A. The general solution of the differential equations $\frac{d y}{d x}=\frac{1+y^2}{1+x^2}$ is $\tan ^{-1} y=\tan ^{-1} x+c$; Where c is a constant
B. The general solution of the differential equation $\frac{d y}{d x}+\sqrt{\frac{1-y^2}{1-x^2}}=0$ is $\sin ^{-1} y-\sin ^{-1} x=c$; Where c is a constant
C. The general solution of the differential equation $\frac{y d x-x d y}{y}=0$ is y = cx; Where c is a constant
D. The general solution of the differential equation $\frac{d y}{d x}=e^{x+y}$ is $e^x+e^y=c$; Where c is a constant
E. The differential equation $\left(x^2+x y\right) d y=\left(x^2+y^2\right) d x$ is homogeneous differential equation.
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) - A, C, E only
(A) $\frac{d y}{d x}=\frac{1+y^2}{1+x^2}⇒\int\frac{1}{1+y^2}dy=\int\frac{1}{1+x^2}dx$
$\tan ^{-1} y=\tan ^{-1} x+c$ → True
(B) $\frac{d y}{d x}+\sqrt{\frac{1-y^2}{1-x^2}}=0⇒\int\frac{+1}{\sqrt{1-y^2}}dy=\int\frac{1}{\sqrt{1-x^2}}dx$
$⇒\sin ^{-1} y-\sin ^{-1} x=c$ → False
(C) $\frac{y d x-x d y}{y}=0⇒\frac{ydx-xdy}{y^2}=0$
so $\int d(\frac{x}{y})=0$
$\frac{x}{y}=k$ so $y = cx$ → True
(D) $\frac{d y}{d x}=e^xe^y$
so $\int e^{-y}dy=\int e^xdx$
$-e^{-y}=e^x+c$ → False
(E) $\left(x^2+x y\right) d y=\left(x^2+y^2\right) d x$
$\frac{dy}{dx}=\frac{x^2+y^2}{x^2+xy}$
$\frac{dy}{dx}=\frac{1+(\frac{y}{x})^2}{1+(\frac{y}{x})}=f(\frac{y}{x})$ → True