Differential Equations

$ \begin{aligned} & \lim _{1 \rightarrow x} \frac{t^{10} f(x)-x^{10} f(t)}{t^9-x^9}=1 \\\\ & \Rightarrow \lim _{1 \rightarrow x} \frac{10 t^9 f(x)-x^{10} f^{\prime}(t)}{9 t^8}=1 \\\\ & \Rightarrow 10 x f(x)-x^2 f^{\prime}(x)=9 \\\\ & \Rightarrow x^2 f^{\prime}(x)=10 x f(x)-9 \\\\ & \Rightarrow f^{\prime}(x)=\frac{10 f(x)}{x}-\frac{9}{x^2} \\\\ & \Rightarrow \frac{d y}{d x}-\frac{10}{x} y=-\frac{9}{x^2} \\\\ & \Rightarrow y \cdot \frac{1}{x^{10}}=\int-\frac{9}{x^2} \cdot \frac{1}{x^{10}} d x \\\\ & \Rightarrow \frac{y}{x^{10}}=\frac{9}{11 x^{11}}+c \\\\ & \because f(1)=2 \Rightarrow \frac{2}{1}=\frac{9}{11}+c \Rightarrow c=\frac{13}{11} \\\\ & \therefore f(x)=\frac{9}{11 x}+\frac{13}{11} x^{10} \end{aligned} $

$\Rightarrow$ Option (B) is correct.

$ \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.

To determine which option can form a differential equation of order two, consider each geometric configuration:

Option A: All circles passing through the origin

The general equation for a circle passing through the origin is:

$ x^2 + y^2 - 2hx - 2py = 0 \,\,\,\,\,\,\,\,\,\,(i) $

In this equation, there are two arbitrary constants, $ h $ and $ p $. The presence of two arbitrary constants implies that the differential equation derived from this circle's equation would be of order two, since each arbitrary constant contributes to the order of the differential equation.

Conclusion

Among the given options, the circles passing through the origin allow for a differential equation of order two. This is due to the presence of two arbitrary constants in the circle's equation, which dictate the order of the corresponding differential equation.

To find the differential equation that has $ ax + by = 1 $ as its general solution, follow these steps:

Differentiate the Equation: Start by differentiating both sides of $ ax + by = 1 $ with respect to $ x $. This gives:

$ a + b \frac{dy}{dx} = 0 $

Solve for $\frac{dy}{dx}$: From the differentiated equation, solve for $\frac{dy}{dx}$:

$ \frac{dy}{dx} = \frac{-a}{b} $

Differentiate Again: Differentiate the expression $\frac{dy}{dx} = \frac{-a}{b}$ with respect to $ x $:

$ \frac{d^2y}{dx^2} = 0 $

Therefore, the differential equation that has $ ax + by = 1 $ as its general solution is $\frac{d^2y}{dx^2} = 0$.

Given differential equation

$ \begin{array}{ll} e^x y d x+e^x d y+x d x=0 \\ \Rightarrow \left(e^x y+x\right) d x+e^x d y=0 \\ \Rightarrow e^x \frac{d y}{d x}+e^x y+x=0 \\ \Rightarrow \frac{d y}{d x}+y=-\frac{x}{e^x} \end{array} $

which is linear differential equation

$ \therefore \mathrm{IF}=e^{\int P d x}=e^{\int d x}=e^x $

$\therefore$ Required solution

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

The differential equation for a family of hyperbolas with centers at the origin and axes along the coordinate axes is derived from the standard hyperbola equation:

$ \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 $

Differentiating this equation with respect to $ x $, we get:

$ \frac{2x}{a^2} - \frac{2y}{b^2} \cdot \frac{dy}{dx} = 0 $

This simplifies to:

$ \Rightarrow \frac{2x}{a^2} = \frac{2y}{b^2} \cdot \frac{dy}{dx} \Rightarrow \frac{y}{x} \cdot \frac{dy}{dx} = \frac{b^2}{a^2} $

Differentiating both sides with respect to $ x $ again, we obtain:

$ \frac{y}{x} \cdot \frac{d^2y}{dx^2} + \frac{dy}{dx} \cdot \frac{x \cdot \frac{dy}{dx} - y \cdot 1}{x^2} = 0 $

Which can be rewritten as:

