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Home  >>  CBSE XII  >>  Math  >>  Differential Equations
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Find the general solution of the differential equation $\large\frac{dy}{dx}$$+2y=\sin x$

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Toolbox:
  • To solve the first order linear differential equation of the form $\large\frac{dy}{dx}$$ + Py = Q$
  • (i) Write the given equation in the form of $\large\frac{dy}{dx}$$ + Py = Q$
  • (ii) Find the integrating factor (I.F) = $e^{\int Pdx}$.
  • (iii) Write the solution as y(I.F) = integration of $Q(I.F) dx + C$
Step 1:
Given $\large\frac{dy}{dx}$$ + 2y = \sin x$
The given equation is of the form $\large\frac{dy}{dx }$$+ Py = Q$, where $P = 2$ and $Q = \sin x$
$I.F = e^{\int 2.dx} = e^{2x}$
Hence the solution is $y.e^{2x }= \int\sin x.e^{2x} dx + C$
This can be integrated by parts where integration of $udv = uv - \int vdu$
Step 2:
let $u = e^{2x},du = 2e^{2x}dx$ and $dv = \sin x$; $v = - \cos x$
$I = - (e^{2x}\cos x) - \int(-\cos x).2e^{2x}dx$
Again Let $u = e^x; du = e^{2x}dx$ and $dv = \cos xdx ;v = \sin x$
$I = - (e^x\cos x) + 2e^{2x}(\sin x) - 2I[\int \sin x 2e^{2x}.dx]$
$\Rightarrow e^{2x}[2\sin x-\cos x]-4I$
$5I = e^{2x}[2\sin x - \cos x]$
$I = \large\frac{e^{2x}[2\sin x - \cos x]}{5}$
$\therefore y e^{2x} =\large\frac{ e^{2x}[2\sin x - \cos x]}{5}$$ + C$
$y = (\large\frac{1}{5})$$[2\sin x - \cos x] + C$ is the required solution.
answered Aug 1, 2013 by sreemathi.v
 

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