Differential Equations Test - 10

Given,

\(d y=x \sec \frac{y}{x} d x+\frac{y}{x} d x\)

\(\frac{d y}{d x}=x \sec \frac{y}{x}+\frac{y}{x} \)

\(\text { Put } y=v x \text {, and } \frac{d y}{d x}=v+x \frac{d v}{d x} \)

\(v+x \frac{d v}{d x}=x \sec v+v\)

\(\frac{d v}{d x}=\sec v\)

\(\cos v d v=d x\)

By integrating both sides we get,

\(\int \cos v d v=\int d x\) \(\sin v=x+c\)

Put \(v=\frac{y}{x}\) in the above equation we get

\(\sin \frac{y}{x}=x+c\)

Put \(y(1)=\frac{\pi}{ 2}\) in the above equation we get \(c=0\)

\(y=x \sin ^{-1} x\)

Put \(x=\frac{1}{2}\) in the above equation we get

\(y=\frac{1}{2} \sin ^{-1} \frac{1}{2}\)

\(=\frac{\pi}{12}\)

Hence, the correct option is (A).

Given,

\(\frac{x d y}{d x}+3 y=4 x^{3}\)...(i)

\(\Rightarrow \frac{ dy }{ d x}+\frac{3 y }{ x }=4 x ^{2}\)

By comparing equation (i) with \(\frac{ dy }{ dx }+ Py = Q\) we get,

\(P=\frac{3}{x}\) and \(Q=4 x^{2}\)

\(\Rightarrow I.F. =e^{\int P d x}=e^{\int \frac{3}{x} dx }\)\(\quad\quad(\because \int\frac{1}{x}=\log  x)\)

\(\Rightarrow I. F .=e^{3 \log x}\)

\(\Rightarrow I. F. =e^{\log x^{3}}\)

\(\Rightarrow I.F. =x^{3} (\because e^{\log x}=x)\)

Now general solution will be,

\( y . \text { (I. F. })=\int( Q .( I . F )) dx + c\)

\(\Rightarrow y . \left( x ^{3}\right)=\int\left(4 x ^{2} ×\left( x ^{3}\right)\right) d x + c\)

\(\Rightarrow x ^{3} . y =\int 4 x ^{5} dx + c\)

\(\Rightarrow x ^{3} . y = \frac{4 x ^{6}}{6}+ c\)

\(\Rightarrow x^{3} . y=\frac{2}{3} x^{6}+c\)

Hence, the correct option is (C).

Given,

\(\left(1-x^{2}\right) \frac{d y}{d x}-x y=1\)....(i)

Dividing both sides in equation (i) by \(1-x^{2}\), we will be able to get it in the standard form.

\(\frac{\mathrm{dy}}{\mathrm{dx}}+\left(\frac{-\mathrm{x}}{1-\mathrm{x}^{2}}\right) \mathrm{y}=\frac{1}{1-\mathrm{x}^{2}} \)

\(\mathrm{P}=\frac{-\mathrm{x}}{1-\mathrm{x}^{2}}\)...(ii)

Let's calculate \(\int P d x\).

\(\int P d x=\int \frac{-x}{1-x^{2}} d x\)

Substituting \(1-x^{2}=t\) in equation (ii), so that \(-2 x d x=d t\), we get:

\(\int P d x=\frac{1}{2} \int \frac{1}{t} d t\)

Using \(\int \frac{1}{x} \mathrm{~d} \mathrm{x}=\log \mathrm{x}+\mathrm{C}\) and ignoring the constant \(\mathrm{C}\), we get:

\(\int \mathrm{P} \mathrm{dx}=\frac{1}{2} \log \mathrm{t}\)...(iii)

By substituting \(1-x^{2}=t\) in equation(iii) we get:

\(\int P d x=\frac{1}{2} \log \left(1-x^{2}\right)\)

\(\Rightarrow \int P d x=\log \sqrt{1-x^{2}}\)

Now, the integrating factor will be:

\(F=e^{\int P d x}\)

\(\Rightarrow F=e^{\log \sqrt{1-x^{2}}}\)

\(\Rightarrow F=\sqrt{1-x^{2}}\), which is the required integrating factor.

