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Solve the following. $\large\frac{dy}{dx}+\frac{4x}{x^{2}+1}$$y=\large \frac{1}{(x^{2}+1)^{2}}$

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Toolbox:
  • Linear Differential equation.
  • This is of the form $\large\frac{dy}{dx}$$+Py=Q$ where P and Q are functions of x only.
  • The integrating factor $I= e^{\int \large pfd}$ and the $G.S$ is$ ye^{\large pdx}=\int Q e^{\large pdx} dx+c$
Step 1:
$\large\frac{dy}{dx}+\frac{4x}{x^2+1}$$y=\large\frac{1}{(x^2+1)^2}$
$p(x)=\large\frac{4x}{x^2+1}, $$Q(x)=\large\frac{1}{(x^2+1)^2}$
$e^{\large\int pdx}=e^{\int \large\frac{4x}{x^2+1}dx}$
$e^{2 \int \large \frac{2x}{x^2+1}dx}=e^{2 \log (x^2+1)}$
$\qquad=(x^2+1)^2$
Step 2:
The GS is $ ye^{\int pdx}=\int Q e^{\int pdx}+c$
$y(x^2+1)^2=\int \frac{1}{(x^2+1)^2} $$(x^2+1)^2 dx +c$
$y(x^2+1)^2=x+c$
answered Sep 5, 2013 by meena.p
 
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