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

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,
$\large\frac{x}{y}=\frac{2y^2}{2}+c$
=>$x=y^3+cy$
answered May 14, 2013 by