Hence, the correct option is (D).

Given,

\(\frac{ d ^{3} y}{d x^{3}}=\frac{d^{2} y}{d x^{2}}+\sin 60^{\circ}\)\(\quad(\because \sin 60^{\circ} = \frac{\sqrt{3}}{2})\)

\(\Rightarrow \frac{d^{3} y}{d x^{3}}=\frac{d^{2} y}{d x^{2}}+\frac{\sqrt{3}}{2}\)

Here, the highest derivative is \(\frac{ d ^{3} y}{ d x^{3}}\) so the order is 3.

The power of the highest derivative is one so the degree is 1.

\(\therefore\) Degree and order is \(1,3\).

Hence, the correct option is (A).

Given,

\(\left(x^{2}+y^{2}\right) d x=2 x y d y\)

\(\Rightarrow \frac{d y}{ d x}=\frac{(x^{2}+y^{2})}{2 x y}\) is homogeneous.

Put \(y=v x\)

\(\Rightarrow \frac{\mathrm{d} y}{d x}=v+\frac{x d v }{d x}\)

\(\Rightarrow v+x \frac{d v}{d x}=\frac{x^{2}+v^{2} x^{2}}{2 v x^{2}}=\frac{x^{2}\left(1+v^{\circ}\right)}{2 v x^{2}} \)

\(\Rightarrow x \frac{d v}{d x}=\frac{1+v^{2}}{2 v}-v=\frac{1-v^{2}}{2 v} \)

\(\Rightarrow \frac{2 v}{1-v^{2}} d v=\frac{1}{x} d x\)

\(\Rightarrow-\log \left(1-v^{2}\right)=\log x+\log c \)

\(\Rightarrow \log [\frac{1 }{(1-v^{2})}]=\log x+\log c \)

\(\Rightarrow [\frac{1}{(1-v^{2})}]=c x\)

\(\Rightarrow [\frac{1}{(1-\frac{y^{2}}{x^{2}})}]=c x\)

\(\Rightarrow [\frac{x^{2}}{(x^{2}-y^{2})}]=c x\)

\(\Rightarrow x^{2}=c x\left(x^{2}-y^{2}\right)\) is the general solution of the differential equation.

Hence, the correct option is (C).

Given,

\(\frac{\mathrm{d} x}{\mathrm{~d} y}=\left(1+\mathrm{x}^{2}\right)\left(1+\mathrm{y}^{2}\right)\)

\(\Rightarrow \frac{\mathrm{d} \mathrm{x}}{\left(1+\mathrm{x}^{2}\right)}=\left(1+\mathrm{y}^{2}\right) \mathrm{dy} \)

\(\Rightarrow\left(1+\mathrm{y}^{2}\right) \mathrm{dy}=\frac{\mathrm{d} \mathrm{x}}{\left(1+\mathrm{x}^{2}\right)}\)....(i)

Integrating both sides in equation (i) we get,

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

\(\Rightarrow y+\frac{y^{3}}{3}=\tan ^{-1} x+c\)

Hence, the correct option is (B).

Given equation is \(y=a e^{3 x} \cos (x+b)\)

There are two constants a and b so differentiate two times Differentiating wr.t \(\mathrm{x}\), we get

\(\Rightarrow y^{\prime}=3 a e^{3 x} \cos (x+b)-a e^{3 x} \sin (x+b)\)

\(\Rightarrow y^{\prime}=3 y-a e^{3 x} \sin (x+b)\)

\(\Rightarrow a e^{3 x} \sin (x+b)=3 y-y^{\prime}\)

Differentiating again w.r.t \(\mathrm{x}\), we get

\(\Rightarrow 3 \mathrm{ae}^{3 x} \sin (x+b)+a e^{3 x} \cos (x+b)=3 y^{\prime}-y^{\prime \prime}\)

\(\Rightarrow 3\left(3 y-y^{\prime}\right)+y-3 y^{\prime}-y^{n}=0\)

\(\Rightarrow 9 y-3 y^{\prime}+y-3 y^{\prime}-y^{n}=0\)

\(\Rightarrow y^{\prime \prime}+6 y^{\prime}-10 y=0\)

Hence, the correct option is (C).

