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Explain solution RD Sharma class 12 Chapter 21 Differential Equation exercise 21.10 question 32

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Answer: $y=\sin x+\frac{2 \cos x}{x}-\frac{2 \sin x}{x}+\frac{c}{x^{2}}$

Hint: To solve this equation we use $\frac{d y}{d x}+P y=Q$ where $P,Q$ + are constants.

Give: $\begin{aligned} &x \frac{d y}{d x}+2 y=x \cos x \\ & \end{aligned}$

Solution: $\frac{d y}{d x}+\frac{2 y}{x}=\frac{x \cos x}{x}$

$\begin{aligned} &=\frac{d y}{d x}+\frac{2}{x} y=\cos x \\ & \end{aligned}$

$=\frac{d y}{d x}+P y=Q \\$

$P=\frac{2}{x^{\prime}} Q=\cos x$

$If$ of differential equation is

$\begin{aligned} &I f=e^{\int P d x} \\ & \end{aligned}$

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

$=e^{2 \log x} \\$

$=e^{\log x^{2}} \\$

$=x^{2} \\$

$y I f=\int \text { QIf } d x+C$

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

$=x^{2}(\sin x)-\int 2 \sin x d x+C \\$

$=x^{2}(\sin x)-2\left[x(\cos x)-\int 1(\cos x) d x+C\right] \\$

$=x^{2}(\sin x)-2 x(\cos x)+2 \int \cos x d x+C \\$

$=x^{2}(\sin x)-2 x(\cos x)-2 \sin x+C \\$

$=y x^{2}=x^{2}(\sin x)-2 x(\cos x)-2 \sin x+C$

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

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

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