } y=x a \cos (\log x)+x b \sin (\log x) \\ & \frac{d y}{d x}=a \cos (\log x)+a x \frac{(-\sin (\log x))}{x} +b \sin (\log x)+b x \frac{\cos (\log x)}{x} \end{aligned} $

$ \begin{array}{r} \Rightarrow \frac{d y}{d x}=\frac{y}{x}+(-a \sin (\log g x))+b \cos (\log x) \\ \Rightarrow \frac{d^2 y}{d x^2}=\frac{d y}{d x} \cdot \frac{1}{x}+y\left(\frac{-1}{x^2}\right) -\frac{a \cos (\log x)}{x}-\frac{b \sin (\log x)}{x} \end{array} $

$ \begin{aligned} & \Rightarrow \frac{d^2 y}{d x^2}=\frac{d y}{d x} \cdot \frac{1}{x}-\frac{y}{x^2}-\frac{1}{x^2}(y) \\ & \Rightarrow x^2 \frac{d^2 y}{d x^2}-x \frac{d y}{d x}+2 y=0 \end{aligned} $

The given differential equation is

$ \begin{aligned} & y^2 d x+\left(x^2-x y-y^2\right) d y=0 \\ \Rightarrow \quad & \frac{d y}{d x}=\frac{-y^2}{x^2-x y-y^2}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i) \end{aligned} $

above equation is in the form of homogeneous equation.

So, we put $y=v x$

$ \begin{aligned} & \Rightarrow \quad \frac{d y}{d x}=v+x \frac{d v}{d x} \text { in Eq. (i), We get, } \\ & v+x \frac{d v}{d x}=\frac{-v^2 x^2}{x^2-x(v x)-v^2 x^2} \\ & \Rightarrow x \frac{d v}{d x}=\frac{-v^2}{1-v-v^2}-v \end{aligned} $

$ \Rightarrow x \frac{d v}{d x}=\frac{-v\left(v^2-1\right)}{v^2+v-1}-\left[\frac{1}{v^2-1}+\frac{1}{v}\right] d v=\frac{d x}{x} $

On integrating both sides, we get

$ \begin{aligned} & -\left[\frac{1}{2 \times 1} \log \left|\frac{v-1}{v+1}\right|+\log v\right]=(\log x)+\log C \\ & \Rightarrow-\log \left(\sqrt{\frac{v-1}{v+1}} \cdot v\right)=\log C x \\ & \Rightarrow-\log \sqrt{\frac{y-x}{y+x}} \cdot \frac{y}{x}=\log C x \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)\end{aligned} $

On putting $x=2, y=1$

$ \begin{aligned} -\log \sqrt{\frac{1}{3}} \cdot \frac{1}{2} & =\log 2 C \\ \Rightarrow \quad C & =\sqrt{3} \end{aligned} $

So, solution of above differential equation is (from Eq. (ii)

$ \begin{aligned} & \sqrt{\frac{y-x}{y+x}} \cdot \frac{y}{x}=\frac{1}{\sqrt{3} x} \\ & 3 y^2(y-x)=y+x \\ & \quad-3 y^2(x-y)=x+y \end{aligned} $

On comparing it with given solution, we get

$ k=-3 $

Given, differential equation

$ \frac{d y}{d x}+\frac{y}{x}=x^2 $

This is linear differential equation

$ \begin{aligned} & P=1 / x, Q=x^2 \\ & \mathrm{IF}=e^{\int \frac{1}{x} d x}=e^{\ln x}=x \end{aligned} $

Solution $y \cdot x=\int x^2 \cdot x d x+C$

$ \Rightarrow \quad x y=\frac{x^4}{4}+C $

$y^2=4 a(x-a)$

Differentiating, $2 y \frac{d y}{d x}=4 a \Rightarrow y \frac{d y}{d x}=2 a$

Again differentiating,

$ y \frac{d^2 y}{d x^2}+\frac{d y}{d x} \cdot \frac{d y}{d x}=0 \Rightarrow y \frac{d^2 y}{d x^2}+\left(\frac{d y}{d x}\right)^2=0 $