Given,

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

\(\text { Let } \frac{y}{x}=t \)

\(\Rightarrow y=x t\)

Differentiating with respect to \(\mathrm{x}\), we get

\(\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=\mathrm{x} \frac{\mathrm{dt}}{\mathrm{dx}}+\mathrm{t}\)

Now,

\(\mathrm{x} \frac{\mathrm{d} t}{\mathrm{dx}}+\mathrm{t}=\sec \mathrm{t}+\mathrm{t}\)

\(\Rightarrow \mathrm{x} \frac{\mathrm{dt}}{\mathrm{dx}}=\sec \mathrm{t}\)

\(\Rightarrow \frac{\mathrm{dt}}{\sec \mathrm{t}}=\frac{\mathrm{dx}}{\mathrm{x}}\)

Integrating both sides, we get

\(\Rightarrow \int \frac{\mathrm{dt}}{\sec t}=\int \frac{\mathrm{dx}}{\mathrm{x}}\)

\(\Rightarrow \int \cos \mathrm{t} \mathrm{dt}=\int \frac{\mathrm{dx}}{\mathrm{x}} \)

\(\Rightarrow \sin \mathrm{t}=\log \mathrm{x}+\log \mathrm{c}\)

\(\Rightarrow \sin \mathrm{t}=\log (\mathrm{cx}) \quad(\because \log \mathrm{m}+\log \mathrm{n}=\log (\mathrm{mn})) \)

\(\therefore \sin \left(\frac{\mathrm{y}}{\mathrm{x}}\right)=\log (\mathrm{cx})\)

Hence, the correct option is (C).

Given,

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

For the given differential equation the highest order derivative is 2.

The given differential equation is not a polynomial equation because it involved a logarithmic term in its derivatives so, its degree is not defined.

Hence, the correct option is (D).

Given,

\(y=a \sin (\lambda x+a)\)....(i)

Now differentiating both sides equation (i) we get,

\(\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}\)=\(\frac{d}{dx}(a \sin (\lambda x+a))\)

\(\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}\) = \(\mathrm{a} \cos (\lambda \mathrm{x}+\alpha)\)×\(\frac{d}{dx}(\lambda \mathrm{x}+\alpha)\)

\(\Rightarrow \frac{\mathrm{dy}}{\mathrm{d} \mathrm{x}}=\mathrm{a} \lambda \cos (\lambda \mathrm{x}+\alpha)\)

Again differentiating both sides we get,

\(\Rightarrow \frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}=-\mathrm{a} \lambda^{2} \sin (\lambda x+\mathrm{a})\)

From equation (i) we get,

\(\Rightarrow \frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{d} \mathrm{x}^{4}}=-\lambda^{2} \mathrm{y} \)

\(\therefore \frac{\mathrm{d}^{2} y}{\mathrm{~d} \mathrm{x}^{2}}+\lambda^{2} \mathrm{y}=0\)

Hence, the correct option is (A).

Given,

Equation of parabola with \(\mathrm{y}\)-axis and vertex \((0, \mathrm{k})\) is:

\((x-0)^{2}=4 a(y-k)\)

\(\Rightarrow x^{2}=4 a y-4 a k\)

Taking derivative on both side, we get,

\(\frac{d}{dx}(x^{2})=\frac{d}{dx}(4 a y-4 a k)\)

\(\Rightarrow 2 {x}=4 \mathrm{a} \frac{\mathrm{dy}}{\mathrm{dx}} \)

\(\Rightarrow \frac{1}{\mathrm{x}} \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{1}{2 \mathrm{a}}\)

Again taking derivative on both side, we get,

\(\Rightarrow \frac{\mathrm{d}}{\mathrm{dx}}\left(\frac{1}{\mathrm{x}} \frac{\mathrm{dy}}{\mathrm{dx}}\right)=\frac{\mathrm{d}}{\mathrm{dx}}\left(\frac{1}{2 \mathrm{a}}\right) \)

\(\Rightarrow \frac{1}{\mathrm{x}} \frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}+\frac{\mathrm{dy}}{\mathrm{dx}}\left(\frac{-1}{\mathrm{x}^{2}}\right)=0\)\

\(\Rightarrow \mathrm{x} × \frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}-\frac{\mathrm{dy}}{\mathrm{dx}}=0\)

Hence, the correct option is (A).

