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

Question

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

✓

Answer

We have,
$\frac{\text{dy}}{\text{dx}}+\text{y}\tan\text{x}=\text{x}^2\cos^2\text{x}$
Comparing with $\frac{\text{dy}}{\text{dx}}+\text{Py}=\text{Q},$ we get
$\text{P}=\tan\text{x}$
$\text{Q}=\text{x}^2\cos^2\text{x}$
Now,
I.F. $=\text{e}^{\int\text{Pdx}}$
$=\text{e}^{\int\tan\text{xdx}}$
$=\text{e}^{\log|\sec\text{x}|}=\sec\text{x}$
Therefore, solution is given by,
$\text{y}\times\text{I.F.}=\int\text{x}^2\cos^2\text{x}\times\text{I.F.}\text{dx + C}$
$\Rightarrow\ \text{y}\sec\text{x}=\int\text{x}^2\cos\text{x dx + C}$
$\Rightarrow\ \text{y}\sec\text{x}=\text{I + C}$
Where,
$\text{I}=\int\text{x}^2\cos\text{xdx + C}$
$\Rightarrow\ \text{I}=\text{x}^2\int\cos\text{xdx}-\int\Big[\frac{\text{d}}{\text{dx}}(\text{x}^2)\int\cos\text{xdx}\Big]\text{dx}$
$\Rightarrow\ \text{I}=\text{x}^2\sin\text{x}-2\int\text{x}\sin\text{xdx}$
$\Rightarrow\ \text{x}^2\sin\text{x}-2\text{x}\int\sin\text{xdx}+2\int\Big[\frac{\text{d}}{\text{dx}}(\text{x})\int\sin\text{xdx}\Big]\text{dx}$
$\Rightarrow\ \text{I}=\text{x}^2\sin\text{x}+2\text{x}\cos\text{x}-2\int\cos\text{xdx}$
$\Rightarrow\ \text{I}=\text{x}^2\sin\text{x}+2\text{x}\cos\text{x}-2\sin\text{x}$
$\therefore\ \text{y}\sec\text{x}=\text{x}^2\sin\text{x}+2\text{x}\cos\text{x}-2\sin\text{x + C}$

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