Mark (√) against the correct answer in the following: The solution of the $\mathrm{DE} \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{-2 \mathrm{xy}}{\left(\mathrm{x}^{2}+1\right)}$ is A. $y^{2}(x+1)=c$ B. $y\left(x^{2}+1\right)=\mathrm{C}$ C. $x^{2}(y+1)=\mathrm{c}$ D. None of these

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

Mark (√) against the correct answer in the following: The solution of the $\mathrm{DE} \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{-2 \mathrm{xy}}{\left(\mathrm{x}^{2}+1\right)}$ is
A. $y^{2}(x+1)=c$
B. $y\left(x^{2}+1\right)=\mathrm{C}$
C. $x^{2}(y+1)=\mathrm{c}$
D. None of these

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

Mark (√) against the correct answer in the following: The solution of the $\mathrm{DE} \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{-2 \mathrm{xy}}{\left(\mathrm{x}^{2}+1\right)}$ is
A. $y^{2}(x+1)=c$
B. $y\left(x^{2}+1\right)=\mathrm{C}$
C. $x^{2}(y+1)=\mathrm{c}$
D. None of these

Solution:

$\begin{array}{l} \frac{d y}{d x}=\frac{-2 x y}{x^{2}+1} \\ \frac{d y}{y}=\frac{-2 x d x}{x^{2}+1} \end{array}$
Let $\mathrm{x}^{2}+1=\mathrm{t}$
On differentiating on both sides we get $2 \mathrm{xdx}=\mathrm{dt}$
$\frac{d y}{y}=\frac{-d t}{t}$
On integrating on both sides, we get
$\begin{array}{l} \operatorname{Iogy}=-\log t+C \\ \operatorname{Logy}+\operatorname{Logt}=C \\ \text { Logyt }=\mathrm{C} \\ \mathrm{yt}=\mathrm{C} \end{array}$
As $\mathrm{t}=x^{2}+1$
Therefore, $y\left(x^{2}+1\right)=c$ is the solution of $\frac{d y}{d x}=\frac{-2 x y}{x^{2}+1}$