Target Exam

CUET

Subject

-- Mathematics - Section B1

Chapter

Differential Equations

Question:

Solution of the differential equation

$x=1+x y \frac{d y}{d x}+\frac{(x y)^2}{2 !}\left(\frac{d y}{d x}\right)^2+\frac{(x y)^3}{3 !}\left(\frac{d y}{d x}\right)^3+...$ is

Options:

$y=\log _e(x)+C$

$y=\left(\log _e x\right)^2+C$

$y= \pm \sqrt{\left(\log _e x\right)^2+2 C}$

$x y=x^y+k$

Correct Answer:

$y= \pm \sqrt{\left(\log _e x\right)^2+2 C}$

Explanation:

We have,

$x=e^{x y \frac{d y}{d x}}$

$\Rightarrow \log x=x y \frac{d y}{d x}$

$\Rightarrow y d y=\frac{\log x}{x} d x \Rightarrow y d y=\log x d(\log x)$

On integrating, we get

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

$\Rightarrow y^2 =\left(\log _e x\right)^2+2 C \Rightarrow y= \pm \sqrt{\left(\log _e x\right)^2+2 C}$