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

This results in the differential equation:

$ xyy_2 + y_1^2x - yy_1 = 0 $

To solve the given differential equation:

$ (x y + y^2) \, dx - (x^2 - 2 x y) \, dy = 0 $

we start by rewriting it in the form of a differential equation:

$ \frac{dy}{dx} = \frac{x y + y^2}{x^2 - 2 x y} $

Using the substitution $ y = ux $, where $ u $ is a function of $ x $, we differentiate $ y $ with respect to $ x $:

$ \frac{dy}{dx} = u + x \frac{du}{dx} $

Substituting these into the differential equation provides:

$ u + x \frac{du}{dx} = \frac{u + u^2}{1 - 2u} $

Rearranging terms, we get:

$ x \frac{du}{dx} = \frac{3u^2}{1 - 2u} $

This leads to:

$ \frac{1 - 2u}{3u^2} \, du = \frac{dx}{x} $

Integrating both sides yields:

$ \int \left(\frac{1}{3u^2} - \frac{2}{3u}\right) \, du = \int \frac{dx}{x} $

Calculating the integrals gives:

$ -\frac{1}{3u} - \frac{2}{3} \log u = \log x + \log c $

Substituting back $ u = \frac{y}{x} $:

$ -\frac{x}{3y} - \frac{2}{3} \log \frac{y}{x} = \log xc $

After simplification, the general solution is:

$ c x y^2 e^{\frac{x}{y}} = 1 $

To solve the differential equation $ (1+\tan y)(dx-dy)+2x \, dy = 0 $, we'll start by rewriting the equation:

$ dx + \tan y \, dx - dy - \tan y \, dy + 2x \, dy = 0. $

This simplifies to:

$ -dy(1+\tan y-2x) + dx(1+\tan y) = 0. $

Rewriting the equation, we separate variables:

$ \frac{dy}{dx} = \frac{1+\tan y}{1+\tan y-2x}. $

Alternatively, in terms of $ \frac{dx}{dy} $:

$ \frac{dx}{dy} = 1 - \frac{2x}{1+\tan y}. $

Now, this can be rewritten as:

$ \frac{dx}{dy} + 2 \frac{x}{1+\tan y} = 1. $

This is a first-order linear differential equation of the form:

$ \frac{dx}{dy} + Px = Q $

where $ P = \frac{2}{1+\tan y} $ and $ Q = 1 $.

To solve it, we calculate the integrating factor (IF):

$ \mathrm{IF} = e^{\int P \, dy} = e^{\int \frac{2}{1+\tan y} \, dy}. $

Evaluating the integral, we get:

$ \mathrm{IF} = (\sin y + \cos y) e^y. $

The general solution formula is:

$ x(\mathrm{IF}) = \int Q \cdot (\mathrm{IF}) \, dy + C. $

Substitute the values:

$ x(\sin y+\cos y) e^y = \int 1 \cdot (\sin y+\cos y) e^y \, dy + C. $

Splitting the integral:

$ = \int \sin y \, e^y \, dy + \int \cos y \, e^y \, dy + C. $

Therefore:

$ x(\sin y+\cos y) e^y = \frac{e^y}{2}(\sin y - \cos y) + \frac{e^y}{2}(\cos y + \sin y) + C. $

Simplifying gives:

$ x(\sin y+\cos y) e^y = e^y \cdot \sin y + C. $

This simplifies to:

$ e^y[x \cos y + x \sin y - \sin y] = C. $

Hence, the general solution is:

$ e^y(x \cos y + x \sin y - \sin y) = C. $

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

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

$ \begin{aligned} & v+x \frac{d v}{d x}=\frac{v x+\sqrt{(v x)^2+x^2}}{x} \\ &\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,=v+\sqrt{1+v^2} \\ & \Rightarrow \quad x \frac{d v}{d x}=\sqrt{1+v^2} \\ & \Rightarrow \frac{1}{\sqrt{1+v^2}} \frac{d v}{d x}=\frac{1}{x} \Rightarrow \int \frac{d v}{\sqrt{1+v^2}}=\int \frac{d x}{x} \end{aligned} $

