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Please Solve RD Sharma Class 12 Chapter 21 Differential Equation Exercise 21.10 Question 13 Maths Textbook Solution.

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

Hint: To solve this equation we use e^{\int Pdx}  formula.

Give:  \frac{d y}{d x}+y=\cos x

Solution:  \frac{d y}{d x}+P(x) y=Q(x)

\begin{aligned} &\text { If }=e^{\int P(x) d x} \\ &\end{aligned}

y \text { If }=\int Q(x) \text { If }d x \\

\frac{d y}{d x}+y=\cos x \\

P(x)=1, Q(x)=\cos x \\

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

=e^{x} \\

y e^{x}=\int \cos x e^{x} d x \ldots(i) \\

y\: \text{If}=\int Q(x)\text{If} \: d x

Suppose   \begin{aligned} & I=\int \cos x e^{x} d x\\ & \end{aligned}

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

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

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

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

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

=I=\frac{e^{x}}{2}(\cos x+\sin x) \ldots(i i) \\

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



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