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Please Solve RD Sharma Class 12 Chapter 21 Differential Equation Exercise 21.10 Question 14 Maths Textbook Solution.

Answers (1)

Answer:  y=C e^{-x}+\frac{1}{5}(2 \sin x-\cos x)

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

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

Solution:  P=2, Q=\sin x

\begin{aligned} &\text{If}=e^{\int f d x} \\ & \end{aligned}

=e^{\int 1 d x} \\

=e^{x} \\

=y I f=\int Q\: \text{If} \: d x+C \\

 \begin{aligned} &=y e^{x}=\int e^{2 x} \sin x d x+C \\ & \end{aligned}

=\int e^{a x} \sin b x=\frac{e^{a x}}{a^{2}+b^{2}}(a \sin b x-b \cos b x) \\

=y e^{2 x}=\frac{e^{2 x}}{5}(2 \sin x-\cos x)+C \\

=y=\frac{2 \sin x-\cos x}{5}+C e^{-2 x}

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