Question
Solve the following differential equation:
$\frac{\text{dy}}{\text{dx}}+2\text{y}=\sin\text{x}$
$\frac{\text{dy}}{\text{dx}}+2\text{y}=\sin\text{x}$
✓
Answer
We have,
$\frac{\text{dy}}{\text{dx}}+2\text{y}=\sin\text{x}\ \dots(1)$
Clearly, it is a linear differential equation of the form
$\frac{\text{dy}}{\text{dx}}+\text{Py}=\text{Q}$
where
$\text{P}=1$
and
$\text{Q}=\sin\text{x}$
$\therefore$ I.F. $=\text{e}^{\int\text{Pdx}}$
$=\text{e}^{\int2\text{dx}}$
$=\text{e}^{2\text{x}}$
Multiplying both sides of (1) by I.F. = e2x, we get
$\text{e}^{2\text{x}}\Big(\frac{\text{dy}}{\text{dx}}+2\text{y}\Big)=\text{e}^{2\text{x}}\sin\text{x}$
$\Rightarrow\ \text{e}^{2\text{x}}\frac{\text{dy}}{\text{dx}}+2\text{e}^{2\text{x}}\text{y}=\text{e}^{2\text{x}}\sin\text{x}$
Integrating both sides with respect to x, we get
$\text{ye}^{2\text{x}}=\int\text{e}^{2\text{x}}\sin\text{x dx + C}$
$\Rightarrow\ \text{ye}^{2\text{x}}=\frac{1}5\int[2\text{e}^{2\text{x}}(2\sin\text{x}-\cos\text{x}+\text{e}^{2\text{x}}(2\cos\text{x}+\sin\text{x})]$
Putting $\text{e}^{2\text{x}}(2\sin\text{x}-\cos\text{x})=\text{t}$
$\Rightarrow[2\text{e}^{2\text{x}}(2\sin\text{x}-\cos\text{x})+\text{e}^{2\text{x}}(2\cos\text{x}+\sin\text{x})]\text{dx}=\text{dt}$
$\therefore\ \text{ye}^{2\text{x}}=\frac{1}5\int\text{dt + C}$
$\Rightarrow\text{ye}^{2\text{x}}=\frac{\text{t}}5+\text{C}$
$\text{ye}^{2\text{x}}=\frac{\text{e}^{2\text{x}}}5(2\sin\text{x}-\cos\text{x})+\text{C}$
$\Rightarrow\ \text{y}=\frac{1}5(2\sin\text{x}-\cos\text{x})+\text{Ce}^{-2\text{x}}$
Hence, $\text{y}=\frac{1}5(2\sin\text{x}-\cos\text{x})+\text{Ce}^{-2\text{x}}$ is the required solution.
$\frac{\text{dy}}{\text{dx}}+2\text{y}=\sin\text{x}\ \dots(1)$
Clearly, it is a linear differential equation of the form
$\frac{\text{dy}}{\text{dx}}+\text{Py}=\text{Q}$
where
$\text{P}=1$
and
$\text{Q}=\sin\text{x}$
$\therefore$ I.F. $=\text{e}^{\int\text{Pdx}}$
$=\text{e}^{\int2\text{dx}}$
$=\text{e}^{2\text{x}}$
Multiplying both sides of (1) by I.F. = e2x, we get
$\text{e}^{2\text{x}}\Big(\frac{\text{dy}}{\text{dx}}+2\text{y}\Big)=\text{e}^{2\text{x}}\sin\text{x}$
$\Rightarrow\ \text{e}^{2\text{x}}\frac{\text{dy}}{\text{dx}}+2\text{e}^{2\text{x}}\text{y}=\text{e}^{2\text{x}}\sin\text{x}$
Integrating both sides with respect to x, we get
$\text{ye}^{2\text{x}}=\int\text{e}^{2\text{x}}\sin\text{x dx + C}$
$\Rightarrow\ \text{ye}^{2\text{x}}=\frac{1}5\int[2\text{e}^{2\text{x}}(2\sin\text{x}-\cos\text{x}+\text{e}^{2\text{x}}(2\cos\text{x}+\sin\text{x})]$
Putting $\text{e}^{2\text{x}}(2\sin\text{x}-\cos\text{x})=\text{t}$
$\Rightarrow[2\text{e}^{2\text{x}}(2\sin\text{x}-\cos\text{x})+\text{e}^{2\text{x}}(2\cos\text{x}+\sin\text{x})]\text{dx}=\text{dt}$
$\therefore\ \text{ye}^{2\text{x}}=\frac{1}5\int\text{dt + C}$
$\Rightarrow\text{ye}^{2\text{x}}=\frac{\text{t}}5+\text{C}$
$\text{ye}^{2\text{x}}=\frac{\text{e}^{2\text{x}}}5(2\sin\text{x}-\cos\text{x})+\text{C}$
$\Rightarrow\ \text{y}=\frac{1}5(2\sin\text{x}-\cos\text{x})+\text{Ce}^{-2\text{x}}$
Hence, $\text{y}=\frac{1}5(2\sin\text{x}-\cos\text{x})+\text{Ce}^{-2\text{x}}$ is the required solution.
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