Get Answers to all your Questions

header-bg qa

Explain solution RD Sharma class 12 Chapter 21 Differential Equation exercise 21.10 question 10

Answers (1)

Answer:  y=e^{x}+C x

Hint: To solve this equation we will use differentiate separately.

Give:  \begin{aligned} &x \frac{d y}{d x}-y=(x-1) e^{x} \\ & \end{aligned}

Solution:  \frac{d y}{d x}-\frac{1}{x} y \frac{(x-1) e^{x}}{x}

\begin{aligned} &\quad \frac{d y}{d x}+P y=Q \\ & \end{aligned}

P=\frac{-1}{x}, Q=\frac{(x-1) e^{x}}{x} \\

I \! f=e^{\int P d x} \\

=e^{-\int \frac{1}{x} d x}

\begin{aligned} &=e^{\log x^{-1}} \\ & \end{aligned}

=x^{-1} \\

y I \! f=\int Q I\! f+C \\

=y \frac{1}{x}=\int(x-1) e^{x} \frac{1}{x} d x+C \\

=\frac{y}{x}=\int \frac{x e^{x}}{x^{2}} d x-\int \frac{e^{x}}{x^{2}} d x+C

\begin{aligned} &=\frac{1}{x} e^{x}-\int \frac{1}{x^{2}} e^{x} d x-\int \frac{1}{x^{2}} e^{x} d x \\ \end{aligned}

=\frac{1}{x} e^{x} \\

=\frac{y}{x}=\frac{e^{x}}{x}+C \\

=y=e^{x}+C x

Posted by


View full answer

Crack CUET with india's "Best Teachers"

  • HD Video Lectures
  • Unlimited Mock Tests
  • Faculty Support