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Please Solve RD Sharma Class 12 Chapter 21 Differential Equation Exercise 21.10 Question 2 Maths Textbook Solution.

Answers (1)

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

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

Give: 4 \frac{d y}{d x}+8 y=5 e^{-3 x}

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

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

I\! f=e^{\int f(x) d x} \\

=e^{\int 2 d x} \\

=e^{2 x} \\

=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}


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