Solving a DE

Printable View

February 22nd 2010, 06:04 PM
arbolis

Solving a DE

I must solve $(1+x^2)y'+4xy=\frac{1}{(1+x^2)^2}$ .
I think it's a first order linear non homogeneous DE which can be solved by the integrating factor method.
So far I've rewritten the equation as $\frac{dy}{dx}+ \underbrace {\left ( \frac{4x}{1+x^2}\right ) }_{p(x)} y=\underbrace{\frac{1}{(1+x^2)^3}}_{q(x)}$ .
I found the integrating factor to be $2 \ln (1+x^2)$ .
So multiplying the DE by it and integrating with respect to x, I found out that $2 \ln (1+x^2)=2 \int \frac{\ln (1+x^2)}{(1+x^2)^3}dx$ .
I thought of using integration by parts to solve the integral, but I'm stuck at finding a primitive of $\ln (1+x^2)$ , as stupid as it may sound.
Am I in the right direction? If so, could you provide a tip to find a primitive of $\ln (1+x^2)$ ?
Thanks in advance.
February 22nd 2010, 07:05 PM
Prove It

Quote:

Originally Posted by arbolis

I must solve $(1+x^2)y'+4xy=\frac{1}{(1+x^2)^2}$ .
I think it's a first order linear non homogeneous DE which can be solved by the integrating factor method.
So far I've rewritten the equation as $\frac{dy}{dx}+ \underbrace {\left ( \frac{4x}{1+x^2}\right ) }_{p(x)} y=\underbrace{\frac{1}{(1+x^2)^3}}_{q(x)}$ .
I found the integrating factor to be $2 \ln (1+x^2)$ .
So multiplying the DE by it and integrating with respect to x, I found out that $2 \ln (1+x^2)=2 \int \frac{\ln (1+x^2)}{(1+x^2)^3}dx$ .
I thought of using integration by parts to solve the integral, but I'm stuck at finding a primitive of $\ln (1+x^2)$ , as stupid as it may sound.
Am I in the right direction? If so, could you provide a tip to find a primitive of $\ln (1+x^2)$ ?
Thanks in advance.

You are definitely on the right track, but you have the wrong integrating factor.

The integrating factor is actually

$e^{\int{\frac{4x}{1+x^2}\,dx}} = e^{2\ln{(1 + x^2)}} = e^{\ln{(1 + x^2)^2}} = (1 + x^2)^2$ .

So multiplying through by the integrating factor:

$(1 + x^2)^2\frac{dy}{dx} + 4x(1 + x^2)y = \frac{1}{1 + x^2}$

$\frac{d}{dx}[(1 + x^2)^2y] = \frac{1}{1 + x^2}$

$(1 + x^2)^2y = \int{\frac{1}{1 + x^2}\,dx}$

$(1 + x^2)^2y = \arctan{x} + C$

$y = \frac{\arctan{x} + C}{(1 + x^2)^2}$ .
February 22nd 2010, 07:10 PM
arbolis

Oops, I forgot to take the exponential of "my" integrating factor.
Thanks for the rest.