$ \begin{aligned} & \Rightarrow \log \left|v+\sqrt{1+v^2}\right|=\log x+\log c \\ & \Rightarrow \log \left|\frac{y}{x}+\sqrt{1+\frac{y^2}{x^2}}\right|=\log c x \\ & \Rightarrow \log \left|\frac{y+\sqrt{x^2+y^2}}{x}\right|=\log c x \\ & \Rightarrow \log \left|y+\sqrt{\left(x^2+y^2\right)}\right|-\log x=\log c x \\ & \Rightarrow \log \left|y+\sqrt{x^2+y^2}\right|=\log c^2 \\ & \text { or } y+\sqrt{y^2+x^2}=\alpha^2 \end{aligned} $

$x\left(\frac{d^2 y}{d x^2}\right)^{1 / 2}=\left(1+\frac{d y}{d x}\right)^{4 / 3}$

Simplifying, on squaring both sides, we get

$ \begin{aligned} \left(x \frac{d^2 y}{d x^2}\right)^{1 / 2 \times 2} & =\left(1+\frac{d y}{d x}\right)^{4 / 3 \times 2} \\ x^2\left(\frac{d^2 y}{d x^2}\right) & =\left(1+\frac{d y}{d x}\right)^{8 / 3} \end{aligned} $

On cubic both sides, we get

$ \begin{aligned} & \left(x^2 \frac{d^2 y}{d x^2}\right)^3=\left(1+\frac{d y}{d x}\right)^{8 / 3 \times 3} \\ & x^6\left(\frac{d^2 y}{d x^2}\right)^3=\left(1+\frac{d y}{d x}\right)^8 \end{aligned} $

Clearly, here the highest derivative with respect to $x$ is 2 , so order is 2 .

Moreover the highest power of highest derivative is 3 i.e. $\left(\frac{d^2 y}{d x^2}\right)^3$, so degree is 3 .

$ \begin{aligned} & \text { Sum }=\text { order }+ \text { degree }=2+3 \\ & \text { Sum }=5 \end{aligned} $

We have,

$ y=A \cos 3 x+B \sin 3 x \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)$

On differentiate $y$ w.r.t $x$, we get

$ \begin{aligned} & \frac{d y}{d x}=-3 A \sin 3 x+3 B \cos 3 x \\ \Rightarrow \quad & \frac{d^2 y}{d x^2}=-9 A \cos 3 x-9 B \sin 3 x \\ \Rightarrow \quad & \frac{d^2 y}{d x^2}=-9(A \cos 3 x+B \sin 3 x) \\ \Rightarrow \quad & \frac{d^2 y}{d x^2}=-9 y \quad \text { [from Eq. (i)] } \\ \Rightarrow \quad & \frac{d^2 y}{d x^2}+9 y=0 \end{aligned} $

We start with the given differential equation:

$ \cos x \frac{d y}{d x} - y \sin x = 6x, \left(0 < x < \frac{\pi}{2}\right) $

and the initial condition:

$ y\left(\frac{\pi}{3}\right)=0 $

Rewriting the differential equation:

$ \frac{d y}{d x} - y \frac{\sin x}{\cos x} = \frac{6x}{\cos x} $

This simplifies to:

$ \frac{d y}{d x} - y \tan x = 6x \sec x $

This is in the form of the linear differential equation:

$ \frac{d y}{d x} + Ay = B $

where $ A = -\tan x $ and $ B = 6x \sec x $.

The integrating factor (IF) is found using:

$ \mathrm{IF} = e^{\int \operatorname{Adx}} = e^{-\int \tan x \, dx} = e^{-\log \sec x} = \frac{1}{\sec x} $

Applying the integrating factor, the solution becomes:

$ y \cdot \frac{1}{\sec x} = \int 6x \sec x \cdot \frac{1}{\sec x} \, dx $

Simplifying, we get:

$ \frac{y}{\sec x} = 3x^2 + C $

where $ C $ is a constant. Using the initial condition $ y\left(\frac{\pi}{3}\right)=0 $:

$ 0 = 3\left(\frac{\pi}{3}\right)^2 + C \implies C = -\frac{\pi^2}{3} $

Thus, substituting back, we have:

$ \frac{y}{\sec x} = 3x^2 - \frac{\pi^2}{3} $

Solving at $ x = \frac{\pi}{6} $:

