Get Answers to all your Questions

header-bg qa

Need solution for RD Sharma maths Class 12 Chapter 21 Differential Equation Exercise Revision Exercise (RE) Question 64 Subquestion (vi) textbook solution.

Answers (1)

Answer : y=1+C e^{-x}

Hint              : You must know the rules of solving differential equation and integration

Given           : \frac{d y}{d x}+y=1

Solution : \frac{d y}{d x}+y=1

\begin{aligned} &\frac{d y}{d x}=1-y \\ &\frac{1}{1-y} d y=d x \end{aligned}

Integrating both sides

\begin{aligned} &\int \frac{1}{(1-y)} d y=\int d x \\ &-\int \frac{1}{y-1} d y=\int d x \\ &\int \frac{1}{y-1} d y=-\int d x \end{aligned}

\begin{aligned} &\log |y-1|=-x+\log c \\ &\log |y-1|-\log c=-x \\ &\log \left|\frac{y-1}{c}\right|=-x \end{aligned}

\begin{aligned} &\frac{y-1}{c}=e^{-x} \\ &y=1+C e^{-x} \end{aligned}

Posted by

infoexpert23

View full answer

Crack CUET with india's "Best Teachers"

  • HD Video Lectures
  • Unlimited Mock Tests
  • Faculty Support
cuet_ads