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Need solution for RD Sharma Maths Class 12 Chapter 21 Differential Equation Excercise 21.10 Question 4

Answers (1)

Answer:  y=-e^{-2 x}+C e^{-x}

Hint: To solve this equation we use  ylf  formula

Give:  \frac{d y}{d x}+y=e^{-2 x}

Solution:  \frac{d y}{d x}+P y=Q

\begin{aligned} &P=1, Q=e^{-2 x} \\ \end{aligned}

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

y \times I f=\int Q \times I\! f d x+C \\

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

\begin{aligned} &y e^{x}=\int e^{-x} d x+C \\ \end{aligned}

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

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

y=-e^{2 x}+C e^{-x}



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