$ y = \sec x \cdot \left(3x^2 - \frac{\pi^2}{3}\right) $

Calculating:

$ \sec\left(\frac{\pi}{6}\right) = \frac{2}{\sqrt{3}} $

$ 3\left(\frac{\pi}{6}\right)^2 = \frac{\pi^2}{12} $

Therefore,

$ y = \frac{2}{\sqrt{3}} \cdot \left(\frac{\pi^2}{12} - \frac{\pi^2}{3}\right) = \frac{2}{\sqrt{3}} \cdot \left(-\frac{\pi^2}{4}\right) $

Simplifying gives:

$ y\left(\frac{\pi}{6}\right) = \frac{-\pi^2}{2\sqrt{3}} $

We have,

$ \frac{d y}{d x}=\frac{y+x \tan \left(\frac{y}{x}\right)}{x} $

Put $y=v x$

$ \begin{array}{ll} \Rightarrow & \frac{d}{d x} \cdot(v x)=\frac{v x+x \tan \left(\frac{v x}{x}\right)}{x} \\ \Rightarrow & v+x \frac{d v}{d x}=\frac{x(v+\tan v)}{x} \\ \Rightarrow & \frac{x d v}{d x}=v+\tan v-v \\ \Rightarrow & x \frac{d v}{d x}=\tan v \Rightarrow \frac{d v}{\tan v}=\frac{d v}{x} \\ \Rightarrow & \frac{\cos v d v}{\sin v}=\frac{d x}{x} \end{array} $

$ \begin{aligned} &\text { On taking integration both sides, we get }\\ &\begin{aligned} & \Rightarrow \quad \int \frac{\cos v d v}{\sin v}=\int \frac{d x}{x} \\ & \text { Let } \quad \sin v=u \\ & \ \quad \,\,\,\,\,\,\,\,\cos v d v=d u \\ & \Rightarrow \quad \int \frac{1}{u} d u=\int \frac{d x}{x} \\ & \Rightarrow \quad \log u=\log x+\log C \\ & \Rightarrow \quad \log (\sin v)=\log x+\log C \\ & \Rightarrow \quad \log \left(\frac{\sin v}{x}\right)=\log C \Rightarrow \frac{\sin v}{x}=C \end{aligned} \end{aligned} $

$ \begin{aligned} &\begin{array}{ll} \Rightarrow & \sin v=x C \\ \Rightarrow & \sin \frac{y}{x}=x C \end{array}\\ &\begin{array}{ll} \Rightarrow & v^2=1.44 \\ \Rightarrow & v=1.2 \mathrm{~m} / \mathrm{s} \end{array} \end{aligned} $

To find the differential equation by eliminating the parameters $ a $ and $ b $ from the equation $ y = a e^{2x} + bx e^{2x} $, we take the derivatives with respect to $ x $ and express them in terms of $ y $.

First Derivative:

$ y' = \frac{d}{dx} (a e^{2x} + bx e^{2x}) = 2a e^{2x} + (2bx e^{2x} + b e^{2x}) $

$ = 2a e^{2x} + 2bx e^{2x} + b e^{2x} = 2y + b e^{2x} $

Second Derivative:

$ y'' = \frac{d}{dx} (2a e^{2x} + 2bx e^{2x} + b e^{2x}) $

$ = 4a e^{2x} + 2b e^{2x} (2x + 1) + b (2 e^{2x}) $

$ = 4a e^{2x} + 4bx e^{2x} + 2b e^{2x} + 2b e^{2x} $

$ = 4a e^{2x} + 4bx e^{2x} + 4b e^{2x} $

$ = 4(a e^{2x} + bx e^{2x}) + 4b e^{2x} $

$ = 4y + 4b e^{2x} $

Constructing the Differential Equation:

Substitute into the expression $ y'' - 4y' + 4y $:

$ y'' - 4y' + 4y = [4y + 4b e^{2x}] - 4[2y + b e^{2x}] + 4y $

$ = 4y + 4b e^{2x} - 8y - 4b e^{2x} + 4y $

$ = 0 $

The expression simplifies to zero, confirming that the differential equation $ y'' - 4y' + 4y = 0 $ is satisfied, supporting the correctness of option (D).

