# Randy's question at Yahoo! Answers (linear differential quation)

#### Fernando Revilla

##### Well-known member
MHB Math Helper
Randy's question at Yahoo! Answers (linear differential equation)

Here is the question:

What is the general Solution of
(1 + a^2)q' + 6aq = a/(1 + a2)^3,
Where q is a function of a? you can switch q with y and a with x if that is easier. Please show work
thank you
Here is a link to the question:

What is the general solution of the differential Equation? - Yahoo! Answers

I have posted a link there to this topic so the OP can find my response.

Last edited:

#### Fernando Revilla

##### Well-known member
MHB Math Helper
It is a linear equation. Let's solve the homogeneous $$(1+x^2)\dfrac{dy}{dx}+6xy=0$$:
$$\dfrac{dy}{y}+\dfrac{6x}{1+x^2}\;dx=0\\ \log |y|+3\log (1+x^2)=K\\ \log |y|=K-3\log (1+x^2)\\ y=C(1+x^2)^{-3}$$

Now, we use the variation of parameters method. Substituing $$y=C(x)(1+x^2)^{-3}$$ in the original equation:

$$(1+x^2)[C'(x)(1+x^2)^{-3}+C(x)(-3)(1+x^2)^{-3}2x]+6xC(x)(1+x^2)^{-3}=x(1+x^2)^{-3}$$

Symplifying: $$C'(x)=\dfrac{x}{1+x^2}$$ hence $$C(x)=\dfrac{1}{2}\log (1+x^2)+c$$. As a consequence the general solution of the given equation is

$$y=\dfrac{c}{(1+x^2)^3}+\dfrac{\log (1+x^2)}{(1+x^2)^3}$$