Order $m=1$

Degree $n=2$

$ \therefore \quad m+n^2=1+(2)^2=1+4=5 $

$\frac{d y}{d x}=\frac{2 x+3 y}{3 x-2 y}$

On putting, $y=v x$

$ \begin{aligned} & \frac{d y}{d x}=v+x \frac{d v}{d x} \\ \Rightarrow & v+x \frac{d v}{d x}=\frac{2 x+3 v x}{2 x-2 v x} \\ \Rightarrow & v+x \frac{d v}{d x}=\frac{2+3 v}{3-2 v} \\ \Rightarrow & v-\frac{2+3 v}{3-2 v}=\frac{-x d v}{d x} \\ \Rightarrow & \frac{3 v-2 v^2-2-3 v}{3-2 v}=-x \frac{d v}{d x} \\ \Rightarrow & \frac{-2 v^2-2}{3-2 v}=-x \frac{d v}{d x} \\ \Rightarrow & \frac{-d x}{x}=\frac{3-2 v}{-2 v^2-2} d v \end{aligned} $

$ \begin{aligned} &\text { On integrating both sides, we get }\\ &\begin{aligned} &-\int \frac{d x}{x}=\int \frac{2 v-3}{2 v^2+2} d v \\ & \Rightarrow-\log |x|=\int \frac{2 v}{2 v^2+2} d v-\int \frac{3}{2 v^2+2} d v \\ & \Rightarrow-\log |x|=\frac{1}{2} \log \left|v^2+1\right|-\frac{3}{2} \tan ^{-1} v+C \\ & \Rightarrow-\log |x|=\frac{1}{2} \log \left|(y / x)^2+1\right| -\frac{3}{2} \tan ^{-1}(y / x)+C \\ & \Rightarrow \frac{3}{2} \tan ^{-1}(y / x)=\log \left(\frac{y^2+x^2}{x^2}\right)^{1 / 2}+\log x+C \end{aligned} \end{aligned} $

$ \begin{aligned} &\begin{aligned} & \Rightarrow \frac{3}{2} \tan ^{-1}(y / x)=\log \left(y^2+x^2\right)^{1 / 2} -\log x+\log x+C \\ & \Rightarrow 3 \tan ^{-1}(y / x)=2 \log \left(y^2+x^2\right)^{1 / 2} \\ & \Rightarrow \quad 3 \tan ^{-1}(y / x)=\log \left(y^2+x^2\right)^{\frac{1}{2} \times 2} \\ & \Rightarrow \quad \tan ^{-1}(y / x)=\frac{1}{3} \log \left(y^2+x^2\right) \\ & \Rightarrow \quad y / x=\tan \left[\frac{1}{3} \log \left(y^2+x^2\right)\right] \\ & \Rightarrow \quad y=x \tan \left[\frac{1}{3} \log \left(y^2+x^2\right)\right] \end{aligned}\\ &\text { Thus, } f(x)=\frac{1}{3} \log \left(x^2+y^2\right) \end{aligned} $

$\left(x^2+2\right) d y+2 x y d x=e^x\left(x^2+2\right) d x$

Dividing on both sides by ' $d x^{\prime}$ ',

$ \begin{aligned} & \left(x^2+2\right) \frac{d y}{d x}+2 x y=e^x\left(x^2+2\right) \\ \Rightarrow & \left(x^2+2\right) \cdot \frac{d y}{d x}+y \cdot \frac{d}{d x}\left(x^2+2\right)=e^x\left(x^2+2\right) \\ \Rightarrow & \frac{d}{d x}\left[y \cdot\left(x^2+2\right)\right]=e^x\left(x^2+2\right) \end{aligned} $

Integrating on both sides, we get

$ \begin{aligned} & \int \frac{d}{d x}\left[y \cdot\left(x^2+2\right)\right] \cdot d x=\int e^x \cdot\left(x^2+2\right) \cdot d x \\ & \begin{array}{r} \Rightarrow \quad\left(x^2+2\right) y=\left(x^2+2\right) \\ \quad \int e^x \cdot d x-\int\left[\frac{d}{d x}\left(x^2+2\right) \cdot \int e^x \cdot d x\right] d x \end{array} \\ & =\left(x^2+2\right) e^x-\int 2 x \cdot e^x \cdot d x \\ & =\left(x^2+2\right) e^x-2 \int x \cdot e^x d x \\ & =\left(x^2+2\right) e^x-2\left[x \cdot \int e^x d x\right. \left.\quad-\left[\int \frac{d}{d x}(x) \cdot \int e^x d x\right] d x\right] \end{aligned} $

