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

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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}

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