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

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  • A linear differential equation of the form $ \large\frac{dy}{dx}+py=Q$ has the generlal equation as $y e^{\int pdx}=\int Q e^{\int pdx}.dx+c$,when $ e^{\int pdx}$ is the integrating factor (I.F)
  • $\int e^{ax}.\sin bx\;dx=\large\frac{e^{\Large ax}}{a^2+b^2}$$[a \sin bx-b \cos bx]$
Given $\large\frac{dy}{dx}$$+3y=\sin 2x$
This is clearly a linear differential equation of the form $\large\frac{dy}{dx}$$+Py=Q$
Here $P=3$ and $Q=\sin 2x$
Let us first find the integrating factor I.F which is $e^{\int pdx}$
$\int pdx=\int 3 dx=3x$
Hence $I.F=e^{3x}$
Hence the required solution is
$y \times I.F= \int Q \times I.F dx +c$
$y e^{3x}=\int e^{3x}.\sin 2xdx+c$
=>$y.e^{3x}=\int e^{3x}.\sin 2xdx+c$
Step 2:
$y .e^{3x}=\int e^{3x}. \sin 2xdx+c$
$\int e^{3x}. \sin 2xdx$ is of the form
$\int e^{ax}. \sin bx dx=\large\frac{e^{\Large ax}}{a^2+b^2} $$[a \sin bx -b \cos bx]$
Here $a=3$ and $b=2$
Hence $\int e^{3x}. \sin 2xdx=\large\frac{\Large e^3x}{3^2+2^2}$$ [3 \sin 2x-2 \cos 2x]$
$=\large\frac{e^{\Large 3x}}{13}$$[3 \sin 2x -2 \cos 2x]$
Hence the required solution is
$ye^{3x}=\large\frac{\Large e^{3x}}{13}$$[3 \sin 2x-2 \cos 2x]+c$
Dividing throughout by $e^{3x}$ we get
$y= \large\frac{3 \sin 2x-2 \cos 2x}{13}$$+ce^{-3x}$
answered May 13, 2013 by meena.p