Originally Posted by

**BigC** I've been working on this one differential equation for an hour and a half now and I can't for the life of me figure out where I'm going wrong: $\displaystyle xy' -15 y = (9 x^2 -5 x + 2) y^{\frac{4}{5}},\qquad y(1)=0$

First I divide by $\displaystyle x$. Then by $\displaystyle y^{\frac{4}{5}}$.

This leaves me with:

f(x) = (9x^ - 5x + 2)

y^(-4/5)y' - (15/x)y^(1/5) = f(x)/x

I substitute:

v = y^(1/5)

v' = (y^(-4/5)y')/5

==> 5v' - (15/x)*v = f(x)/x

==> v' - (3/x)*v = f(x)/(5x) [Linear Eqn]

Find integrating factor:

mu = e^(I(-3/x)) = x^-3

So I have:

(v*x^-3)' = 9/(5x^2) - (x^-3) + 2/(5x^4)

Integrate:

v*x^-3 = -9/5x + 1/(2x^2) - 2/(15x^3) + c

v = (-9x^2)/5 + x/2 - 2/15 + Cx^3 ===> g(x)

y^(1/5) = g(x)

y = g(x) ^ 5

Now to solve the IVP I simply y = 0 and x = 1[

Leaves me with:

0 = -9/5 + 1/2 -2/15 + C

C = -43/30

y = ((-9x^2)/5 + x/2 - 2/15 - (43x^3/30))^5 is my final solution

Which when I plug it into the answer box I am told is incorrect. I feel like I must be missing something horribly obvious, and will probably injure my forehead once someone explains to me where I'm going wrong.