Differential Equations

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} $

$\sqrt{\frac{d^2 y}{d x^2}}=\sqrt[3]{\left[y \frac{d y}{d x}+x \sin \left(\frac{d y}{d x}\right)\right]^2}$

$ \begin{gathered} \Rightarrow\left(\frac{d^2 y}{d x^2}\right)^{\frac{1}{2}}=\left[y \frac{d y}{d x}+x \sin \left(\frac{d y}{d x}\right)\right]^{\frac{2}{3}} \\ \Rightarrow\left(\frac{d^2 y}{d x^2}\right)^{\frac{1}{2} \times 6}=\left[y \frac{d y}{d x}+x \sin \left(\frac{d y}{d x}\right)\right]^{\frac{2}{3} \times 6} \\ \quad\left(\frac{d^2 y}{d x^2}\right)^{\prime}=\left[y \frac{d y}{d x}+x \sin \left(\frac{d y}{d x}\right)\right]^4 \end{gathered} $

Order $=2$

Degree $=$ Not defined

( ∵ The above differential equation is not a polynomial in $\frac{d y}{d x}$ )

$ \begin{aligned} &\text { } \frac{d y}{d x}=\frac{(x-2)(y+1)}{(x+1)(y-2)}\\ &\begin{aligned} & \Rightarrow \quad \int \frac{y-2}{y+1} d y=\int \frac{x-2}{x+1} d x \\ & \Rightarrow \quad \int\left(1-\frac{3}{y+1}\right) d y=\int\left(1-\frac{3}{x+1}\right) d x \\ & \Rightarrow \quad y-3 \ln |y+1|=x-3 \ln |x+1|+c \\ & \Rightarrow \quad x-y+3 \ln \left|\frac{y+1}{x+1}\right|=c \end{aligned} \end{aligned} $

$ \because\left(x-x_1\right)^2+(y-3)^2=r^2\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)$

On differentiating,

$ \begin{array}{lc} & 2\left(x-x_1\right)+2(y-3) \cdot y^{\prime}=0 \\ \Rightarrow & x-x_1=-(y-3) y^{\prime} \end{array} $

$ \begin{aligned} &\text { On putting in Eq. (i), }\\ &\begin{aligned} (y-3)^2\left(y^{\prime}\right)^2+(y-3)^2 & =r^2 \\ 1+\left(\frac{d y}{d x}\right)^2 & =\frac{r^2}{(y-3)^2} \end{aligned} \end{aligned} $

Given equation

$ \begin{aligned} & x\left(\frac{d^2 y}{d x^2}\right)^{\frac{1}{3}}+2 x^2\left(\frac{d^2 y}{d x^2}\right)^{\frac{5}{3}}+7 \frac{d y}{d x}+y=0 \\ & \Rightarrow\left[x\left(\frac{d^2 y}{d x^2}\right)^{\frac{1}{3}}+2 x^2\left(\frac{d^2 y}{d x^2}\right)^{\frac{5}{3}}\right]^3=\left(7 \frac{d y}{d x}+y\right)^3 \end{aligned} $

Degree of differential equation $=5$

Given, $x y d y-\left(1+y^2\right) d x=0$

$ \begin{aligned} & \int \frac{y}{1+y^2} d y=\int \frac{d x}{x} \\ \Rightarrow & \frac{1}{2} \ln \left(1+y^2\right)=\ln x+c \end{aligned} $

Passes through points $(1,0)$

$ \begin{gathered} \Rightarrow \frac{1}{2} \ln 1=\ln 1+c \Rightarrow c=0 \\ \Rightarrow \frac{1}{2} \ln \left(1+y^2\right)=\ln x \\ \Rightarrow \ln \sqrt{1+y^2}=\ln x \sqrt{1+y^2}=x \\ \quad S_1: 1+y^2=x^2 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)\\ \quad S_2: x^2+3 y^2=3 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)\end{gathered} $