Given the general solution $ y = a^3 e^{b^2 x + c} $, where $ a $ and $ c $ are arbitrary constants and $ b $ is a fixed constant, we are tasked with identifying the order of the differential equation related to this solution.

To find the order, let's derive the expression for $ y $:

$ \begin{aligned} y' &= \frac{d}{dx}\left(a^3 e^{b^2 x + c}\right) \\ &= a^3 b^2 e^{b^2 x + c}. \end{aligned} $

Notice that we can express this in terms of $ y $:

$ y' = b^2 y. $

This differential equation, $ y' - b^2 y = 0 $, is a first-order linear differential equation. The order of a differential equation is determined by the highest derivative present. In this case, since the highest derivative is $ y' $, the order is 1.

To solve the differential equation $(x + 2y^3) \frac{dy}{dx} = y$, we begin by rearranging terms:

$ \frac{x + 2y^3}{y} = \frac{dx}{dy} $

This can be rewritten as:

$ \frac{dx}{dy} = \frac{x}{y} + 2y^2 $

This gives us a form where:

$ \frac{dx}{dy} - \frac{x}{y} = 2y^2 $

We recognize this as a linear first-order differential equation, which can be solved using an integrating factor. The integrating factor ($\text{IF}$) is derived as follows:

$ \text{IF} = e^{\int -\frac{1}{y} \, dy} = e^{-\ln|y|} = \frac{1}{y} $

Apply the integrating factor to both sides of the differential equation:

$ \frac{x}{y} = \int \left(2y^2 \cdot \frac{1}{y}\right) \, dy $

This integral becomes:

$ \frac{x}{y} = 2 \int y \, dy $

Calculate the integral:

$ \frac{x}{y} = 2 \left(\frac{y^2}{2}\right) + C $

Simplifying gives:

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

Therefore, multiplying both sides by $y$ results in:

$ x = y(y^2 + C) $

Thus, the solution to the differential equation is:

$ x = y(y^2 + C) $

Given differential equation is

$ \begin{aligned} \frac{d^3 y}{d x^3}=\left[1+\left(\frac{d y}{d x}\right)^2\right]^{\frac{5}{2}} \\ \Rightarrow \quad\left(\frac{d^3 y}{d x^3}\right)^2=\left[1+\left(\frac{d y}{d x}\right)^2\right]^5 \end{aligned} $

Highest derivative is third order and power of highest order derivative is 2 .

Hence, order is 3 and degree is 2 .

To find the integrating factor of the given differential equation, we start with the equation:

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

Rewriting the equation, we have:

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

This simplifies to:

$ \frac{d y}{d x} - \cot x \cdot y = \operatorname{cosec} x $

The standard form for integrating factors is:

$ I F = e^{\int -\cot x \, dx} $

Calculating the integral, we find:

$ I F = e^{-\log |\sin x|} $

This can further be expressed as:

$ I F = e^{\log (\sin x)^{-1}} = \frac{1}{\sin x} $

Therefore, the integrating factor for the given differential equation is:

$ \frac{1}{\sin x} = \operatorname{cosec} x $

To solve the given differential equation:

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

we start by expressing it in differential form:

$ \frac{d y}{d x} = \frac{y \sin \frac{y}{x} - x}{x \sin \frac{y}{x}} $

Introduce a new variable $ v $ such that $ y = vx $. This implies:

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

Substitute $ y = vx $ into the differential equation:

$ v + x \frac{d v}{d x} = \frac{vx \sin v - x}{x \sin v} $

Simplify:

$ x \frac{d v}{d x} = \frac{vx \sin v - x - vx \sin v}{x \sin v} = \frac{-x}{x \sin v} = \frac{-1}{\sin v} $

This results in the following separable equation:

$ \sin v \, d v = -\frac{d x}{x} $

Integrate both sides:

$ \int \sin v \, d v = -\int \frac{d x}{x} $

The integration yields:

$ -\cos v = -\log x + C $

Rearranging gives:

$ \log x - \cos v = C $

Substituting back $ v = \frac{y}{x} $, we arrive at the solution:

$ \cos \left(\frac{y}{x}\right) = \log_e x + C $

To determine the sum of the order and degree of the given differential equation:

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

First, rewrite the equation in a form where the degree is clear:

$ \left(\frac{d^4 y}{d x^4}\right)^2 = \left(C + \left(\frac{d y}{d x}\right)^2\right)^3 $

From this, we observe:

The order of the differential equation is the highest derivative present, which is 4.

