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Questions  >>  CBSE XII  >>  Math  >>  Differential Equations
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Q)

Find the general solution of $(x+2y^3)\Large \frac{dy}{dx}\normalsize =y \;dx$

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A)
Toolbox:
  • 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:
$\large\frac{dx}{dy}$$+Px=Q$
Where $P=\large\frac{-1}{y}$ and $Q=2y^2$
The integrating factor $I.F=e^{\large\int pdx}=e^{\int \Large\frac{-1}{y} \large dy}$
$e^{\int \Large\frac{-1}{y} \large dy}=e^{\large-\log y}=e^{\log \Large(\frac{1}{y})}$
But $e^{\large\log y}=y$
Hence $I.F=\large\frac{1}{y}$
Now the solution to the equation is
$x \times I.F= \int Q \times I.F\;dy+c$
=>$x.\large\frac{1}{y}$$=\int 2 y^2 .\large\frac{1}{y}\;$$dy+c$
$=\large\frac{x}{y}$$=\int 2ydy+c$
On integrating we get,
$\large\frac{x}{y}=\frac{2y^2}{2}+c$
=>$x=y^3+cy$
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