Solve the following differential equation: dy dx +y =x^2 +2x

Question

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

✓

Answer

We have,
$\frac{\text{dy}}{\text{dx}}+\text{y}\cot\text{x}=\text{x}^2\cot\text{x}+2\text{x}\ \dots(1)$
Clearly, it is a linear differential equation of the form
$\frac{\text{dy}}{\text{dx}}+\text{Py}=\text{Q}$
where
$\text{P}=\cot\text{x}$
$\text{Q}=\text{x}^2\cot\text{x}+2\text{x}$
$\therefore$ I.F. $=\text{e}^{\int\text{Pdx}}$
$=\text{e}^{\int\cot\text{xdx}}$
$=\text{e}^{\log|\sin\text{x}|}=\sin\text{x}$
Multiplying both sides of (1) by $\sin\text{x},$ we get
$\sin\text{x}\Big(\frac{\text{dy}}{\text{dx}}+\text{y}\cot\text{x}\Big)=\sin\text{x}(\text{x}^2\cot\text{x}+2\text{x})$
$\Rightarrow\ \sin\text{x}\frac{\text{dy}}{\text{dx}}+\text{y}\cos\text{x}=\text{x}^2\cos\text{x}+2\text{x}\sin\text{x}$
Integrating both sides with respect to x, we get
$\text{y}\sin\text{x}=\int\text{x}^2\cos\text{xdx}+\int2\text{x}\sin\text{xdx + C}$
$\Rightarrow\ \text{y}\sin\text{x}=\text{x}^2\int\cos\text{xdx}-\int\Big[\frac{\text{d}}{\text{dx}}(\text{x}^2)\int\cos\text{xdx}\Big]\text{dx}+\int2\text{x}\sin\text{x dx + C}$
$\Rightarrow\ \text{y}\sin\text{x}=\text{x}^2\sin\text{x}-\int2\text{x}\sin\text{xdx}+\int2\text{x}\sin\text{xdx + C}$
$\Rightarrow\ \text{y}\sin\text{x}=\text{x}^2\sin\text{x +C}$
Hence, $\Rightarrow\ \text{y}\sin\text{x}=\text{x}^2\sin\text{x +C}$ is the required solution.

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