Question
Solve the following differential equation:
$\text{x}\frac{\text{dy}}{\text{dx}}+2\text{y}=\text{x}\cos\text{x}$

Answer

Here, $\frac{\text{dy}}{\text{dx}}+\frac{2\text{y}}{\text{x}}=\cos\text{x}$

It is a linear differential equation. Comparing it with,

$\frac{\text{dy}}{\text{dx}}+\text{Py}=\text{Q}$

$\text{P}=\frac{2}{\text{x}},\text{Q}=\cos\text{x}$

I.F. $=\text{e}^{\int\text{Pdx}}$

$=\text{e}^{2\int\frac{1}{\text{x}}\text{dx}}$

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

$=\text{x}^2$

Solution of the equation is given by,

$\text{y}\times(\text{I.F.}=\int\text{Q}\times(\text{I.F.})\text{dx + C}$

$\text{y}(\text{x}^2)=\int\cos\text{x}(\text{x}^2)\text{dx + C}$

$\text{yx}^2=\int\text{x}^2\cos\text{xdx + C}$

$=\text{x}^2\int\cos\text{x}-\int(2\text{x}\times\int\cos\text{xdx})\text{dx + C}$

Usind integration by parts

$\text{yx}^2=\text{x}^2\sin\text{x}-\int2\text{x}\sin\text{xdx + C}$

$=\text{x}^2\sin\text{x}-2\big[\text{x}\times\int\sin\text{xdx}-\int(1\times\int\sin\text{xdx})\text{dx}\big]+\text{C}$

$=\text{x}^2\sin\text{x}+2\text{x}\cos\text{x}-2\int\cos\text{xdx + C}$

$\text{yx}^2=\text{x}^2\sin\text{x}+2\text{x}\cos\text{x}-2\sin\text{x + C}$

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

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