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Home  >>  CBSE XII  >>  Math  >>  Model Papers
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Solve the following differential equation : $\large\frac{dy}{dx}$$+y\cot\: x = x^2\cot x+2x $

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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:
$\large\frac{dy}{dx}$$+y\cot x=2x+x^2\cot x$
Clearly the given equation is of the form
$\large\frac{dy}{dx}$$+Py=Q$
The solution for the above equation is
$y.e^{\int Pdx}=\int Q.e^{\int Pdx}dx+c$
Here $P=\cot x$
$Q=2x+x^2\cot x$
$\int Pdx=\log \sin x$
$e^{\large\int Pdx}=e^{\large\log \sin x}=\sin x$
Step 2:
$y\sin x=\int (2x+x^2\cot x).\sin xdx+c$
$\qquad\;\;=\int 2x\sin xdx+\int x^2\cot x.\sin xdx+c$
$\qquad\;\;=\int 2x\sin xdx+\int x^2\cos x.dx+c$
Step 3:
Consider $I=2\int x\sin xdx$
This is of the form $\int u dv=uv-\int vdu$
Let $du=x$
$u=\large\frac{x^2}{2}$$dx$
$v=\sin x$
$dv=\cos xdx$
$I=-\sin x\big(\large\frac{2x^2}{2}\big)$$-\int\cos x\big(\large\frac{2x^2}{2}\big)$$dx+\int x^2\cos xdx+c$
$y\sin x=x^2\sin x-\int x^2\cos xdx+\int x^2\cos xdx+c$
$y\sin x=x^2\sin x+c$
This is the required solution.
answered Sep 25, 2013 by sreemathi.v
 

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