Given,

\(\mathrm{y}^{2}+(\mathrm{x}-\mathrm{b})^{2}=\mathrm{c}\)

There are two constants \(\mathrm{b}\) and \(\mathrm{c}\) so differentiate two times

Differentiating w.r.t \(\mathrm{x}\)

\(2 y \frac{d y}{d x}+2(x-b)=0 \)

\(y \frac{d y}{d x}=b-x\)

Differentiating again w.r.t \(\mathrm{x}\)

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

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

Hence, the correct option is (C).

Given,

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

on seprating the variables we get,

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

On integrating, we get,

\(\int\frac{ dy }{1- y }= \int dx\)

\(\Rightarrow-\log (1-y)=x+C\)

\(\Rightarrow \log (1-y)^{-1}=x+C \quad\left[m \log n=\log n^{m}\right]\)

\(\Rightarrow \log \left|\frac{1}{1-y}\right|=x+C\)

Hence, the correct option is (A).

Given,

\(\left(y \frac{d y}{d x}-\frac{1}{x}\right)=0\)

\(\Rightarrow y \frac{d y}{d x}=\frac{1}{x}\)

\(\Rightarrow y d y=\frac{d x}{x}\)...(i)

Integrating both sides in equation (i) we get,

\(\Rightarrow \frac{y^{2}}{2}=\log x+\mathrm{c}\)

Hence, the correct option is (C).

Given,

\(y^{\prime \prime \prime}-\sin \left(y^{\prime}\right)+y=0\)

In the given equation the polynomial equation cannot be formed in \(y ^{\prime}\), so order and degree cannot be defined.

Hence, the correct option is (D).

Given,

\(y=c_{1} e^{x}+c_{2} e^{-x}\)

Differentiating w.r.t. \(x\) we get,

\(\frac{d y}{d x}=\frac{d}{d x} c_{1} e^{x}+\frac{d}{d x} c_{2} e^{-x}\)

\(\Rightarrow \frac{d y}{d x}=c_{1} e^{x}-c_{2} e^{-x}\)

\(\Rightarrow \frac{d}{d x}\left(\frac{d y}{d x}\right)=\frac{d}{d x}\left(c_{1} e^{x}-c_{2} e^{-x}\right)\)

\(\Rightarrow \frac{d^{2} y}{d x^{2}}=c_{1} e^{x}+c_{2} \mathrm{e}^{-x}=y\)

\(\Rightarrow \frac{d^{2} y}{d x^{2}}-y=0\)

Hence, the correct option is (B).

Given,

\(\frac{\mathrm{dy}}{\mathrm{dx}}=2^{\mathrm{x}-1}\)

\(\Rightarrow \frac{d y}{d x}=\frac{2^{x}}{2} \)

\(\Rightarrow 2 d y=2^{x} \mathrm{dx}\)

Now, variables are separated.

Integrating both sides, we get

\(\Rightarrow 2 \int d y=\int 2^{x} d x\) \(\quad(\because\int a^xdx = \frac{a^x}{loga})\)

\(\Rightarrow 2 y=\frac{1}{\log2} 2^{x}+c\)

Hence, the correct option is (C).