$ \begin{aligned} &\begin{aligned} & =\left(x^2+2\right) e^x-2\left[x \cdot e^x-\int e^x \cdot d x\right] \\ & =\left(x^2+2\right) e^x-2\left[e^x(x-1)\right] \\ & =e^x\left[x^2+2-2 x+2\right] \\ & =e^x\left(x^2-2 x+4\right) \end{aligned}\\ &\text { Hence, }\left(x^2+2\right) y=e^x\left(x^2-2 x+4\right)+C \end{aligned} $

Given,

$ \begin{aligned} & (3 x-4 y)(d x-3 d y)+(6 d x-4 d y)=0 \\ & \Rightarrow(3 x-4 y+6) d x+(-9 x+12 y-4) d y=0 \\ & \Rightarrow \quad \frac{d y}{d x}=\frac{3 x-4 y+6}{9 x-12 y+4} \end{aligned} $

Let $\quad 3 x-4 y=z$

$ \therefore \quad 3-4 \frac{d y}{d x}=\frac{d z}{d x} \Rightarrow \frac{d y}{d x}=\frac{1}{4}\left(3-\frac{d z}{d x}\right) $

Now, $\frac{1}{4}\left(3-\frac{d z}{d x}\right)=\frac{z+6}{3 z+4}$

$ \begin{aligned} \Rightarrow \quad 3-\frac{d z}{d x} & =\frac{4 z+24}{3 z+4} \\ \Rightarrow \quad \frac{d z}{d x} & =3-\frac{4 z+24}{3 z+4} \\ & =\frac{9 z+12-4 z-24}{3 z+4} \\ & =\frac{5 z-12}{3 z+4} \end{aligned} $

$ \begin{aligned} &\Rightarrow \frac{3 z+4}{5 z-12} d z=d x \Rightarrow \frac{1}{5}\left[3+\frac{56}{5 z-12}\right] d z=d x\\ &\text { On integrating both sides, we get }\\ &\frac{1}{5}\left[3 z+\frac{56}{5} \ln |5 z-12|\right]=x+A \end{aligned} $

$ \begin{array}{ll} \Rightarrow {\left[3(3 x-4 y)+\frac{56}{5} \ln |5(3 x-4 y)-12|\right]} \\ =5 x+5 A \\ \Rightarrow 9 x-12 y+\frac{56}{5} \ln |15 x-20 y-12| \\ =5 x+5 A \\ \Rightarrow 45 x-60 y+56 \ln |15 x-20 y-12| \\ \Rightarrow \begin{array}{l} 25 x+25 A \\ 20 x-60 y+56 \ln |15 x-20 y-12| \\ =25 A \end{array} \\ \Rightarrow 5 x-15 y+14 \ln |15 x-20 y-12|=C \\ {\left[\text { where, } C=\frac{25}{4} A\right]} \end{array} $

$ \begin{aligned} & \text { Given, }(\sec x+\tan x) \frac{d y}{d x}+\left(\sec ^2 x\right. \\ & +\sec x \tan x) y=1 \\ & \Rightarrow \quad(\sec x+\tan x)\left[\frac{d y}{d x}+y \cdot \sec x\right]=1 \\ & \Rightarrow \quad \frac{d y}{d x}+y \cdot \sec x=\frac{1}{(\sec x+\tan x)} \end{aligned} $

Eq. (i) is the linear differential equation of the form $\frac{d y}{d x}+P y=Q$.

where $P=\sec x$ and $Q=\frac{1}{\sec x+\tan x}$

$ \text { } \begin{array}{rlrl} & \text { Now, IF } =e^{\int P d x}=e^{\int \sec x d x} \\ & = e^{\log |\sec x+\tan x|} \\ \Rightarrow & \text { IF } =\sec x+\tan x \end{array} $

∴ Solution is $y \times$ IF $=\int$ IF $\times Q d x+C$

$ \begin{aligned} & \Rightarrow y(\sec x+\tan x)=\int 1 d x+C \\ & \Rightarrow y(\sec x+\tan x)=x+C \end{aligned} $

we have, $y=A e^x+B \sin 2 x$

$ \begin{aligned} d y / d x & =A e^x+2 B \cos 2 x \\ d^2 y / d x^2 & =A e^x-4 B \sin 2 x \end{aligned} $

