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Home  >>  CBSE XII  >>  Math  >>  Differential Equations
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Find a particular solution satisfying the given condition $\large\frac{dy}{dx}-$$3y\cot x=\sin 2x;\;y=2\;when\;x=\large\frac{\pi}{2}$

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Toolbox:
  • To solve 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:
The given equation is of the form $\large\frac{dy}{dx}$$ - 3y\cot x = \sin 2x$
Here $P = - 3\cot x$ and $Q = \sin 2x$
Let us find the integration factor
$\int pdx = \int 3\cot xdx = -3[\log|\sin x|] = \log\big(\large\frac{1}{\sin^3x}\big)$
Hence $I.F = e^{\large\log(1/sin^3x)} = \log cosec^3x$
Step 2:
The required solution is $y(cosec^3x) = \int [\sin2x.cosec^3x]dx + C$
$\int \sin 2x.cosec^3x dx = \int (2\sin x\cos x).1/\sin^3x)dx = 2\int (\cot x cosecx)dx = - 2cosecx + C $
Hence the required solution is $y cosec^3x = 2 cosecx + C$
Step 3:
To evaluate the value of $C$ let us substitute the given values of x and y in the above equation
$2cosec^3\big(\large\frac{\pi}{2}\big) =$$- 2cosec\big(\large\frac{\pi}{2}\big)$$ + C$
$Cosec\big(\large\frac{\pi}{2}\big)$$ = 1$
$\therefore 2 = -2 + C$
Or $C = 4$
Substituting this value we get
$y cosec^3x = -2 cosecx + 4$
dividng throughout by $cosec^3x$ we get
$y = \large\frac{-2}{cosec^2x} +\frac{ 4}{cosec^3x }$
$y = - 2\sin^2x + 4 \sin^3x$
This is the required solution.
answered Jul 31, 2013 by sreemathi.v
 

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