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Differential Equations — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsDifferential Equations5 Marks
Question
Find the general solution of $\frac{d y}{d x}+2 y=\sin x$
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Answer

It is given that $\frac{d y}{d x}+2 y=\sin x$
This is equation in the form of $\frac{d y}{d x}+p y=Q$ (where, p = 2 and Q = sin x)
Now, I.F = $e^{\int p d x}=e^{\int 2 d x}=e^{2 x}$
Thus, the solution of the given differential equation is given by the relation:
$\mathrm{y}(\mathrm{I}. \mathrm{F} .)=\int(\mathrm{Q} \times \mathrm{I.F.}) \mathrm{d} \mathrm{x}+\mathrm{C}$
$\Rightarrow \mathrm{ye}^{2 \mathrm{x}}=\int \sin \mathrm{x} \cdot \mathrm{e}^{2 \mathrm{x}} \mathrm{d} \mathrm{x}+\mathrm{C}$ .....(i)
Let $I=\int \sin x \cdot e^{2 x} d x$
$\Rightarrow \mathrm{I}=\sin \mathrm{x} \int \mathrm{e}^{2 \mathrm{x}} \mathrm{d} \mathrm{x}-\int\left(\frac{\mathrm{d}}{\mathrm{dx}}(\sin \mathrm{x}) \cdot \int \mathrm{e}^{ 2x }\mathrm{d} \mathrm{x}\right) \mathrm{d} \mathrm{x}$
= $\sin x \cdot \frac{e^{2 x}}{2}-\int\left(\cos x \cdot \frac{e^{2 x}}{2}\right) d x$
= $\frac{\mathrm{e}^{2 \mathrm{x}} \sin \mathrm{x}}{2}-\frac{1}{2}\left[\cos \mathrm{x} \int \mathrm{e}^{2 \mathrm{x}}-\int\left(\frac{\mathrm{d}}{\mathrm{d} \mathrm{x}}(\cos \mathrm{x}) \cdot \int \mathrm{e}^{2 \mathrm{x}} \mathrm{d} \mathrm{x}\right) \mathrm{d} \mathrm{x}\right]$
= $\frac{\mathrm{e}^{2 \mathrm{x}} \sin \mathrm{x}}{2}-\frac{1}{2}\left[\cos \mathrm{x} \frac{\mathrm{e}^{2 \mathrm{x}}}{2}-\int\left[(-\sin \mathrm{x}) \cdot \frac{\mathrm{e}^{2 \mathrm{x}}}{2}\right] \mathrm{d} \mathrm{x}\right]$
= $\frac{\mathrm{e}^{2 \mathrm{x}} \sin \mathrm{x}}{2}-\frac{\mathrm{e}^{2 \mathrm{x}} \cos \mathrm{x}}{4}-\frac{1}{4} \int\left(\sin \mathrm{x} \cdot \mathrm{e}^{2 \mathrm{x}}\right) \mathrm{d} \mathrm{x}$
= $\frac{e^{2 x}}{4}(2 \sin x-\cos x)-\frac{1}{4} I$
$\Rightarrow \frac{5}{4} \mathrm{I}=\frac{\mathrm{e}^{2 \mathrm{x}}}{4}(2 \sin \mathrm{x}-\cos \mathrm{x})$
$\Rightarrow \mathrm{I}=\frac{\mathrm{e}^{2 \mathrm{x}}}{5}(2 \sin \mathrm{x}-\cos \mathrm{x})$
Now, putting the value of I in (i), we get,
$\Rightarrow \mathrm{ye}^{2 \mathrm{x}}=\frac{\mathrm{e}^{2 \mathrm{x}}}{5}(2 \sin \mathrm{x}-\cos \mathrm{x})+\mathrm{C}$
$\Rightarrow \mathrm{y}=\frac{1}{5}(2 \sin \mathrm{x}-\cos \mathrm{x})+\mathrm{Ce}^{-2 \mathrm{x}}$
Therefore, the required general solution of the given differential equation is
$y=\frac{1}{5}(2 \sin x-\cos x)+C e^{-2 x}$

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