On substituting these values in

$ \begin{aligned} & \text { option (a) } \\ & \begin{array}{r} \text { LHS }=(\cos 2 x-\sin 2 x)\left(A e^x-4 B \sin 2 x\right) \\ +4 \sin 2 x\left(A e^x+2 B \cos 2 x\right) \\ -4(\sin 2 x+\cos 2 x)\left(A e^x+B \sin 2 x\right) \\ =A e^x(\cos 2 x)-4 B \sin 2 x \cos 2 x-A e^x \sin 2 x \\ +4 B \sin ^2 2 x+4 A e^x \sin 2 x+8 B \sin 2 x \cos 2 x \\ -4 A e^x \sin 2 x-4 B \sin ^2 2 x-4 A e^x \cos 2 x \\ -4 B \sin 2 x \cos 2 x \neq 0 \end{array} \end{aligned} $

No other option is also satisfying the equation.

$\frac{d y}{d x}=\sin (x-y)+\cos (x-y)$

$ \begin{aligned} & \Rightarrow 1-\frac{d y}{d x}=\frac{d v}{d x} \Rightarrow \frac{d y}{d x}=1-\frac{d v}{d x} \\ & \therefore 1-\frac{d v}{d x}=\sin v+\cos v \\ & \Rightarrow(1-\sin v-\cos v)=\frac{d v}{d x} \\ & \Rightarrow d x=\frac{d v}{1-\sin v-\cos v} \\ & \Rightarrow d x=\frac{d v}{1-\frac{2 \tan \frac{v}{2}}{1+\tan ^2 \frac{v}{2}}-\frac{1-\tan ^2 \frac{v}{2}}{1+\tan ^2 \frac{v}{2}}} \end{aligned} $

$ \begin{aligned} & \Rightarrow d x=\frac{\sec ^2 \frac{v}{2} d v}{1+\tan ^2 \frac{v}{2}-2 \tan \frac{v}{2}-1+\tan ^2 \frac{v}{2}} \\ & \Rightarrow \quad d x=\frac{\sec ^2 \frac{v}{2} d v}{2 \tan ^2 \frac{v}{2}-2 \tan \frac{v}{2}} \\ & \Rightarrow \quad \int d x=\int \frac{\sec ^2 \frac{v}{2} d v}{2 \tan ^2 \frac{v}{2}-2 \tan \frac{v}{2}} \\ & \begin{aligned} \text { Let } & \tan \frac{v}{2}=t \\ \Rightarrow & \sec \mathrm{c}^2 \frac{v}{2} d v=2 d t \\ \Rightarrow & x+C=\int \frac{d t}{t(t-1)}=\int\left(\frac{-1}{t}+\frac{1}{t-1}\right) d t \\ & =\log |(t-1)|-\log |t|=\log \left|\frac{(t-1)}{t}\right| \\ & =\log \left|\frac{\tan \frac{v}{2}-1}{\tan \frac{v}{2}}\right| \\ & x+C=\log \left|\frac{\tan \frac{x-y}{2}-1}{\tan \frac{x-y}{2}}\right| \end{aligned} \end{aligned} $

$ \begin{aligned} &\text {Given, differential equation }\\ &\begin{aligned} & x^2 d y-\left(x y-y^2\right) d x=0 \\ & \quad \frac{d y}{d x}=\frac{x y-y^2}{x^2} \quad \quad[\text { put } y=v x] \\ & \Rightarrow \quad \frac{d y}{d x}=v+x \frac{d v}{d x} \\ & \therefore \quad v+x \frac{d v}{d x}=\frac{x^2 v-x^2 v^2}{x^2} \\ & \Rightarrow \quad x \frac{d v}{d x}=v-v^2-v \Rightarrow x \frac{d v}{d x}=-v^2 \\ & \Rightarrow \quad \frac{d x}{x}=-\frac{1}{v^2} d v \Rightarrow \log x=\frac{1}{v}+C \\ & \Rightarrow \quad \log x=\frac{x}{y}+C \Rightarrow y \log x=x+C y \end{aligned} \end{aligned} $

Now, equation of family of parabola whose axis is the $X$-axis i.e $y^2=4 a(x+\alpha)$ where, $a$ and $\alpha$ are constant.

Now, Differentiate on both sides w.r.t.

$ \begin{aligned} 2 y \frac{d y}{d x} & =4 a \Rightarrow y \frac{d y}{d x}=2 a \\ \Rightarrow y \frac{d^2 y}{d x^2}+\left(\frac{d y}{d x}\right)^2 & =0 \end{aligned} $

Thus, order 2 and degree 1 .

