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

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Toolbox:
  • Linear Differential equation.
  • This is of the form $\large\frac{dy}{dx}$$+Py=Q$ where P and Q are functions of x only.
  • The integrating factor $I= e^{\int \large pfd}$ and the $G.S$ is$ ye^{\large pdx}=\int Q e^{\large pdx} dx+c$
Step 1:
$(1+x^2) \large\frac{dy}{dx}$$+2xy =\cos x $ divided by $1+x^2$
$\large\frac{dy}{dx}+\frac{2xy}{1+x^2}=\frac{\cos x}{1+x^2}$
$p(x)=\large\frac{2x}{1+x^2}$$,Q(x)=\large\frac{\cos x}{1+x^2}$
$e^{\int pdx}$$=e^{\int \large\frac{2y}{1+x^2}}$$dx$
$\qquad=e^{\large \log (1+x^2)}$
$\qquad=1+x^2$
Step 2:
The solution is $y(1+x^2)=\int \large\frac{\cos x}{1+x^2}.$$1+x^2 dx+c$
$y(1+x^2)=\sin x+c$
answered Sep 6, 2013 by meena.p
 
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