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Home  >>  CBSE XII  >>  Math  >>  Differential Equations
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Find a particular solution of the differential equation $\frac{dy}{dx} + y\cot x=4x\:cosec\:x$, $(x\neq0)$, given that $y = 0$ when $x = \large\frac{\pi}{2}$

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Toolbox:
  • To solve the first order linear differential equation:
  • (i) First write the equation of the form $\large\frac{dy}{dx}$$ + py = Q$
  • (ii) Find the integral factor(I.F) =$ e^{\int pdx}$
  • (iii) Write the solution as y.(I.F) - integration of Q.(I.F) + C
Step 1:
Given $\large\frac{dy}{dx }$$+y\cot x = 4xcosec x $
This is a first order linear differential equation.
Here $p = \cot x$ and $Q = 4xcosec\;x$
$\int pdx=\int \cot x=\log\mid\sin x\mid$
Hence $I.F=e^{\large\log\mid\sin x\mid}=\sin x$
Step 2:
Hence the solution is $y\sin x = \int4x cosec\;x.\sin x dx + C$
$cosec x=\large\frac{1}{\sin x}$
$y\sin x=\int 4xdx+C$
$y\sin x=\large\frac{4x^2}{2}$$+C$
$y\sin x=2x^2+C$-----(1)
Step 3:
To evaluate the value of C, let us substitute $x = \large\frac{\pi}{2}$ and $y = 0 $
$0.\sin\large\frac{\pi}{2}=2.\big(\large\frac{\pi}{2}\big)^2$$+C$
$0=\large\frac{\pi^2}{2}$$+C$
$C=-\large\frac{\pi^2}{2}$
Substituting the value of C in equ(1) we get,
$y\sin x=2x^2-\large\frac{\pi^2}{2}$
This is the required solution.
answered Jul 29, 2013 by sreemathi.v
 

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