$ \begin{aligned} &\text { Given, differential equation is }\\ &\begin{aligned} & \left(x \sin \frac{y}{x}\right) d y=\left(y \sin \frac{y}{x}-x\right) d x \\ & \Rightarrow \quad \frac{d y}{d x}=\frac{y \sin \frac{y}{x}-x}{x \sin \frac{y}{x}} \\ & \text { Let } \quad y=v x \Rightarrow \frac{d y}{d x}=v+x \frac{d v}{d x} \\ & \Rightarrow \quad v+x \frac{d v}{d x}=\frac{v x \sin v-x}{x \sin v} \\ & \Rightarrow \quad v+x \frac{d v}{d x}=\frac{v \sin v-1}{\sin v} \\ & \Rightarrow \quad x \frac{d v}{d x}=\frac{v \sin v-1}{\sin v}-v=-\frac{1}{\sin v} \\ & \Rightarrow \quad \int \sin v d v=-\int \frac{d x}{x}-C \\ & \Rightarrow \quad-\cos v=-\log |x|-C \\ & \Rightarrow \quad \cos \left(\frac{y}{x}\right)=\log |x|+C \end{aligned} \end{aligned} $

The given differential equation is

$ \begin{aligned} & \begin{array}{l} \Rightarrow \quad\left(2 x-10 y^3\right) d y+y d x=0 \\ \Rightarrow \quad \frac{d x}{d y}=-\frac{2 x}{y}+10 y^2 \end{array} \\ & \Rightarrow \quad \frac{d x}{d y}+\frac{2 x}{y}=10 y^2 \\ & \text { Now, IF }=e^{\int \frac{2}{y} d y}=e^{2 \log y}=y^2 \end{aligned} $

solution is

$ \begin{aligned} x \cdot y^2 & =\int\left(10 y^2\right) \cdot\left(y^2\right) d y+C \\ & =\int 10 \cdot y^4 d y+C \\ & =2 y^5+C \\ \Rightarrow x y^2-2 y^5 & =C \end{aligned} $

Let $a$ be any arbitrary constant

$ \begin{aligned} P F & =P A \\ \Rightarrow \quad \sqrt{x^2+y^2} & =(x-a) \end{aligned} $

$ \begin{aligned} \Rightarrow \quad x^2+y^2 & =x^2+a^2-2 a x \\ y^2 & =a^2-2 a x \\ 2 y \frac{d y}{d x} & =-2 a \\ a & =-y \frac{d y}{d x} \end{aligned} $

On putting the value of $a$ (ii) in Eq. (i), we get

$ \begin{aligned} y^2 & =y^2\left(\frac{d y}{d x}\right)^2+2 x y \frac{d y}{d x} \\ m & =1, n=2 \\ m n-m+n & =3 \end{aligned} $

$\begin{aligned} & \ \frac{d y}{d x}+y f^{\prime}(x)-f(x) f^{\prime}(x)=0 \\ & \Rightarrow \frac{d y}{d x}+y \frac{d f(x)}{d x}-f(x) \frac{d f(x)}{d x}=0\end{aligned}$

$\begin{aligned} & \begin{array}{l}\Rightarrow \frac{d y}{d f(x)}+y=f(x) \\ \text { Integrating factor }=e^{\int d f(x)}=e^{f(x)} \\ e^{f(x)} y=\int f(x) e^{f(x)} d f(x) \\ e^{f(x)} y=f(x) \int e^{f(x)} d f(x) \\ \quad-\int\left(\int e^{f(x)} d x \cdot \frac{d f(x)}{d f(x)}\right) d f(x)\end{array} \\ & \begin{aligned} e^{f(x)} y & =f(x) e^{f(x)}-\int e^{f(x)} d f(x)+c\end{aligned} \\ & \begin{aligned} e^{f(x)} y & =f(x) e^{f(x)}-e^{f(x)}+c \\ y & =f(x)-1+c e^{-f(x)}\end{aligned}\end{aligned}$

Given, $x^k+y^k=c^2$, since two constant $c_1$ and $c_2$

$ \begin{aligned} & \text { So, } & k & =2 \\ & & x^2+y^2 & =c^2 \\ \Rightarrow & & 2 x+2 y \frac{d y}{d x} & =0 \\ \Rightarrow & & x+y \frac{d y}{d x} & =0 \\ \Rightarrow & & \frac{d y}{d x}+\frac{x}{y} & =0 \end{aligned} $

