Ask Questions, Get Answers

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

Find the general solution of $\large \frac{dy}{dx}$$+3y=\sin 2x$

Can you answer this question?

1 Answer

0 votes
  • A linear differential equation of the form $ \large\frac{dy}{dx}+py=Q$ has the generlal equation as $y e^{\int pdx}=\int Q e^{\int pdx}.dx+c$,when $ e^{\int pdx}$ is the integrating factor (I.F)
  • $\int e^{ax}.\sin bx\;dx=\large\frac{e^{\Large ax}}{a^2+b^2}$$[a \sin bx-b \cos bx]$
Given $\large\frac{dy}{dx}$$+3y=\sin 2x$
This is clearly a linear differential equation of the form $\large\frac{dy}{dx}$$+Py=Q$
Here $P=3$ and $Q=\sin 2x$
Let us first find the integrating factor I.F which is $e^{\int pdx}$
$\int pdx=\int 3 dx=3x$
Hence $I.F=e^{3x}$
Hence the required solution is
$y \times I.F= \int Q \times I.F dx +c$
$y e^{3x}=\int e^{3x}.\sin 2xdx+c$
=>$y.e^{3x}=\int e^{3x}.\sin 2xdx+c$
Step 2:
$y .e^{3x}=\int e^{3x}. \sin 2xdx+c$
$\int e^{3x}. \sin 2xdx$ is of the form
$\int e^{ax}. \sin bx dx=\large\frac{e^{\Large ax}}{a^2+b^2} $$[a \sin bx -b \cos bx]$
Here $a=3$ and $b=2$
Hence $\int e^{3x}. \sin 2xdx=\large\frac{\Large e^3x}{3^2+2^2}$$ [3 \sin 2x-2 \cos 2x]$
$=\large\frac{e^{\Large 3x}}{13}$$[3 \sin 2x -2 \cos 2x]$
Hence the required solution is
$ye^{3x}=\large\frac{\Large e^{3x}}{13}$$[3 \sin 2x-2 \cos 2x]+c$
Dividing throughout by $e^{3x}$ we get
$y= \large\frac{3 \sin 2x-2 \cos 2x}{13}$$+ce^{-3x}$
answered May 13, 2013 by meena.p
Ask Question
student study plans
JEE MAIN, CBSE, NEET Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App