(C) : We have differential equation, \(4 x y \frac{d y}{d x}=\frac{3(1+x)^{2}\left(1+y^{2}\right)}{\left(1+x^{2}\right)}\) Now separating the variable, we get- \(\frac{4 y d y}{\left(1+y^{2}\right)}=\frac{3(1+x)^{2}}{x\left(1+x^{2}\right)} \Rightarrow \frac{4 y d y}{1+y^{2}}=\left(\frac{3}{x}+\frac{6}{1+x^{2}}\right)\) Now integrating both side, we get- \(\int \frac{4 y d y}{1+y^{2}}=\int\left(\frac{3}{x}+\frac{6}{1+x^{2}}\right) d x\) Let, \(\quad 1+\mathrm{y}^{2}=\mathrm{t}\) \(2 \mathrm{ydy}=\mathrm{dt}\) \(2 \int \frac{\mathrm{dt}}{\mathrm{t}}=\int\left(\frac{3}{\mathrm{x}}+\frac{6}{1+\mathrm{x}^{2}}\right) \mathrm{dx}\) \(2 \log \mathrm{t}=3 \log |\mathrm{x}|+6 \tan ^{-1} \mathrm{x}+\mathrm{c}\) \(2 \log \left(1+y^{2}\right)=3 \log |x|+6 \tan ^{-1} x+c\)
UPSEE-2011
Differential Equation
87230
If \(y^{2}=\mathbf{P}(x)\) be a cubic polynomial, then \(2 \frac{d}{d x}\left(y^{3} \frac{d^{2} y}{d x^{2}}\right)\) is equal to
(A) : Thę differential ęquation- \(x \cos ^{2} y d x=y \cos ^{2} x d y\) Separating the variable, we get- \(\frac{x}{\cos ^{2} x} d x=\frac{y}{\cos ^{2} y} d y\) Integrating both side, we get- \(\int x \sec ^{2} x d x=\int y \sec ^{2} y d y\) \(x \int \sec ^{2} x d x-\int \frac{d}{d x}(x) \cdot \int \sec ^{2} x d x=y \int \sec ^{2} y d y\) \(\mathrm{x} \tan \mathrm{x}-\log |\sec \mathrm{x}|=\mathrm{y} \tan \mathrm{y}-\log |\sec \mathrm{y}|+\mathrm{c}\) \(-\int \frac{d}{d x}(y) \cdot \int \sec ^{2} y d y\) \(x \tan x-y \tan y=\log (\sec x / \sec y)+c\) \(\mathrm{x} \tan \mathrm{x}-\mathrm{y} \tan \mathrm{y}-\log (\sec \mathrm{x} / \mathrm{sec} \mathrm{y})=\mathrm{c}\)
UPSEE-2010]**#
Differential Equation
87232
The solution of the differential equation \(\frac{d y}{d x}=y \tan x-2 \sin x \text { is }\)
(D) : We have differential equation- \(\frac{d y}{d x}=y \tan x-2 \sin x \Rightarrow \frac{d y}{d x}-y \tan x=-2 \sin x\) Which is form of \(\frac{d y}{d x}+P y=Q\) \(P=-\tan x\) and \(Q=-2 \sin x\) \(I . F=e^{\int P d x}=e^{\int-\tan x d x}=e^{-\log \sec x}=\cos x\) Now particular solution, we get- \(y \cos x=\int \cos x(-2 \sin x) d x+c\) \(y \cos x=-\int \sin 2 x+c\) \(y \cos x=\frac{1}{2} \cos 2 x+c\)
(C) : We have differential equation, \(4 x y \frac{d y}{d x}=\frac{3(1+x)^{2}\left(1+y^{2}\right)}{\left(1+x^{2}\right)}\) Now separating the variable, we get- \(\frac{4 y d y}{\left(1+y^{2}\right)}=\frac{3(1+x)^{2}}{x\left(1+x^{2}\right)} \Rightarrow \frac{4 y d y}{1+y^{2}}=\left(\frac{3}{x}+\frac{6}{1+x^{2}}\right)\) Now integrating both side, we get- \(\int \frac{4 y d y}{1+y^{2}}=\int\left(\frac{3}{x}+\frac{6}{1+x^{2}}\right) d x\) Let, \(\quad 1+\mathrm{y}^{2}=\mathrm{t}\) \(2 \mathrm{ydy}=\mathrm{dt}\) \(2 \int \frac{\mathrm{dt}}{\mathrm{t}}=\int\left(\frac{3}{\mathrm{x}}+\frac{6}{1+\mathrm{x}^{2}}\right) \mathrm{dx}\) \(2 \log \mathrm{t}=3 \log |\mathrm{x}|+6 \tan ^{-1} \mathrm{x}+\mathrm{c}\) \(2 \log \left(1+y^{2}\right)=3 \log |x|+6 \tan ^{-1} x+c\)