Given that $y=a x^2+b \log x$

$ \Rightarrow \quad \frac{y}{x^2}=a+b \log x $

On differentiating w.r.t. $x$,

$ \begin{array}{ll} & \frac{1}{x^2} y^{\prime}-\frac{2}{x^3} y=\frac{b}{x} \\ \Rightarrow & \frac{1}{x} y^{\prime}-\frac{2}{x^2} y=b \end{array} $

Again differentiating w.r.t. $x$,

$ \begin{aligned} & \frac{1}{x} y^{\prime \prime}-\frac{y^{\prime}}{x^2}+\frac{4 y}{x^3} \cdot \frac{2}{x^2} y^{\prime} =0 \\ \Rightarrow & -\frac{1}{x} y^{\prime \prime}+\frac{3 y^{\prime}}{x^2} =\frac{4 y}{x^3} \\ \Rightarrow & -x y^{\prime \prime}+3 y^{\prime} =\frac{4 y}{x} \end{aligned} $

On comparing with

$ \begin{aligned} & \quad f(x) y^{\prime \prime}+g(x) y^{\prime}=\frac{4 y}{x} \\ & \therefore \quad f(x)=-x \\ & \text { and } g(x)=3 \end{aligned} $

an $m=1$ degree of differential equation

$ \therefore \quad f(1)+g(1)=-1+3=2 $

Order of differential equation i.e.

$ f(m)+g(m)=l $

Given,

$ \begin{array}{rlrl} & d x =(2 x+3 y-4) d y \\ \Rightarrow \quad & \frac{d x}{d y} =2 x+3 y-4 \end{array} $

On putting

$ \begin{array}{rlrl} 2 x+3 y-4 =z & \\ 2 \frac{d x}{d y}+3 =\frac{d z}{d y} \\ \Rightarrow \quad \frac{d x}{d y} =\frac{1}{2} \frac{d z}{d y}-\frac{3}{2} \end{array} $

Now, $\frac{1}{2} \frac{d z}{d y}-\frac{3}{2}=z$

$ \begin{array}{lc} \Rightarrow & \frac{d z}{d y}=2 z+3 \\ \Rightarrow & \frac{d z}{2 z+3}=d y \\ \Rightarrow & \frac{1}{2} \log (2 z+3)+c_1=y \end{array} $

$ \begin{array}{lr} \Rightarrow & \log (4 x+6 y-8+3)+2 c_1=2 y \\ \Rightarrow & 6 y-3 \log (4 x+6 y-5)=6 c_1 \\ \Rightarrow & 6 y-3 \log (4 x+6 y-5)=c \end{array} $

Given, differential equation

$ \begin{aligned} \left(\frac{d^4 y}{d x^4}+\frac{d^2 y}{d x^2}\right)^{3 / 2} & =5 \frac{d^3 y}{d x^3} \\ \Rightarrow \quad\left(\frac{d^4 y}{d x^4}+\frac{d^2 y}{d x^2}\right)^3 & =25\left(\frac{d^3 y}{d x^3}\right)^2 \end{aligned} $

Order $=4$

So, number of arbitrary constants $=4$

Assertion (A) :

$ y^{\prime}+2 x y^{\prime}+\log _e\left(\frac{d y}{d x}\right)=0 $

Here, degree is not defined. So, Assertion is false but Reason (R) is true.

Given, $S$ be the family of curves given by general solution of differential equation,

$ \begin{aligned} & \frac{y^2 e^{-1 / y}}{\sqrt{x}} d x-2 \sec \sqrt{x} d y=0 \\ \Rightarrow & \quad \frac{d x}{d y}=\frac{2 \sec \sqrt{x} \cdot \sqrt{x}}{y^2} \cdot\left(\frac{1}{e^y}\right) \\ \Rightarrow & \quad \int \frac{e^{1 / y}}{y^2} d y=\int \frac{\cos \sqrt{x}}{2 \sqrt{x}} d x \\ \Rightarrow & -e^{1 / y}=\sin \sqrt{x}+c \end{aligned} $

Curve passes through $\left(\pi^2, 1\right)$

$ \begin{aligned} & & -e & =\sin \pi+c \\ \Rightarrow & & c & =-e \\\therefore & & S:-e^{1 / y} & =\sin \sqrt{x}-e \\ \Rightarrow& & \sin \sqrt{x} & +e^{1 / y}=e \end{aligned} $

Statement I: If the centre of circle lies on $Y$-axis and radius $k$, then equation of circle is $(x)^2+(y-h)^2=k^2$, where $(0, h)$ is the centre of circle.

