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Explain solution RD Sharma class 12 Chapter 21 Differential Equation exercise 21.10 question 12

Answers (1)

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

Hint: To solve this equation we use  \int u v dx  formula.

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

Solution:  \frac{d y}{d x}+P y=Q \\

P=1, Q=\sin x \\

\text { If }=e^{\int P d x} \\

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


\begin{aligned} &y I\! f=\int Q I\! f d x+C \\ & \end{aligned}

y e^{x}=\int \sin x e^{x} d x+C \\

=I=\sin \int e^{x}-\int \frac{d}{d x} \sin x \int e^{x} d x d x \\

=I=\sin e^{x}-\int \cos x e^{x} d x

 \begin{aligned} &=I=\sin e^{x}-\int \cos x-e^{x} d x+\int \sin x e^{x} d x \\ & \end{aligned}

=I=e^{x}[\sin x-\cos x]-I \\

=2 I=e^{x}(\sin x-\cos x) \\

=I=\frac{e^{x}}{2}(\sin x-\cos x)

Put in original equation

 \begin{aligned} &y e^{x}=\frac{e^{x}}{2}(\sin x-\cos x)+C \\ & \end{aligned}

y=c e^{-x}+\frac{1}{2}(\sin x-\cos x)

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