Given:

\(y=a \sin (\lambda x+a)\)...(i)

Now differentiating both sides, we get

\(\Rightarrow \frac{d y}{d x}=\frac{a d \sin (\lambda x+\alpha)}{d(\lambda x+\alpha)} \times \frac{\mathrm{d}\left(\lambda_{x}+\alpha\right)}{d x}\)

\(\Rightarrow \frac{\mathrm{dy}}{\mathrm{d} \mathrm{x}}=\mathrm{a} \lambda \cos (\lambda \mathrm{x}+\alpha)\)

Again differentiating both sides, we get

\(\Rightarrow \frac{\mathrm{d}^{2} y}{\mathrm{dx}^{2}}=-\mathrm{a} \lambda^{2} \sin (\lambda x+\mathrm{a})\)

From equation (i)

\(\Rightarrow \frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}=-\lambda^{2} \mathrm{y} \)

\(\therefore \frac{\mathrm{d}^{2} y}{\mathrm{~d} x^{2}}+\lambda^{2} \mathrm{y}=0\)

Hence, the correct option is (A).

Given,

\(y-b=a \sin x\)

There are two constants a and b so differentiate two times

Differentiating w.r.t \(\mathrm{x}\)

\(y^{\prime}=a \cos x\)....(i)

Again differentiating w.r.t \(x\)

\(y^{\prime \prime}=-a \sin x\)

From equation (i), \(a=\frac{y^{\prime}}{\cos x}\)

\(y^{\prime \prime}=-\frac{y^{\prime}}{\cos x} \times \sin x\)

\(y^{\prime \prime}+y^{\prime} \tan x=0\)

Hence, the correct option is (B).

Given,

\(\mathrm{x}^{2} \frac{\mathrm{dy}}{\mathrm{dx}}=\mathrm{x}^{2}+\mathrm{xy}+\mathrm{y}^{2} \)

\(\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=1+\frac{\mathrm{y}}{\mathrm{x}}+\left(\frac{y}{x}\right)^{2}\)

\(\text { Substituting } \mathrm{y}=v \mathrm{x} \text { and } \frac{\mathrm{dy}}{\mathrm{d} x}=\mathrm{v}+\mathrm{x} \frac{\mathrm{dv}}{\mathrm{dx}} \)

\(\Rightarrow \mathrm{v}+\mathrm{x} \frac{\mathrm{dv}}{\mathrm{d} \mathrm{x}}=1+\mathrm{v}+\mathrm{v}^{2}\)

\(\Rightarrow \mathrm{x} \frac{\mathrm{dv}}{\mathrm{d} x}=1+\mathrm{v}^{2}\)

Integrating both sides we get,

\(\int \frac{\mathrm{d} \mathbf{x}}{\mathbf{x}}=\int \frac{\mathrm{dv}}{1+\mathrm{v}^{2}} \)

\(\Rightarrow \log \mathrm{x}=\tan ^{-1} \mathrm{v}+\mathrm{c}, \mathrm{c}=\text { constant of integration }\)

Putting the value of \(\mathrm{v}\) we get,

\(\therefore \log \mathrm{x}=\tan ^{-1} \frac{y}{x}+c\)

Hence, the correct option is (A).

Given,

\(x d y-2 y d x=0\) 

\(\Rightarrow x d y=2 y d x\)

On separating variable

\(\Rightarrow \frac{d y}{y}=2 \frac{d x}{x}\)

Integrating both sides we get,

\(\Rightarrow \int \frac{d y}{y}=2 \int \frac{\mathrm{d} x}{x}\)

\(\Rightarrow \log \mathrm{y}=2 \log \mathrm{x}+\log \mathrm{c}\)

\(\Rightarrow \log \mathrm{y}=\log \mathrm{x}^{2}+\log \mathrm{c}\)

\(\Rightarrow \log \mathrm{y}-\log\mathrm{x}^{2}=\log\mathrm{c}\)

\(\Rightarrow \log \frac{y}{x^{2}}=\log \mathrm{c}\)

\(\therefore \mathrm{y}=\mathrm{x}^{2} \mathrm{c}\)

Hence, the correct option is (C).

Given,

\(\mathrm{x} \frac{\mathrm{d} y}{\mathrm{~d} x}-\mathrm{y}=\mathrm{x}^{2}\)

\(\Rightarrow \frac{\mathrm{dy}}{\mathrm{d} \mathrm{x}}-\frac{\mathrm{y}}{\mathrm{x}}=\mathrm{x}\)

It is linear differential equation is of first order.