The degree of the differential equation is the power of the highest derivative when the equation is free of fractions and any radical signs. Here, the degree is 2 (as the highest derivative, $ \frac{d^4 y}{d x^4} $, is squared).

Therefore, the sum of the order and degree is $ 4 + 2 = 6 $.

To solve the differential equation:

$ (x+y) y \, dx + (y-x) x \, dy = 0 $

we can start by rewriting the equation:

$ \frac{dy}{dx} = \frac{(x+y) y}{(x-y) x} $

Substitute $ y = vx $, where $ v $ is a function of $ x $. Then, the derivative $ \frac{dy}{dx} $ becomes:

$ \frac{dy}{dx} = v + x \frac{dv}{dx} $

Substitute $ y = vx $ into the differential equation:

$ v + x\frac{dv}{dx} = \frac{(x + vx) vx}{(x - vx) x} $

Simplify the equation:

$ v + x\frac{dv}{dx} = \frac{v + v^2}{1-v} $

Rearranging gives:

$ x\frac{dv}{dx} = \frac{v + v^2 - v + v^2}{1-v} $

Next, separate variables and integrate:

$ \int \left(\frac{1-v}{v^2}\right) dv = \int 2x \, dx $

This can be rewritten as:

$ \int \left(\frac{1}{v^2} - \frac{1}{v}\right) dv = \int 2x \, dx $

Integrate both sides:

$ -\frac{1}{v} - \log v = 2 \log x + \log c $

Substitute back $ v = \frac{y}{x} $:

$ -\frac{x}{y} = \log \frac{y}{x} + \log x^2 + \log c $

Which simplifies to the equation:

$ -\frac{x}{y} = \log (cxy) $

Thus, the general solution is:

$ x + y \log (cxy) = 0 $

We are given the differential equation:

$ \left(y^2 + x + 1\right) \, dy = (y + 1) \, dx $

This can be rewritten as:

$ \frac{dx}{dy} - \frac{1}{y+1} x = \frac{y^2 + 1}{y+1} $

This is a linear differential equation. To solve it, we first find the integrating factor (IF):

$ \mathrm{IF} = e^{\int P \, dy} = e^{\int \frac{-1}{y+1} \, dy} = e^{-\log(y+1)} = \frac{1}{y+1} $

Multiplying through by the integrating factor gives:

$ x \cdot \frac{1}{y+1} = \int \frac{1}{y+1} \cdot \frac{y^2 + 1}{y+1} \, dy = \int \frac{y^2 + 1}{y^2 + 2y + 1} \, dy $

This simplifies as:

$ = \int \left(1 - \frac{2y}{y^2 + 2y + 1}\right) \, dy + C $

Carrying out the integration:

$ = y - \int \left[\frac{2y + 2}{y^2 + 2y + 1} - \frac{2}{(y+1)^2}\right] \, dy + C $

$ = y - \log(y+1)^2 + 2 \int \frac{1}{(y+1)^2} \, dy + C $

$ = y - \log(y+1)^2 - \frac{2}{y+1} + C $

Thus, the general solution is:

$ \frac{x+2}{y+1} + \log(y+1)^2 = y + C $

We have,

$ \left(\frac{d^2 y}{d x^2}\right)^{\frac{-7}{2}}\left(\frac{d^3 y}{d x^3}\right)^2-\left(\frac{d^2 y}{d x^2}\right)^{\frac{-5}{2}}\left(\frac{d^4 y}{d x^4}\right)=0 $

$ \Rightarrow\left(\frac{d^3 y}{d x^3}\right)^2\left(\frac{d^2 y}{d x^2}\right)^{\frac{5}{2}}=\left(\frac{d^4 y}{d x^4}\right)\left(\frac{d^2 y}{d x^2}\right)^{\frac{7}{2}} $

On squaring both sides, we get

$ \Rightarrow\left(\frac{d^3 y}{d x^3}\right)^4\left(\frac{d^2 y}{d x^2}\right)^5=\left(\frac{d^4 y}{d x^4}\right)^2\left(\frac{d^2 y}{d x^2}\right)^7 $

$\therefore$ Order is 4 and degree is 2 .

