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 $\Large \frac{dy}{dx}\normalsize =ay-e^{mx}$

Can you answer this question?

1 Answer

0 votes
  • A linear differential equation is of the form $\large\frac{dy}{dx}$$+Py=Q$
  • The required solution is $ ye^{\large\int pdx}$$=\int Q \times e^{\large\int pdx}+c$
  • Where $e^{\large\int pdx}$ is the integrating factor
  • $ \int e^{ax}dx=\large\frac{1}{a} $$e^{ax}+c$
Step 1:
Given $\large\frac{dy}{dx}$$=ay-e^{mx}$
This can be written as
This is of the form $\large\frac{dy}{dx}$$+Py=Q$
Clearly this is a linear differential equation, Where $P=-a$ and $Q=-e^{mx}$
The solution for a linear differential equation,
$y \times I.F=\int Q.I.F dx+c$
Let us first find the integrating factor I.F which is $e^{\large\int pdx}$
$\int pdx=\int -adx=-ax$
Hence $e^{\large\int pdx}=e^{-ax}$
Substituting this for I.F we get
Step 2:
$y.e^{-ax}=\int -e^{-mx}.e^{-ax}dx+c$
$=-\int e^{-(m+a)x}dx+c$
On integrating this we get
Multiplying throughout by $(a+m)e^{ax}$
But $(a+m)$ is also a constant, hence let us write $c(a+m)$ as K. we get $(a+m)y=e^{mx}+Ke^{-ax}$
Hence the required solution is


answered May 8, 2013 by meena.p

Related questions

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