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Home  >>  CBSE XII  >>  Math  >>  Differential Equations
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The solution of $(1+x^2)\frac{dy}{dx}+2xy-4x^2=0$ is ___________.

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

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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 meena.p
 
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