Hence, required difference $=2$

To solve the given differential equation:

$ x \, dy + \left(y + y^2 x \right) \, dx = 0 $

Rearrange it as:

$ \frac{dy}{dx} + \frac{y}{x} + y^2 = 0 $

Let's make a substitution to simplify this equation. Set $ \frac{1}{y} = t $, which implies $ -\frac{1}{y^2} \frac{dy}{dx} = \frac{dt}{dx} $.

This transforms the equation into:

$ \frac{dt}{dx} - \frac{1}{x} t = 1 $

This is a first-order linear differential equation of the form:

$ \frac{dt}{dx} -\frac{1}{x} t = 1 $

To solve this, we use an integrating factor (IF):

$ \mathrm{IF} = e^{\int -\frac{1}{x} \, dx} = \frac{1}{x} $

Multiplying through by the integrating factor gives:

$ \frac{1}{x} t = \int \frac{1}{x} \, dx + C $

Integrating the right side:

$ \frac{1}{x} t = \log x + C $

Substituting back in terms of $ y $:

$ \frac{1}{xy} = \log x + C $

Given the initial condition $ y = 1 $ when $ x = 1 $:

$ \frac{1}{1 \cdot 1} = \log 1 + C \Rightarrow 1 = 0 + C \Rightarrow C = 1 $

So the equation becomes:

$ \frac{1}{xy} = \log x + 1 $

Rearranging, we find:

$ xy(1 + \log x) = 1 $

Finally, solving for $ y $:

$ y = \frac{1}{x(1 + \log x)} $

To solve the differential equation $ x \, dy - y \, dx = \sqrt{x^2 + y^2} \, dx $, with the initial condition $ y(\sqrt{3}) = 1 $, follow these steps:

Start by rewriting the equation:

$ \frac{dy}{dx} = \frac{\sqrt{x^2 + y^2} + y}{x} $

Apply the substitution $ y = vx $, where $ v $ is a function of $ x $. Then, differentiate to get:

$ \frac{dy}{dx} = v + x \frac{dv}{dx} $

Substitute back into the equation:

$ v + x \frac{dv}{dx} = \frac{\sqrt{x^2 + (vx)^2} + vx}{x} $

Simplify:

$ v + x \frac{dv}{dx} = \frac{\sqrt{x^2(1 + v^2)} + vx}{x} $

$ x \frac{dv}{dx} = \sqrt{1 + v^2} $

Separate the variables:

$ \frac{1}{\sqrt{1 + v^2}} dv = \frac{1}{x} dx $

Integrate both sides:

$ \log|v + \sqrt{1 + v^2}| = \log x + \log c $

Combine the logs:

$ |v + \sqrt{1 + v^2}| = cx $

Substituting back for $ v = \frac{y}{x} $:

$ \frac{y}{x} + \frac{\sqrt{x^2 + y^2}}{x} = cx $

$ y + \sqrt{x^2 + y^2} = cx^2 $

Use the initial condition $ y(\sqrt{3}) = 1 $:

$ 1 + \sqrt{(\sqrt{3})^2 + 1^2} = c (\sqrt{3})^2 $

$ 1 + 2 = 3c $

$ 3 = 3c \implies c = 1 $

The required solution is:

$ y + \sqrt{x^2 + y^2} = x^2 $

To find the differential equation representing a family of circles with centers on the Y-axis, let's consider the circle with center $(0, K)$ and radius $r$. The general equation for such a circle is:

$ x^2 + (y - K)^2 = r^2 \quad \ldots \text{(i)} $

Differentiate equation (i) with respect to $x$:

$ 2x + 2(y - K)\frac{dy}{dx} = 0 $

Which simplifies to:

$ y - K = \frac{-x}{\frac{dy}{dx}} \quad \text{or} \quad K = y + \frac{x}{\frac{dy}{dx}} $

Let's denote $\frac{dy}{dx}$ as $y_1$. So, from the above, we have:

$ y - K = \frac{-x}{y_1} \quad \Rightarrow \quad K = y + \frac{x}{y_1} $

Now, differentiate this equation with respect to $x$ again:

