Question

Solve the initial-value problem (x^2-1)y'-2xy=x(x^2-1), y(0)=4.

Answer 1

I don't know why the answer is y=(x^2-1)[(1/2)ln abs(x^2-1)-4] when I got the answer y=x^4/(4(x^2-1))-x^2/(2(x^2-1))-4/(x^2-1).

Answer 2

Here's the work:

y'-2x/(x^2-1)y=x

u=e^(ln abs(x^2-1))=x^2-1

(x^2-1)y'-2xy=x(x^2-1)

(x^2-1)y=x^4/4-x^2/2+C

y=x^4/(4(x^2-1))-x^2/(2(x^2-1))+C/(x^2-1)

y(0)=-C=4

C=-4

y=x^4/(4(x^2-1))-x^2/(2(x^2-1))-4/(x^2-1)

Answer 3

@goformit100 @paki @SithsAndGiggles @charlotte123

Answer 4

\[\begin{align*}(x^2-1)y'-2xy&=x(x^2-1)\\
y'-\frac{2x}{x^2-1}y&=x
\end{align*}\]
The integrating factor is
\[\begin{align*}\ln\mu(x)&=-\int\frac{2x}{x^2-1}~dx\\
\ln\mu(x)&=-\ln|x^2-1|\\
\mu(x)&=\frac{1}{x^2-1}
\end{align*}\]

Answer 5

I think I forgot the negative sign. So the integrating factor u=e^-ln abs(x^2-1), not e^ln abs(x^2-1), right?

Answer 6

Yes, then log properties say the integrating factor is the reciprocal of what you previously got.

Answer 7

Thank you. I must be careful later.

Answer 8

You're welcome!