Hello,

I cannot for the life of me remember how to do this. Can someone help please?

Solve the following differential equation xy''+y'=12x^2 given that yh1 = 1 is a solution of the homogeneous equation.

Thanks.

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- May 15th 2009, 11:52 PMpberardisolve given a yh1
Hello,

I cannot for the life of me remember how to do this. Can someone help please?

Solve the following differential equation xy''+y'=12x^2 given that yh1 = 1 is a solution of the homogeneous equation.

Thanks. - May 16th 2009, 01:08 AMchisigma
The procedure to arrive, once you know a solution http://www.mathhelpforum.com/math-he...40896746-1.gif of this type of equation, to a second solution http://www.mathhelpforum.com/math-he...f8bae241-1.gif independent from it, is illustrated here...

http://www.mathhelpforum.com/math-help/calculus/82519-variation-parameter-2nd-order-ode.html

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$ - May 16th 2009, 09:57 AMCalculus26
Normally this could be handled using reduction of order where we would make the substitution:

y(x) = yh*z(x)

However since yh = 1 we don't need to do this

Let z = y ' directly and we have

xz' + z =12x^2

(xz) ' =12x^2

xz = 4x^3 + C1

z = 4x^2 +C1/x Note 1/x is the other homogeneous solution

now solve y ' = 4x^2 +C1/x