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Please solve RD Sharma Class 12 Chapter 21 Differential Equation Exercise 21.10 Question 1 maths textbook solution.

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Answer :  y=\frac{1}{5} e^{3 x}+C e^{-2 x}

Hint : To solve this equation we use I  formula

Given : \frac{dy}{dx}+2y=e^{3x}

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

                \begin{aligned} &p=2, Q=e^{3 x} \\ &\text { If } e^{\int p d x} \\ &=e^{\int 2d x} \\ &=e^{2 x} \end{aligned}

                \begin{aligned} &y \times I f=\int Q \times I f d x+C \\ &y \cdot e^{2 x}=\int e^{3 x} \cdot e^{2 x} d x+C \\ &=\int e^{5 x} \end{aligned}

               \begin{aligned} &=y e^{2 x}=\frac{e^{5 x}}{5}+C \\ &=y=\frac{1}{e^{2 x}}\left[\frac{e^{5 x}}{5}+C\right] \\ &=y=\frac{e^{3 x}}{5}+C e^{-2 x} \end{aligned}

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