# The solution of $(1+x^2)\frac{dy}{dx}+2xy-4x^2=0$ is ___________.

$\begin{array}{1 1} 3y(1+x^2)=4x^3+c \\ 3y(1-x^2)=4x^3+c \\ 3y(1+x^2)=4y^3+c \\ 3y(1+y^2)=4x^3+c \end{array}$

Toolbox:
• A linear differential equation of the form $\large\frac{dy}{dx}$$+Py=Q has general solution as ye^{\int pdx}=\int Q.e^{\int pdx}dx+c • Where e^{\int pdx} is the integral factor (I.F) Given (1+x^2)\large\frac{dy}{dx}$$+2xy-4x^2=0$
Divide throughout by $(1+x^2)$
$\large\frac{dy}{dx}+\frac{2xy}{1+x^2}=\frac{4x^2}{1+x^2}$
Clearly this is a linear differential equation of the form $\large\frac{dy}{dx}$$+Py=Q Where P=\large\frac{2x}{1+x^2} and Q=\large\frac{4x^2}{1+x^2} \int pdx=\int \large\frac{2x}{1+x^2} To integrate this let us put 1+x^2=t.Differentiating w.r. x we get, 2xdx=dt Substituitng this , \int \large\frac{dt}{t} On integrating we get \log |t| Substitute for t, then \int pdx=\log |1+x^2| The integrating factor (I.F) e^{\int pdx}=e^{\log |1+x^2|} =(1+x^2) Hence the required solution is y \times I.F=\int Q \times I.F dx+c =>y.(1+x^2)=\int \large\frac{4x^2}{1+x^2}$$ \times (1+x^2)+c$
$=>y(1+x^2)=\int 4x^2dx+c$
(or) $y(1+x^2)=\large\frac{4x^3}{3}$$+c 3y(1+x^2)=4x^3+c is the required solution The solution of (1+x^2)\large\frac{dy}{dx}$$+2xy-4x^2=0$ is $3y(1+x^2)=4x^3+c$
answered May 22, 2013 by