UPSEE-2011
Differential Equation
87230
If \(y^{2}=\mathbf{P}(x)\) be a cubic polynomial, then \(2 \frac{d}{d x}\left(y^{3} \frac{d^{2} y}{d x^{2}}\right)\) is equal to
(A) : Thę differential ęquation- \(x \cos ^{2} y d x=y \cos ^{2} x d y\) Separating the variable, we get- \(\frac{x}{\cos ^{2} x} d x=\frac{y}{\cos ^{2} y} d y\) Integrating both side, we get- \(\int x \sec ^{2} x d x=\int y \sec ^{2} y d y\) \(x \int \sec ^{2} x d x-\int \frac{d}{d x}(x) \cdot \int \sec ^{2} x d x=y \int \sec ^{2} y d y\) \(\mathrm{x} \tan \mathrm{x}-\log |\sec \mathrm{x}|=\mathrm{y} \tan \mathrm{y}-\log |\sec \mathrm{y}|+\mathrm{c}\) \(-\int \frac{d}{d x}(y) \cdot \int \sec ^{2} y d y\) \(x \tan x-y \tan y=\log (\sec x / \sec y)+c\) \(\mathrm{x} \tan \mathrm{x}-\mathrm{y} \tan \mathrm{y}-\log (\sec \mathrm{x} / \mathrm{sec} \mathrm{y})=\mathrm{c}\)
UPSEE-2010]**#
Differential Equation
87232
The solution of the differential equation \(\frac{d y}{d x}=y \tan x-2 \sin x \text { is }\)
(D) : We have differential equation- \(\frac{d y}{d x}=y \tan x-2 \sin x \Rightarrow \frac{d y}{d x}-y \tan x=-2 \sin x\) Which is form of \(\frac{d y}{d x}+P y=Q\) \(P=-\tan x\) and \(Q=-2 \sin x\) \(I . F=e^{\int P d x}=e^{\int-\tan x d x}=e^{-\log \sec x}=\cos x\) Now particular solution, we get- \(y \cos x=\int \cos x(-2 \sin x) d x+c\) \(y \cos x=-\int \sin 2 x+c\) \(y \cos x=\frac{1}{2} \cos 2 x+c\)
(C) : We have differential equation, \(4 x y \frac{d y}{d x}=\frac{3(1+x)^{2}\left(1+y^{2}\right)}{\left(1+x^{2}\right)}\) Now separating the variable, we get- \(\frac{4 y d y}{\left(1+y^{2}\right)}=\frac{3(1+x)^{2}}{x\left(1+x^{2}\right)} \Rightarrow \frac{4 y d y}{1+y^{2}}=\left(\frac{3}{x}+\frac{6}{1+x^{2}}\right)\) Now integrating both side, we get- \(\int \frac{4 y d y}{1+y^{2}}=\int\left(\frac{3}{x}+\frac{6}{1+x^{2}}\right) d x\) Let, \(\quad 1+\mathrm{y}^{2}=\mathrm{t}\) \(2 \mathrm{ydy}=\mathrm{dt}\) \(2 \int \frac{\mathrm{dt}}{\mathrm{t}}=\int\left(\frac{3}{\mathrm{x}}+\frac{6}{1+\mathrm{x}^{2}}\right) \mathrm{dx}\) \(2 \log \mathrm{t}=3 \log |\mathrm{x}|+6 \tan ^{-1} \mathrm{x}+\mathrm{c}\) \(2 \log \left(1+y^{2}\right)=3 \log |x|+6 \tan ^{-1} x+c\)
UPSEE-2011
Differential Equation
87230
If \(y^{2}=\mathbf{P}(x)\) be a cubic polynomial, then \(2 \frac{d}{d x}\left(y^{3} \frac{d^{2} y}{d x^{2}}\right)\) is equal to
(A) : Thę differential ęquation- \(x \cos ^{2} y d x=y \cos ^{2} x d y\) Separating the variable, we get- \(\frac{x}{\cos ^{2} x} d x=\frac{y}{\cos ^{2} y} d y\) Integrating both side, we get- \(\int x \sec ^{2} x d x=\int y \sec ^{2} y d y\) \(x \int \sec ^{2} x d x-\int \frac{d}{d x}(x) \cdot \int \sec ^{2} x d x=y \int \sec ^{2} y d y\) \(\mathrm{x} \tan \mathrm{x}-\log |\sec \mathrm{x}|=\mathrm{y} \tan \mathrm{y}-\log |\sec \mathrm{y}|+\mathrm{c}\) \(-\int \frac{d}{d x}(y) \cdot \int \sec ^{2} y d y\) \(x \tan x-y \tan y=\log (\sec x / \sec y)+c\) \(\mathrm{x} \tan \mathrm{x}-\mathrm{y} \tan \mathrm{y}-\log (\sec \mathrm{x} / \mathrm{sec} \mathrm{y})=\mathrm{c}\)