$ \begin{aligned} & y= \pm \frac{1}{\sqrt{2}}, x^2=1+\frac{1}{2}=\frac{3}{2} \Rightarrow x= \pm \sqrt{\frac{3}{2}} \\ & S_1: y^2=x^2-1 \quad \Rightarrow 2 y y^{\prime}=2 x \\ & \left(y^{\prime}\right)_1=\frac{x}{y}=\frac{x_1}{y_1} \\ & S_2: x^2+3 y^2=3 \quad \Rightarrow \quad 2 x+6 y y^{\prime}=0 \\ & \left(y^{\prime}\right)_2=\frac{-x_1}{3 y_1} \\ & \because \tan \theta=\left|\frac{\left(y^{\prime}\right)_1-\left(y^{\prime}\right)_2}{1+\left(y^{\prime}\right)_1\left(y^{\prime}\right)_2}\right| \\ & =\left|\frac{\frac{x_1}{y_1}+\frac{x_1}{3 y_1}}{1-\frac{x_1^2}{3 y_1^2}}\right|=\left|\frac{4 x_1 y_1}{3 y_1^2-x_1^2}\right| \\ & =\left|\frac{4 \times \sqrt{\frac{3}{2}} \times \frac{1}{\sqrt{2}}}{3 \cdot \frac{1}{2}-\frac{3}{2}}\right| \tan \theta=\left|\frac{2 \sqrt{3}}{0}\right|=\infty \\ & \theta=\frac{\pi}{2} \quad \therefore \quad \frac{2 \theta}{\pi}=\frac{2 \times \frac{\pi}{2}}{\pi}=1 \end{aligned} $

Here, $\frac{d y}{d x}=\cos ^2(3 x+y)$

On putting $3 x+y=t$

$\begin{array}{ll} 3+\frac{d y}{d x} & =\frac{d t}{d x} \\ \Rightarrow & \frac{d y}{d x}=\frac{d t}{d x}-3 \Rightarrow \frac{d t}{d x}-3=\cos ^2 t \\ \Rightarrow & \frac{d t}{d x}=\cos ^2 t+3 \Rightarrow \frac{d t}{\cos ^2 t+3}=d x \end{array}$

Integrate both side,

$\int \frac{d t}{\cos ^2 t+3}=\int d x$

$\begin{aligned} & \Rightarrow \int \frac{\sec ^2 t d t}{1+3 \sec ^2 t}=\int d x \Rightarrow \int \frac{\sec ^2 t}{1+3+3 \tan ^2 t}=\int d x \\ & \Rightarrow \int \frac{\sec ^2 t d t}{4+3 \tan ^2 t}=\int d x \end{aligned}$

$\begin{aligned} & \text { On putting tant }=m, \\ & \sec ^2 t d x=d m=\frac{1}{4} \int \frac{d m}{1+\left(\frac{\sqrt{3}}{2} m\right)^2} x+C \\ & =\frac{1}{4} \times \frac{2}{\sqrt{3}} \tan ^{-1}\left(\frac{\sqrt{3}}{2} m\right)=x+C \\ & =\frac{1}{2 \sqrt{3}} \tan ^{-1}\left[\frac{\sqrt{3}}{2} \tan t\right]=x+C \quad[\because m=\tan t] \\ & =\tan ^{-1}\left[\frac{\sqrt{3}}{2} \tan (3 x+y)=2 \sqrt{3}(x+C)\right] \quad[\therefore t=3 x+y] \end{aligned}$

So, $f(x)=2 \sqrt{3}(x+C)$

Given,

$\begin{aligned} & \cos ^2 x \frac{d y}{d x}+y=\tan x \\ \Rightarrow & \frac{d y}{d x}+y \sec ^2 x=\tan x \cdot \sec ^2 x \quad \text{... (i)} \end{aligned}$

Here, $p=\sec ^2 x$

$\begin{aligned} & \Rightarrow \quad \int p d p=\int \sec ^2 x d x=\tan x \\ & I F=e^{\tan x} \end{aligned}$

Multiplying Eq. (i) by $I F$, we get

$e^{\tan x} \frac{d y}{d x}+e^{\tan x} y \sec ^2 x=e^{\tan x} \cdot \tan x \cdot \sec ^2 x$

