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  • Find the general solution of frac{dy}{dx}-3y= sin 2x

Find the general solution of \frac{dy}{dx}-3y= \sin 2x

Answers (1)

$$ \\ \frac{\mathrm{dy}}{\mathrm{dx}}-3 \mathrm{y}=\sin 2 \mathrm{x} \\ \text { Compare } \frac{\mathrm{dy}}{\mathrm{dx}}-3 \mathrm{y}=\sin 2 \mathrm{x} \text { , and } \frac{\mathrm{dy}}{\mathrm{dx}}+\mathrm{Py}=\mathrm{Q} $$

We get, P= -3 and Q= sin2x

The equation is a linear differential equation where P and Q are functions of x

For the solution of the linear differential equation, we need to find Integrating factor,

$$ \\ \Rightarrow \mathrm{IF}=\mathrm{e}^{[\mathrm{P} \mathrm{dx}} \\ \Rightarrow \mathrm{IF}=\mathrm{e}^{[(-3) \mathrm{d} \mathrm{x}} $$
\Rightarrow \mathrm{IF}=\mathrm{e}^{-3 \mathrm{x}}$
$$ y(\mathrm{IF})=\int \mathrm{Q}(\mathrm{IF}) \mathrm{d} \mathrm{x}+\mathrm{c} $$
The solution of  the linear differential equation is
Substitute values for Q and IF
$\Rightarrow y e^{-3 x}=\int e^{-3 x} \sin 2 x d x \ldots(1)$
Let I=\int e^{-3 x} \sin 2 x d x$
If \mathrm{u}(\mathrm{x})$ and $\mathrm{v}(\mathrm{x})$ are two functions, then by integration by parts.
$$ \int \mathrm{uv}=\mathrm{u} \int \mathrm{v}-\int \mathrm{u}^{\prime} \int \mathrm{v} $$
\mathrm{v}=\sin 2 \mathrm{x}$ and $\mathrm{u}=\mathrm{e}^{-2 x}$
after applying the formula we get,

$$ \\I= \int \mathrm{e}^{-3 \mathrm{x}} \sin 2 \mathrm{x} \mathrm{dx}=\mathrm{e}^{-3 \mathrm{x}} \int \sin 2 \mathrm{x}dx-\int\left(\mathrm{e}^{-3 \mathrm{x}}\right)^{\prime}dx \int \sin 2 \mathrm{x}dx \\ I=-\mathrm{e}^{-3 \mathrm{x}} \frac{\cos 2 \mathrm{x}}{2}+\int 3 \mathrm{e}^{-3 \mathrm{x}} \frac{\cos 2 \mathrm{x}}{2}dx+\mathrm{c} $$
Again, applying the above stated rule in \int 3 \mathrm{e}^{-3 \mathrm{x} \frac{\cos 2 \mathrm{x}}{2} }\text { we get }

$$ I=-\mathrm{e}^{-3 \mathrm{x}} \frac{\cos 2 \mathrm{x}}{2}-\frac{3}{2}\left[\mathrm{e}^{-3 \mathrm{x}} \frac{\sin 2 \mathrm{x}}{2}+\int 3 \mathrm{e}^{-3 \mathrm{x}} \frac{\sin 2 \mathrm{x}}{2}\right]+\mathrm{c} $$

$$ I=-\mathrm{e}^{-3 \mathrm{x}} \frac{\cos 2 \mathrm{x}}{2}-\frac{3}{4}\mathrm{e}^{-3 \mathrm{x}} \sin 2 \mathrm{x}-\frac{9I}{4}+\mathrm{c} $$

$$ \frac{13I}{4}=-\mathrm{e}^{-3 \mathrm{x}} \frac{\cos 2 \mathrm{x}}{2}-\frac{3}{4}\mathrm{e}^{-3 \mathrm{x}} \sin 2 \mathrm{x}+\mathrm{c} $$

$$ I=\frac{-1}{13}\mathrm{e}^{-3 \mathrm{x}}( 2\cos 2 \mathrm{x}+3 \sin 2 \mathrm{x})+\mathrm{c} $$

Put this value in (1) to get

\\ \text { ye }^{- 3 x}=\int e^{-3 x} \sin 2 x d x \\ \text { ye }^{-3 x}=\frac{-e^{-3 x}(2 \cos 2 x+3 \sin 2 x)}{13}+c \\ \Rightarrow y=-\frac{1}{13}(3 \sin 2 x+2 \cos 2 x)+c e^{3 x}

Posted by

infoexpert22

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