UPSEE-2010]**#
Differential Equation
87232
The solution of the differential equation \(\frac{d y}{d x}=y \tan x-2 \sin x \text { is }\)
(D) : We have differential equation- \(\frac{d y}{d x}=y \tan x-2 \sin x \Rightarrow \frac{d y}{d x}-y \tan x=-2 \sin x\) Which is form of \(\frac{d y}{d x}+P y=Q\) \(P=-\tan x\) and \(Q=-2 \sin x\) \(I . F=e^{\int P d x}=e^{\int-\tan x d x}=e^{-\log \sec x}=\cos x\) Now particular solution, we get- \(y \cos x=\int \cos x(-2 \sin x) d x+c\) \(y \cos x=-\int \sin 2 x+c\) \(y \cos x=\frac{1}{2} \cos 2 x+c\)
(C) : We have differential equation, \(4 x y \frac{d y}{d x}=\frac{3(1+x)^{2}\left(1+y^{2}\right)}{\left(1+x^{2}\right)}\) Now separating the variable, we get- \(\frac{4 y d y}{\left(1+y^{2}\right)}=\frac{3(1+x)^{2}}{x\left(1+x^{2}\right)} \Rightarrow \frac{4 y d y}{1+y^{2}}=\left(\frac{3}{x}+\frac{6}{1+x^{2}}\right)\) Now integrating both side, we get- \(\int \frac{4 y d y}{1+y^{2}}=\int\left(\frac{3}{x}+\frac{6}{1+x^{2}}\right) d x\) Let, \(\quad 1+\mathrm{y}^{2}=\mathrm{t}\) \(2 \mathrm{ydy}=\mathrm{dt}\) \(2 \int \frac{\mathrm{dt}}{\mathrm{t}}=\int\left(\frac{3}{\mathrm{x}}+\frac{6}{1+\mathrm{x}^{2}}\right) \mathrm{dx}\) \(2 \log \mathrm{t}=3 \log |\mathrm{x}|+6 \tan ^{-1} \mathrm{x}+\mathrm{c}\) \(2 \log \left(1+y^{2}\right)=3 \log |x|+6 \tan ^{-1} x+c\)
UPSEE-2011
Differential Equation
87230
If \(y^{2}=\mathbf{P}(x)\) be a cubic polynomial, then \(2 \frac{d}{d x}\left(y^{3} \frac{d^{2} y}{d x^{2}}\right)\) is equal to
(A) : Thę differential ęquation- \(x \cos ^{2} y d x=y \cos ^{2} x d y\) Separating the variable, we get- \(\frac{x}{\cos ^{2} x} d x=\frac{y}{\cos ^{2} y} d y\) Integrating both side, we get- \(\int x \sec ^{2} x d x=\int y \sec ^{2} y d y\) \(x \int \sec ^{2} x d x-\int \frac{d}{d x}(x) \cdot \int \sec ^{2} x d x=y \int \sec ^{2} y d y\) \(\mathrm{x} \tan \mathrm{x}-\log |\sec \mathrm{x}|=\mathrm{y} \tan \mathrm{y}-\log |\sec \mathrm{y}|+\mathrm{c}\) \(-\int \frac{d}{d x}(y) \cdot \int \sec ^{2} y d y\) \(x \tan x-y \tan y=\log (\sec x / \sec y)+c\) \(\mathrm{x} \tan \mathrm{x}-\mathrm{y} \tan \mathrm{y}-\log (\sec \mathrm{x} / \mathrm{sec} \mathrm{y})=\mathrm{c}\)
UPSEE-2010]**#
Differential Equation
87232
The solution of the differential equation \(\frac{d y}{d x}=y \tan x-2 \sin x \text { is }\)
(D) : We have differential equation- \(\frac{d y}{d x}=y \tan x-2 \sin x \Rightarrow \frac{d y}{d x}-y \tan x=-2 \sin x\) Which is form of \(\frac{d y}{d x}+P y=Q\) \(P=-\tan x\) and \(Q=-2 \sin x\) \(I . F=e^{\int P d x}=e^{\int-\tan x d x}=e^{-\log \sec x}=\cos x\) Now particular solution, we get- \(y \cos x=\int \cos x(-2 \sin x) d x+c\) \(y \cos x=-\int \sin 2 x+c\) \(y \cos x=\frac{1}{2} \cos 2 x+c\)