Integrating both sides, we get

$y e^{\tan x}=\int e^{\tan x} \tan x \cdot \sec ^2 x d x$

On putting $\tan x=t$,

$\begin{array}{ll} & \sec ^2 x d x=d t \\ \therefore & y e^t=\int t e^{\prime} d t=e^t(t-1)+C \\ \therefore & y e^{\tan x}=\tan x-1+C e^{-\tan x} \\ \text { If } & y\left(\frac{\pi}{4}\right)=1 \\ \Rightarrow & \quad 1=1-1+C e^{-1} \Rightarrow 1=C e^{-1} \\ \therefore & C=e \end{array}$

Any circle with given radius can be written as, $(x-h)^2+(y-k)^2=a^2$ where $(h, k)$ be the centre of the circle which is variable. So, in above algebraic equation, there are two arbitrary constant $h$ and $k$. Hence, order of differential equation will be second order. Hence, assertion and reason are true and reason is the correct explanation to assertion.

Any straight line which is at a constant distance $p$ from the origin is

$x \cos \alpha+y \sin \alpha=p\quad \text{.... (i)}$

Differentiating Eq. (i) on both sides w.r.t. $x$,

$\cos \alpha+\sin \alpha \frac{d y}{d x}=0 \Rightarrow \tan \alpha=-\frac{d x}{d y}=-\frac{1}{y^{\prime}}$

Now, $\sin \alpha=\frac{1}{\sqrt{1+y^{\prime 2}}}, \cos \alpha=\frac{-y^{\prime}}{\sqrt{1+y^{\prime 2}}}$

On putting Eq. (i), we get

$\frac{x\left(-y^{\prime}\right)}{\sqrt{1+y^{\prime 2}}}+\frac{y}{\sqrt{1+y^{\prime 2}}}=p$

$\begin{aligned} & \Rightarrow\left(y-x y^{\prime 2}\right)^2=p^2\left(l+y^{\prime 2}\right) \\ & \text { Order }(l)=1 \\ & \text { Degree }(m)=2 \\ & \therefore lm^2+l^2 m=(1 \cdot 2)^2+(1)^2 \cdot 2=4+2=6 \end{aligned}$

Given, $\frac{d y}{d x}=\frac{2 x-4 y-5}{x-2 y+2}$

Let $x-2=v$

$\therefore 1-2 \frac{d y}{d x}=\frac{d v}{d x} \Rightarrow \frac{d y}{d x}=\frac{\left(1-\frac{d v}{d x}\right)}{2}$

$\therefore \frac{\left(1-\frac{d v}{d x}\right)}{2}=\frac{2 v-5}{v+2}$

$\begin{aligned} & \Rightarrow 1-\frac{d v}{d x}=\frac{4 v-10}{v+2} \Rightarrow 1-\frac{4 v-10}{v+2}=\frac{d v}{d x} \\ & \Rightarrow \frac{d v}{d x}=\frac{12-3 v}{v+2} \Rightarrow \int \frac{(v+2)}{4-v} d v=3 \int d x+C \\ & \Rightarrow-1 \int \frac{-v+4-6}{(4-v)} d v=3 x+C \\ & \Rightarrow-\int d v+6 \int \frac{d v}{4-v}=3 x+C \end{aligned}$

$\begin{aligned} & \Rightarrow-(v)-6 \log |4-v|=3 x+C \\ & \Rightarrow-(x-2 y)-6 \log |4-x+2 y|=3 x+C \\ & \Rightarrow(4 x-2 y)+6 \log |4-x+2 y|=-C \\ & \Rightarrow(2 x-y)+3 \log |x-2 y-4|=k \quad\left(k=-\frac{C}{2}\right) \end{aligned}$

