Question

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

Answer 1

@SithsAndGiggles

Answer 2

Here's the work:

dy/((y+1)(y-2)^2)=-x^2 dx

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

Now what?

Answer 3

So you're stuck on the partial fractions?

Answer 4

Yes.

Answer 5

Okay, you have a quadratic factor in the denominator, which means you should write it as
\[\frac{By+C}{(y-2)^2}\]
or, because the quadratic is reducible, you could use
\[\frac{B}{y-2}+\frac{C}{(y-2)^2}\]
This is another one of those partial fraction tricks.

Answer 6

So A/(y+1)+(By+C)/(y-2)^2?

Answer 7

That's right.

Answer 8

Wait a minute. Let me work it out.

Answer 9

So I got A(y-2)^2+(By+C)(y+1)=1
A=1/9 when y=-1, but how do I find B, C?

Answer 10

Hmm, why did you use \(A=\dfrac{1}{9}\)?

Answer 11

Since I want to find A, B, C.

Answer 12

Oh sorry I misunderstood what you meant... You got \(A=-\dfrac{1}{9}\) when you plugged in \(y=-1\)... Got it.

To find \(B\) and \(C\), you have to plug in two values of \(y\) (not -1, of course). Then you'll have a system of equations for which you'll need to solve for two unknowns \(B\) and \(C\).

Answer 13

I got A=1/9, B=-1/9, C=1, am I right?

Answer 14

A and B are right. You should have something like this:
\[\begin{align*}
\frac{A}{y+1}+\frac{By+C}{(y-2)^2}&=\frac{1}{(y+1)(y-2)^2}\\
A(y-2)^2+(By+C)(y+1)&=1
\end{align*}\]
Letting \(y=-1\), you have \(A=\dfrac{1}{9}\).

Letting \(y=0\), you have
\[4A+C=1~~\iff~~\frac{4}{9}+C=1~~\iff~~C=\frac{5}{9}\]
Letting \(y=2\), you have
\[0A+(2B+C)(2+1)=1~~\iff~~6B+3C=1~~\iff~~6B+\frac{15}{9}=1\]
and so \(B=-\dfrac{1}{9}\).

Answer 15

Okay, got A, B, C. Now how do I integrate (By+C)/(y-2)^2, I know that A/(y+1)=(1/9) ln abs(y+1).

Answer 16

Okay so we have
\[\int\frac{\dfrac{1}{9}}{y+1}~dy+\int\frac{-\dfrac{1}{9}y+\dfrac{5}{9}}{(y-2)^2}~dy\]
Factor and split up the integrals:
\[\frac{1}{9}\int\frac{dy}{y+1}-\frac{1}{9}\int\frac{y}{(y-2)^2}~dy+\frac{5}{9}\int\frac{dy}{(y-2)^2}\]
First integral is logarithmic, like you have it.

The other two can use a substitution, \(u=y-2\), which means you have \(du=dy\). Also, \(y=u+2\).
\[\frac{1}{9}\ln|y+1|-\frac{1}{9}\int\frac{u+2}{u^2}~du+\frac{5}{9}\int\frac{dy}{u^2}\]

Answer 17

The second integral (u+2)/u^2 can be simplified to 1/u+2/u^2, right?

Answer 18

Yeah

Answer 19

I got (1/9)ln abs(y+1)-(1/9)ln abs(y-2)+2/(9(y-2))-5/(9(y-2)).

Answer 20

This is so hard...Let me simplify.

Answer 21

Sorry about the delay, I have a bit on my plate at the moment. I'll be back in just a minute.

Answer 22

Sure.

Answer 23

So I got ((y+1)/(y-2))^(1/9)-1/(3y-6)=ce^(-x^3/3)

Answer 24

How do I simplify from there?

Answer 25

Okay I'm back. Sorry for the wait.

You should have
\[\frac{1}{9}\ln|y+1|-\frac{1}{9}\int\left(\frac{1}{u}+\frac{2}{u^2}\right)~du+\frac{5}{9}\int\frac{du}{u^2}\\
\frac{1}{9}\ln|y+1|-\frac{1}{9}\ln|u|+\frac{2}{9u}-\frac{5}{9u}+C\\
\frac{1}{9}\left(\ln|y+1|-\ln|u|\right)-\frac{1}{3u}+C\\
\frac{1}{9}\ln\left|\frac{y+1}{u}\right|-\frac{1}{3u}+C\\
\frac{1}{9}\ln\left|\frac{y+1}{y-2}\right|-\frac{1}{3(y-2)}+C\]
So you have
\[\frac{1}{9}\ln\left|\frac{y+1}{y-2}\right|-\frac{1}{3(y-2)}=-\frac{1}{3}x^3+C\]
There's no way to express the general solution explicitly as a function of \(y\) in terms of \(x\), but that's okay.

Given that \(y(4)=2\), you have
\[\frac{1}{9}\ln\left|\frac{2+1}{2-2}\right|-\frac{1}{3(2-2)}=-\frac{1}{3}(4^3)+C\]
which is strange... Are you sure those are the right initial values?

Answer 26

Yes, I copied down the problem correctly. It's also strange to me because there's no answer in the book for this problem and I don't know why.

Answer 27

Well \(y=2\) isn't in the domain for the dependent variable. It must be a typo. I'd say submit the general solution and be done with it :P

Answer 28

Okay, thanks for the help. No wonder the problem is weird.