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Please solve RD Sharma Class 12 Chapter 21 Differential Equation Exercise 21.10 Question 2 maths textbook solution.

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Answer : y=\frac{-5}{4} e^{-3 x}+C e^{-2 x}

Hint : To solve this equation we use e\; \int \; fx\; dx formula

Given : 4\frac{dy}{dx}+8y=5e^{-3x}

Solution : \frac{dy}{dx}+2y=\frac{5}{4}e^{-3x}

               \begin{aligned} &P(x)=2, Q(x)=\frac{5}{4} e^{-3 x} \\ &I f=e^{\int f(x) d x} \\ &=e^{\int 2 d x} \\ &=e^{2 x} \end{aligned}

               \begin{aligned} &=y e^{2 x}=\int e^{2 x} \times \frac{5}{4} e^{-3 x} d x+C \\ &=y e^{2 x}=\frac{-5}{4} e^{-x} d x+C \\ &=y=\frac{-5}{4} e^{-3 x}+C e^{-2 x} \end{aligned}

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