Mark (√) against the correct answer in the following: the general solution of the DE $\left(1+x^{2}\right) d y-x y d x=0$ is A. $y=C\left(1+x^{2}\right)$ B. $y^{2}=\mathrm{C}\left(1+x^{2}\right)$ C. $y \sqrt{1+x^{2}}=C$ D. None of these

Home » Mark (√) against the correct answer in the following: the general solution of the DE is A. B. C. D. None of these

Mark (√) against the correct answer in the following: the general solution of the DE $\left(1+x^{2}\right) d y-x y d x=0$ is
A. $y=C\left(1+x^{2}\right)$
B. $y^{2}=\mathrm{C}\left(1+x^{2}\right)$
C. $y \sqrt{1+x^{2}}=C$
D. None of these

CBSE | Class 12 | Linear Differential Equations | Maths | RS Aggarwal

Mark (√) against the correct answer in the following: the general solution of the DE $\left(1+x^{2}\right) d y-x y d x=0$ is
A. $y=C\left(1+x^{2}\right)$
B. $y^{2}=\mathrm{C}\left(1+x^{2}\right)$
C. $y \sqrt{1+x^{2}}=C$
D. None of these

Solution:

$\left(1+x^{2}\right) d y-x y d x=0$
$\frac{d y}{y}=\frac{x}{1+x^{2}} d x$
Let $1+\mathrm{x}^{2}=\mathrm{t}$
$\begin{array}{r} 2 \mathrm{x} \mathrm{dx}=\mathrm{dt} \\ \frac{d y}{y}=\frac{d t}{2 t} \end{array}$
On integrating on both sides we get
$\log y=\frac{\log t}{2}+C$ $2 \log y=\log t+C$ $\log y^{2}=\log t+C$ $y^{2}=\left(1+x^{2}\right) c$ (the solution of given equation)