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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   formula

Give:  \frac{d y}{d x}+2 y=e^{3 x}

Solution:   \begin{aligned} & \frac{d y}{d x}+P y=Q \\ & \end{aligned}

p=2, Q=e^{3 x} \\ 

\text { If } e^{\int p d x}

\begin{aligned} &=e^{\operatorname{\int xdx}} \\ & \end{aligned}

=e^{2 x} \\

y \cdot e^{2 x}=\int e^{3 x} \cdot e^{2 x} d x+C

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

\begin{aligned} &=\int e^{5 x} \\ & \end{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}


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