\( \operatorname{IF}=e^{\frac{-1}{x}} d x\)

\(\Rightarrow \operatorname{IF}=e^{-\ln x}\)

\(\Rightarrow  \operatorname{IF}=\frac{1}{x} \)

\(\text { Now, } y \times( \operatorname{IF})=\int Q( \operatorname{IF}) d x \)

\(\Rightarrow y \times \frac{1}{x}=\int x \times \frac{1}{x} d x \)

\(\Rightarrow \frac{y}{x}=\int d x\)

Integrating,

\(\Rightarrow \frac{y}{x}=x+c\) (where \(c\) is integration constant)

\(y(1)=1\)

\(\Rightarrow \frac{1}{1}=1+c \)

\(\Rightarrow c=0 \)

\( \frac{y}{x}=x \text { OR } y=x^{2}\)

For \(\mathrm{y}(2)\)

\(y=2^{2} \)

\(\Rightarrow y=4\)

Hence, the correct option is (D).

Given,

\(x d y=y d x+y^{2} d y\)

\(\Rightarrow x d y-y d x=y^{2} d y\)

\(\Rightarrow \frac{x d y-y d x}{y^{2}}=d y\) \(\quad\quad(\because d\left(\frac{x}{y}\right)=\frac{y d x-x d y}{y^{2}}=\frac{x d y-y d x}{y^{2}}=-d\left(\frac{x}{y}\right))\)

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

Integrating both side,

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

\(\Rightarrow-\frac{x}{y}=y+c\)........(i)

Given that \(y(1)=1\)

So by putting \(x=1\) and \(y=1\) in equation (i) we get,

\(\Rightarrow \frac{-1}{1}=1+c\)

\(\Rightarrow c=-2\)

\(\Rightarrow-\frac{x}{y}=y-2\)

\(\Rightarrow y^{2}-2 y+x=0\) (particular solution).......(ii)

Now the value of \(y(-3)\) put \(x=-3\) in equation (ii) we get,

\(y^{2}-2 y-3=0\)

\(\Rightarrow y^{2}-3 y+y-3=0\)

\(\Rightarrow y(y-3)+1(y-3)=0\)

\(\Rightarrow(y-3)(y+1)=0\)

So two values of \(y\) we will get \(y=3, y=-1\)

But in our question given that \(y>0\) Therefore we will take \(y=3\).

Hence, the correct option is (A).

Given,

\(\frac{d y}{d x}+y \tan x=2 \sin x\)...(i)

By comparing with \(\frac{d y}{d x}+P y=Q\) we get,

\(P=tan x\) and \(Q=2 sin x\)

Integrating factor (I.F) of the equation is given by

\(\text { I. } F=e^{\int P d x}\)

\(I . F=e^{\int \tan x d x}\)

\(=e^{\log \sec x}\)

\(=\sec x\)

Its solution is:

\(y \times(I . F)=\int[Q \times(I . F)] d x+C\)

\(y \sec x=\int[2 \sin x \times \sec x] d x+C \)

\(y=\int\left[2 \sin x \times \frac{1}{\cos x}\right] d x+C\)

\(y=\int[2 \tan x] d x+C=2 \log \sec x+C \)

\(y=\frac{2 \log \sec x+C}{\sec x}\)

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

Hence, the correct option is (A).

Given,

\(\frac{d y}{d x}=e^{-3 y}\)

\(\Rightarrow\frac{d y}{e^{-3 y}}=d x\)

\(\Rightarrow e^{3 y} d y=d x\)

on integrating both side we get,

\(\int e^{3 y} d y=\int d x\)

\(\Rightarrow\frac{e^{3 y}}{3}=x+C\).......(i)

\(\Rightarrow\frac{e^{2 (0)}}{3}=5+C\)  \(\quad\quad(\because{e^{0}} = 1)\)

\(\Rightarrow C+5= \frac{1}{3}\)

\(\Rightarrow C= \frac{1}{3}-5\)

\(\Rightarrow C=\frac{-14}{3}\)

Substituting \(y=5\) and \(C = \frac{-14}{5}\), we get,

\(\frac{e^{15}}{3}=x-\frac{14}{3}\)

\(\Rightarrow x=\frac{e^{15}+14}{3}\)

Hence, the correct option is (A).

