The procedure to arrive, once you know a solution of this type of equation, to a second solution independent from it, is illustrated here...
http://www.mathhelpforum.com/math-help/calculus/82519-variation-parameter-2nd-order-ode.html
Kind regards
The procedure to arrive, once you know a solution of this type of equation, to a second solution independent from it, is illustrated here...
http://www.mathhelpforum.com/math-help/calculus/82519-variation-parameter-2nd-order-ode.html
Kind regards
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