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Please solve RD Sharma Class 12 Chapter 21 Differential Equation Exercise 21.10 Question 3 maths textbook solution.

Answers (1)

Answer : ye^{2x}=2e^{3x}+C

Hint : To solve this equation we use \frac{dy}{dx}+Py=Q formula

Give : \frac{dy}{dx}+2y=6e^{x}

Solution : \frac{dy}{dx}+2y=6e^{x}         .....(i)

                \frac{dy}{dx}+Py=Q

                \begin{aligned} &P=2, Q=6 e^{x} \\ &I f=e^{\int Pd x} \\ &=e^{2 \int d x} \\ &=e^{2 x} \end{aligned}

                \begin{aligned} &=y \times I f=\int(I f \times Q) d x \\ &=y \times e^{2 x}=\int e^{2 x} \times 6 e^{x} d x \\ &=y e^{2 x}=6 \int e^{3 x} d x \end{aligned}

                \begin{aligned} &=y e^{2 x}=\frac{6}{3} e^{x}+C \\ &=y e^{2 x}=2 e^{x}+C \end{aligned}

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