$ 0 = 1 + \left(y - K\right) \frac{d^2y}{dx^2} + \left(\frac{dy}{dx}\right)^2 $

Substituting $y - K = \frac{-x}{y_1}$ into this, we get:

$ 1 + \left(-\frac{x}{y_1}\right)\frac{d^2y}{dx^2} + y_1^2 = 0 $

This can be rearranged to:

$ 1 - \frac{x \frac{d^2y}{dx^2}}{y_1} + y_1^2 = 0 $

$ x \frac{d^2y}{dx^2} = y_1(y_1^2 + 1) $

So, the differential equation is:

$ x \frac{d^2y}{dx^2} = y_1(y_1^2 + 1) $

To solve the given differential equation:

$\left(\sin y \cos^2 y - x \sec^2 y\right) dy = (\tan y) dx,$

we start by rewriting it in the standard linear form:

$\frac{dx}{dy} + \frac{x}{\sin y \cos y} = \cos^3 y.$

Recognizing this as a linear differential equation in $x$, we determine the integrating factor (IF):

Compute the Integrating Factor (IF):

$\mathrm{IF} = e^{\int \frac{1}{\sin y \cos y} \, dy}.$

Solve the Integral:

$\int \frac{1}{\sin y \cos y} \, dy = 2 \int \csc 2y \, dy.$

Solving this integral gives:

$e^{\int \csc 2y \, dy} = e^{\log |\csc 2y - \cot 2y|} = \csc 2y - \cot 2y.$

Thus, the integrating factor simplifies to:

$\mathrm{IF} = \frac{1 - \cos 2y}{\sin 2y} = \frac{2 \sin^2 y}{2 \sin y \cos y} = \tan y.$

Apply the Integrating Factor:

Multiply the entire linear differential equation by this integrating factor:

$x \times \tan y = \int \cos^3 y \times \tan y \, dy.$

Integrate the Right-Hand Side:

Substitute $\cos^2 y = 1 - \sin^2 y$. Then:

$\int \cos^3 y \tan y \, dy = \int \cos^2 y \sin y \, dy.$

Perform the integration:

$= -\frac{\cos^3 y}{3} + C.$

Resulting General Solution:

Therefore, the general solution to the differential equation is:

$3x \tan y + \cos^3 y = C.$

To solve the given differential equation, we start by analyzing the equation:

$ (x-y-1) dy = (x+y+1) dx $

This can be rewritten in differential form as:

$ \frac{dy}{dx} = \frac{x+y+1}{x-y-1} $

This expression can also be represented as:

$ \frac{dy}{dx} = \frac{x+(y+1)}{x-(y+1)} $

Introduce a substitution where $ y+1 = z $. Thus, we have:

$ \frac{dy}{dx} = \frac{dz}{dx} $

After substitution, the equation becomes:

$ \frac{dz}{dx} = \frac{x+z}{x-z} $

Next, we use another substitution $ z = vx $. Therefore, we differentiate $ z $ with respect to $ x $:

$ \frac{dz}{dx} = v + x \frac{dv}{dx} $

Substituting this back into our equation results in:

$ v + x \frac{dv}{dx} = \frac{1+v}{1-v} $

Reorganize terms and separate variables:

$ x \frac{dv}{dx} = \frac{1+v-v+v^2}{1-v} $

This simplifies to:

$ \left(\frac{1-v}{1+v^2}\right) dv = \frac{dx}{x} $

Integrating both sides, we have:

$ \int \left( \frac{1}{1+v^2} - \frac{v}{1+v^2} \right) dv = \int \frac{dx}{x} $

The integrals yield:

$ \tan^{-1} v - \frac{1}{2} \log \left| 1+v^2 \right| = \log x + C $

Substitute back $ v = \frac{z}{x} $:

$ \tan^{-1} \frac{z}{x} - \frac{1}{2} \log \left| \frac{z^2}{x^2} \right| = \log x + C $

Simplifying using $ z = y+1 $:

$ \tan^{-1} \left(\frac{y+1}{x}\right) - \frac{1}{2} \log \left| \frac{x^2 + (y+1)^2}{x^2} \right| - \log x = C $

Finally, this simplifies to:

$ \tan^{-1} \left(\frac{y+1}{x}\right) - \frac{1}{2} \log \left| x^2 + y^2 + 2y + 1 \right| = C $