$ \begin{aligned} & & 2 x+2(y-h) \frac{d y}{d x} & =0 \\ \Rightarrow & & (y-h) \frac{d y}{d x} & =-x \\ \Rightarrow & & y-h & =-\frac{x}{d y / d x} \end{aligned} $

Now, substitute value of $y-h$ in equation of circle,

$ \begin{gathered} x^2+\frac{x^2}{\left(\frac{d y}{d x}\right)^2}=k^2 \\ \Rightarrow \quad x^2\left(\frac{d y}{d x}\right)^2+x^2=k^2\left(\frac{d y}{d x}\right)^2 \\ \Rightarrow\left(x^2-k^2\right)\left(\frac{d y}{d x}\right)^2+x^2=0 \end{gathered} $

∴ Statement I is true.

Statement II: If the centre of circle lies on $X$-axis and circle passes through origin is $(x-h)^2+y^2=h^2$, where centre of circle is $(h, 0)$ and radius is $h$ unit. On differentiate both sides w.r.t. $x$, we get

$ \begin{aligned} & & 2(x-h)+2 y \frac{d y}{d x} & =0 \\ \Rightarrow & & x-h+y \frac{d y}{d x} & =0 \\ \Rightarrow & & h & =x+y \frac{d y}{d x} \end{aligned} $

Now, substitute the value of $h$ in the equation of circle

$ \begin{aligned} & \quad\left(x-x-\frac{y d y}{d x}\right)^2+y^2=\left(x+\frac{y d y}{d x}\right)^2 \\ & \Rightarrow \quad y^2\left(\frac{d y}{d x}\right)^2+y^2=x^2+y^2\left(\frac{d y}{d x}\right)^2 +(2 x y) \frac{d y}{d x} \\ & =x^2-y^2+(2 x y) \frac{d y}{d x}=0 \end{aligned} $

∴ Statement II is also true.

$y^2-5 a x-5 a^{\frac{3}{2}}=0$

On differentiating w.r.t. $x$, we get

$ \begin{array}{rlrl} & (2 y) \frac{d y}{d x}-5 a =0 \\ a & =\left(\frac{2}{5} y\right) \frac{d y}{d x} \\ \Rightarrow \quad y^2-5\left(\frac{2}{5} y \frac{d y}{d x}\right) x & =5\left(\frac{2}{5} y \frac{d y}{d x}\right)^{\frac{3}{2}} \\ \Rightarrow \quad & y^2-(2 x y) \frac{d y}{d x} =(2 y) \frac{d y}{d x}\left(\left(\frac{2}{5} y\right) \frac{d y}{d x}\right)^{\frac{1}{2}} \\ \Rightarrow \quad & y-(2 x) \frac{d y}{d x} =\left(\frac{2 d y}{d x}\right)\left(\left(\frac{2}{5} y\right) \frac{d y}{d x}\right)^{\frac{1}{2}} \end{array} $

On squaring both sides, we get

$ y^2+4 x^2\left(\frac{d y}{d x}\right)^2-2 x y \frac{d y}{d x}=4\left(\frac{d y}{d x}\right)^2 $

$ \left(\left(\frac{2}{5} y\right) \frac{d y}{d x}\right) $

$ \Rightarrow y^2+4 x^2\left(\frac{d y}{d x}\right)^2-2 x y\left(\frac{d y}{d x}\right)=\frac{8}{5} y\left(\frac{d y}{d x}\right)^3 $

Order of differential equation is 1 and degree is 3

$ \begin{array}{ll} \therefore & m=1 \text { and } n=3 \\ & m-n=1-3=-2 \end{array} $

$ \begin{aligned} & \text { } \begin{aligned} (1 & \left.\left.+x^2\right) y \sin x-2 x y\right) d x-\log y^{1+x^2} d y =0\\ \Rightarrow \quad \frac{d y}{d x} & =\frac{\left(1+x^2\right) y \sin x-2 x y}{\left(1+x^2\right) \log y} \\ & =\frac{y}{\log y} \sin x-\frac{2 x y}{\left(1+x^2\right) \log y} \\ & =\frac{y}{\log y}\left[\sin x-\frac{2 x}{\left(1+x^2\right)}\right] \end{aligned} \end{aligned} $

