Question

Solve the initial value problem: x(dy/dx)+2y(x)=9y(x)^2, y(1)=-1.
I got y = -(e^((9x^2/2)-9/2))/x^2 but I have my doubts..

Answer 1

\[x \frac{ dy }{ dx }+2y(x)=9y(x)^2, y(1)=-1\]

Answer 2

\[y=-\frac{ e ^{\frac{ 9x^2 }{ 2 }-\frac{ 9 }{ 2 }} }{ x ^{2} }\]

Answer 3

if you have doubts, double check that it works ...

Answer 4

My working out seems right to me, but I have been wrong before so I was just wondering whether anyone else got the same answer that I did

Answer 5

did you try to seperate it?

xy' = 9y^2 - 2y

dy/(9y^2 - 2y) = dx/x

Answer 6

I think I made a mistake in my working out, checking it now, might be a while...

Answer 7

decomp the fraction

1/(y(9y-2)) = A/y + B/(9y-2)

1 = A(9y-2) + By

let y=0; A=-1/2

1 = -(9y-2)/2 + By

let y=2/9; B = 9/2

-1/2 ln(y) + 1/2 ln(9y-2) = ln(x) + C

ln((9y-2)/y) = ln(x^2) + 2C

(9y-2)/y = Cx^2; C=7

9y-2 = 7yx^2

y(9-7x^2) = 2

etc ...

Answer 8

I ended up getting a similar answer to yours, but instead of 9, I had 11... so my answer was \[y=\frac{ 2 }{9-11x ^{2} }\]

Answer 9

Instead of 7*

Answer 10

your initial value work is flawed then, assuming you posted it correctly

Answer 11

x=-1, y=1

Answer 12

lol, these old eyes get tired

Answer 13

i read it wrong ;) your good

Answer 14

Aha, it happens to all of us! Thanks again for your help, much appreciated!! :D