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

Question

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

✓

Answer

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

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