On variable separable,

$ \begin{aligned} & \int \frac{\log y}{4} d y=\int \sin x d x-\int \frac{2 x d x}{1+x^2} \\ \Rightarrow & \frac{(\log y)^2}{2}=-\cos x-\log \left(1+x^2\right)+c_1 \\ \Rightarrow & (\log y)^2+2 \cos x+2 \log \left(1+x^2\right)=2 c_1 \\ \Rightarrow & (\log y)^2+2 \cos x+\log \left(1+x^2\right)^2=c \end{aligned} $

$ \begin{aligned} &\text { Let the equation of the ellipse be }\\ &\frac{x^2}{a^2}+\frac{y^2}{b^2}=1 \end{aligned} $

Differentiating with resp. to $x$,

$ \Rightarrow \quad \frac{2 x}{a^2}+\frac{2 y y^{\prime}}{b^2}=0 \Rightarrow \frac{y y^{\prime}}{x}=\frac{-b^2}{a^2} $

Again, differentiating w.r. to $x$,

$ \begin{aligned} & \Rightarrow \quad \frac{x\left(x y^{\prime \prime}+\left(y^{\prime}\right)^2\right)-\left(y y^{\prime}\right)}{x^2}=0 \\ & \Rightarrow \quad x\left[y y^{\prime}+\left(y^{\prime}\right)^2\right]-y y^{\prime}=0 \\ & \Rightarrow \quad x y y^{\prime}+x\left(y^{\prime}\right)^2-y y^{\prime}=0 \end{aligned} $

$\left(\frac{d^2 y}{d x^2}\right)^{4 / 3}+x\left(\frac{d y}{d x}\right)^2-y \cos \left(\frac{d y}{d x}\right)=0$

The degree of given differential equation is not defined as it is not a polynomial differential equation.

$ \begin{aligned} & \text { } \begin{aligned} \frac{d y}{d x} & =\frac{2 x-3 y+5}{6 x-9 y+7}=\frac{1(2 x-3 y)+5}{3(2 x-3 y)+7} \\Let\,\,\,\,\,\,\,\,\, u & =2 x-3 y \\ \frac{d u}{d x} & =2-3 \frac{d y}{d x} \\ \Rightarrow \quad \frac{d y}{d x} & =\left(2-\frac{d u}{d x}\right) \frac{1}{3} \end{aligned} \end{aligned} $

$ \begin{aligned} & \frac{1}{3}\left(2-\frac{d u}{d x}\right)=\frac{u+5}{3 u+7} \\ \Rightarrow \quad 2-\frac{d u}{d x} & =\frac{3 u+15}{3 u+7} \\ \frac{d u}{d x} & =2-\left(\frac{3 u+15}{3 u+7}\right) \end{aligned} $

$ \begin{aligned} & \frac{d u}{d x}=\frac{6 u+14-3 u-15}{3 u+7} \Rightarrow \frac{d u}{d x}=\frac{3 u-1}{3 u+7} \\ & \int\left(\frac{3 u+7}{3 u-1}\right) d u=\int d x \\ & \Rightarrow \int\left(\frac{3 u-1}{3 u-1}+\frac{8}{3 u-1}\right) d u=x \\ & u+\frac{8}{3} \log |3 u-1|+c=x \\ & 2 x-3 y+\frac{8}{3} \log \\ & |6 x-9 y-1|-x+c=0 \\ & x-3 y+\frac{8}{3} \log |6 x-9 y-1|+c=0 \\ & {[\because u=2 x-3 y]} \end{aligned} $

$ \begin{aligned} &\\ &\begin{aligned} & \text { } a x^2+b y^2=1 \\ & \Rightarrow \quad 2 a x+2 b y \cdot y^{\prime}=0 \\ & \Rightarrow \quad b y \cdot y^{\prime}=-a x \\ & \Rightarrow \quad \quad \frac{y y^{\prime}}{x}=\frac{-a}{b} \\ & \Rightarrow \quad \frac{x\left(\left(y^{\prime}\right)^2+y \cdot y^{\prime \prime}\right)-y y^{\prime}}{x^2}=0 \\ & \Rightarrow \quad x y \cdot y^{\prime \prime}+x \cdot\left(y^{\prime}\right)^2-y y^{\prime}=0 \end{aligned} \end{aligned} $