Thus, C = 3

Given, $y=(a+b) \sin (x+c)-d e^{x+e+f}\quad \text{... (i)}$

$ \begin{aligned} & \frac{d y}{d x}=(a+b) \cos (x+c)-d e^{x+c+f} \quad \text{... (ii)}\\ & \frac{d^2 y}{d x^2}=-(a+b) \sin (x+c)-d e^{x+e+f} \quad \text{... (iii)}\\ & =\frac{d^3 y}{d x^3}=-(a+b) \cos (x+c)-d e^{x+c+f} \quad \text{... (iv)} \end{aligned}$

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

$\left(y+\frac{d^2 y}{d x^2}\right)=-2 d e^{x+e+f}\quad \text{... (v)}$

Adding Eqs (ii) and (iv), we get

$\left(\frac{d y}{d x}+\frac{d^3 y}{d x^3}\right)=-2 d e^{x+e+s} \quad \text{... (vi)}$

On equating Eqs. (v) and (vi) we get

$\begin{aligned} & \frac{d^3 y}{d x^3}+\frac{d y}{d x}=\frac{d^2 y}{d x^2}+y \\ & \Rightarrow \frac{d^3 y}{d x^3}-\frac{d^2 y}{d x^2}+\frac{d y}{d x}-y=0 \end{aligned}$

Order of above differential equation is 3.

$\begin{aligned} & \frac{d y}{d x}-y \log _c 0.5=0 \\ & \Rightarrow \frac{d y}{y}=\log 0.5 d x \\ & \Rightarrow \quad \int \frac{d y}{y}=\int \log 0.5 d x \\ & \Rightarrow \log y=(\log 0.5) x+C \quad \text{... (i)}\\ & \because \quad y(0)=1 \text { i.e at } x=0, y=1 \\ & \therefore \log (1)=0+C \\ & \Rightarrow \quad C=0 \\ \end{aligned}$

Therefore, Eq. (i) becomes $\log y=(\log 0.5) x$

$\begin{aligned} & \Rightarrow \quad \log y=\log (0.5)^x \\ & \Rightarrow \quad y=(0.5)^x \end{aligned}$

When $x \rightarrow \infty, y \rightarrow k$

$\begin{array}{ll} \therefore & k=(0.9)^{\infty} \\ \Rightarrow & k=\left(\frac{1}{2}\right)^{\infty}=\frac{1}{2^{\infty}}=\frac{1}{\infty}=0 \end{array}$

$y=A e^x+B e^{-2 x}$

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

$\frac{d y}{d x}=A e^x-2 B e^{-2 x} \quad \text{... (i)}$

Again on differentiating w.r.t. $x$, we get

$\frac{d^2 y}{d x^2}=A e^x+4 B e^{-2 x} \quad \text{.... (ii)}$

Adding Eqs. (ii) and (iii), we get

$\begin{aligned} & \frac{d^2 y}{d x^2}+\frac{d y}{d x}=2 A e^x+2 B e^{-2 x}=2 y \\ \Rightarrow & \frac{d^2 y}{d x^2}+\frac{d y}{d x}-2 y=0 \end{aligned}$

Given that

$\left( {{{dy} \over {dx}}} \right)\tan x = y{\sec ^2}x + \sin x$

$\tan \left( {{{dy} \over {dx}}} \right) - y{\sec ^2}x = \sin x$

${{dy} \over {dx}} + \left( {{{ - {{\sec }^2}x} \over {\tan x}}} \right)y = {{\sin x} \over {\tan x}}$

Here, $P = - {{{{\sec }^2}x} \over {\tan x}}$ and $Q = \cos x$

$IF = {e^{\int {P\,dx} }} = {e^{\int {{{ - {{\sec }^2}x} \over {\tan x}}dx} }}$

$IF = {e^{ - \log (\tan x)}} = {e^{\log \cot x}} = \cot x$

Now, $y \times IF = \int {(Q \times IF)dx + C} $

$y\,.\,\cot x = \int {\cos x\,.\,\cot x + C} $

$y\,.\,\cot x = \int {{{1 - {{\sin }^2}x} \over {\sin x}} + C} $

$y\,.\,\cot x = \log |{\mathop{\rm cosec}\nolimits} \,x - \cot x| + \cos x + C$