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Please solve RD Sharma Class 12 Chapter 21 Differential Equation Exercise 21.10 Question 4 maths textbook solution.

Answers (1)

Answer : y=-e^{-2x}+Ce^{-x}

Hint : To solve this equation we use y\; I\; f formula

Give : \frac{dy}{dx}+y=e^{-2x}

Solution : \frac{dy}{dx}+Py=Q

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

\begin{aligned} &y \times e^{x}=\int e^{-2 x} e^{x} d x+C \\ &y e^{x}=\int e^{-x} d x+C \\ &y e^{x}=-e^{-x}+C \end{aligned}

\begin{aligned} &y=-e^{-x} e^{x}+C e^{-x} \\ &y=-e^{2 x}+C e^{-x} \end{aligned}

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