Ask Questions, Get Answers

Want to ask us a question? Click here
Browse Questions
Home  >>  CBSE XII  >>  Math  >>  Differential Equations
0 votes

Solve the differential equation $dy=\cos x(2-y\; cosec x)dx$ given that $y=2$ when $x=\Large\frac{\pi}{2}$.

Can you answer this question?

1 Answer

0 votes
  • A linear differential equation of the form $ \large\frac{dy}{dx}$$+Py=Q,$ has the general solution as $ye^{\int pdx}=\int Q. e ^{\int pdx}.dx+c$
  • $\int \cot x dx=\log |\sin x|+c$
Step 1:
Given $dy=\cos x(2-y \;cosec x)dx$
This can be written as
$\large\frac{dy}{dx}$$=2\cos x-y \cos x. cosec x$
$=2 \cos x -y . \large\frac{\cos x}{\sin x}$
$=2 \cos x- y \cot x$
Therefore $\large\frac{dy}{dx}$$+y \cot x= 2\cos x$
Clearly this is a linear differential equation of the form
Where $P=\cot x$ and $Q=2 \cos x$
Let us find the integral factor I.F which is
$e^{\large\int pdx}=e^{\large\int \cot xdx}$
But $\int \cot x dx=\log |\sin x|$
Therefore $ I.F=e^{\large\log |\sin x|}$
But $e^{\large\log x}=x$
$e^{\large\log(\sin x)}=\sin x$
Hence the required solution is
$y \times I.F=\int Q \times I.F dx+c$
$y \sin x=\int 2 \cos x.\sin x dx+c$
But $2 sin x\cos x=\sin 2x$
Therefore $ y \sin x =\int \sin 2x+c$
$ \int \sin 2x=\large\frac{-1}{2}$$\cos 2x$
Hence $ y \sin x=\large\frac{-1}{2}$$\cos 2x+c$
$2y\sin x+\cos 2x=c$
Step 2:
Given $y=2$ when $x=\pi/2$
Substituting this to find the value of c
$2(2)\sin \pi/2+\cos 2.\pi/2=c$
But $\sin \pi/2=1$ and $\cos \pi=-1$
=>$4 \times 1+(-1)=c$
=>$4 -1=c=>c=3$
Hence the required solution is
$2y \sin x+\cos 2x=3$
or $2y \sin x+\cos 2x-3=0$
answered May 9, 2013 by meena.p

Related questions

Ask Question
student study plans
JEE MAIN, CBSE, AIPMT Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App