Solve the following differential equation: dy dx +y = - Vidyadip

Question

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

✓

Answer

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

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