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

Fernando Revilla

Well-known member
MHB Math Helper
Jan 29, 2012
661
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
Jan 29, 2012
661
It is a linear equation. Let's solve the homogeneous [tex](1+x^2)\dfrac{dy}{dx}+6xy=0[/tex]:
[tex]\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}[/tex]

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

[tex](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}[/tex]

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

[tex]y=\dfrac{c}{(1+x^2)^3}+\dfrac{\log (1+x^2)}{(1+x^2)^3}[/tex]