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 $(x+2y^3)\Large \frac{dy}{dx}\normalsize =y \;dx$

Can you answer this question?

1 Answer

0 votes
  • A linear differential equation of the form.
  • $\large\frac{dx}{dy}$$+Px=Q$ has the general solution as $xe^{\int pdy}=\int Q.e^{\int pdy}.dy+c$
  • Where $e ^{\int pdy}$ is the integrating factor(I.F)
Given $ (x+2y^3)dy=y \;dx$
This equation can be written as
$\large\int \frac{dx}{dy}+(-\frac{1}{y})$$x=2y^2$
Clearly this equation is a linear differential equation of the form:
Where $P=\large\frac{-1}{y}$ and $Q=2y^2$
The integrating factor $I.F=e^{\large\int pdx}=e^{\int \Large\frac{-1}{y} \large dy}$
$e^{\int \Large\frac{-1}{y} \large dy}=e^{\large-\log y}=e^{\log \Large(\frac{1}{y})}$
But $e^{\large\log y}=y$
Hence $I.F=\large\frac{1}{y}$
Now the solution to the equation is
$x \times I.F= \int Q \times I.F\;dy+c$
=>$x.\large\frac{1}{y}$$=\int 2 y^2 .\large\frac{1}{y}\;$$dy+c$
$=\large\frac{x}{y}$$=\int 2ydy+c$
On integrating we get,
answered May 14, 2013 by meena.p
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