The family of circle having centre \((0,0)\) and radius \(r\) is:

\(x^{2}+y^{2}=r^{2}\)

There is only one constant \(r\).

Differentiating w.r.t. \(x\) we get,

\(\frac{d}{dx}(x^{2}+y^{2})=\frac{d}{dx}r^{2}\)

\(\Rightarrow 2 x+2 y \frac{d y}{d x}=0\)

\(\Rightarrow \frac{d y}{d x}=-\frac{x}{y}\)

Hence, the correct option is (C).

Given,

\(2 \mathrm{y} \frac{\mathrm{d} \mathrm{x}}{\mathrm{dy}}+\mathrm{x}=5 \mathrm{y}^{2}\)....(i)

Equation (i) can be simplified as:

\(\frac{\mathrm{d} x}{\mathrm{~d} y}+\frac{x}{2 y}=\frac{5}{2} y\)

On comparing eqn \((i)\) with standard eqn,\(\frac{\mathrm{dx}}{\mathrm{dy}}+\mathrm{Px}=\mathrm{Q}\), we get

\(P=\frac{1}{2 y} \text { and } Q=\frac{5}{2} y \)

Therefore,

\(I F=e^{\int P d y}=e^{\int \frac{1}{2 y} d y} \)

\(\Rightarrow I F=e^{\frac{1}{2} \log y}=e^{\log y^{\frac{1}{2}}}\)

\(\Rightarrow I F=\sqrt{y}\)\(\quad\quad(\because \mathrm{e}^{a \log x}=x^{a})\)

Hence, the correct option is (A).

Given,

\(\frac{y^{2}}{x^{2}}=\frac{d y}{d x}\)

Separate the variables,

\(\frac{ dx }{x^{2}}=\frac{d y}{y^{2}}\)

on integrating both side we get,

\(\int \frac{ dx }{ x ^{2}}=\int \frac{d y}{y^{2}}\)\(\quad\quad(\because\int\frac{1}{x^{2}}dx=\frac{-1}{x})\)

\(\Rightarrow-\frac{1}{x}+C=-\frac{1}{y}\)

\(\Rightarrow \frac{1}{y}=\frac{1}{x}+C\)

Hence, the correct option is (A).

Given,

\(y^{2}=a\left(b^{2}-x^{2}\right)\)

Differentiating w.r.t \(x\)

\(\Rightarrow 2 y y^{\prime}=a(-2 x)\)

\(\Rightarrow y y^{\prime}=-a x\)

Differentiating w.r.t \(x\) again

\(\Rightarrow y y^{\prime \prime}+\left(y^{\prime}\right) 2=-a\)

From (i) and (ii) we get,

\(\Rightarrow y y^{\prime}=x\left(y y^{\prime \prime}+\left(y^{\prime}\right)^{2}\right) \)

\(\Rightarrow y y^{\prime}-x y y^{\prime \prime}-x\left(y^{\prime}\right)^{2}=0\)

Hence, the correct option is (D).

Given,

\(\sin x \frac{d y}{d x}+\frac{y}{\sin x}=x \sin x e^{\cot x}\)

\(\frac{d y}{d x}+\frac{y}{\sin ^{2} x}=x . e^{\cot x}\)

It is form of \(\frac{d y}{d x}+P y=Q\)

\(\text { I. } F .=e^{\int p d x}\)

\(\text { I. } F .=e^{\int \operatorname{cosec}^{2} x d x}\)

\(=e^{-\cot x}\)

The solution of the linear equation is given by

\(y(I . F)=\int Q(I . F \cdot) d x+c\)

\(\Rightarrow y e^{-\cot x}=\int x e^{\cot x} \cdot e^{-\cot x} d x+c \)

\(\Rightarrow y e^{-\cot x}=\int x d x+c \)

\(\Rightarrow y e^{-\cot x}=\frac{x^{2}}{2}+c\) x}\)

Hence, the correct option is (A).