Solving the ordinary differential equation: $(1 + x^2)\frac{dy}{dx} + 4xy = \frac{1}{(1+x^2)^2}$.

Question

I tried using integrating factor but was stuck at the integration of $e^2x^2/(1+x^2)^2$

Answer 1

Hint: The integrating factor is $$\exp{\left(\int \frac{4x dx}{1+x^2}\right)}=(1+x^2)^2$$ Note that I assumed you wrote the ODE as its standard form with $1$ as the coefficient of $y'$.

Answer 2

We transform the ODE $$a\cdot\frac{dy}{dx}+b\cdot y=c$$ into: $$\frac{dy}{dx}+Py=Q$$ then the Integrating Factor is $e^{\int P\space dx}$

Here, we divide by $(1+x^2)$ so that $P=\frac{4x}{1+x^2}$ and $Q=(1+x^2)^{-3}$

Then the Integrating Factor is $$e^{\int{\frac{4x}{1+x^2}dx}}$$ Use the substitution $u=1+x^2 \to dx=\frac{du}{2x}$ to evaluate the integral.

Finally, the shortcut: $$I y=\int{IQ \space dx}$$ (where $I$ is the Integrating Factor) will help you the rest of the way.