Rawr Stuck on ODE (linear 1st order)

[Omega_Prime]

Homework Statement

Solve the initial value problem:

y'+$\frac{4y}{x+8}$=(x+8)$^{8}$ , y(0)=8.

The differential equation is linear.

Homework Equations

N/A

The Attempt at a Solution

I can see that the equation is in the form y' +P(x)*y = Q(x) so I'm like "easy, let me get an integration factor going and solve this mofo."

I= e$^{\int P(x)dx}$ = e$^{\int \frac{4}{x+8}dx}$ = e$^{4 ln |x+8|}$ = (x+8)$^{4}$ (I took the positive expression since any number raised to the 4th power is positive...)

Then multiply the equation by I:

(x+8)$^{4}$*y' + (x+8)$^{4}$*4(x+8)$^{-1}$*y = (x+8)$^{4}$*(x+8)$^{8}$

Simplify

y'(x+8)$^{4}$+4y(x+8)$^{3}$ = (x+8)$^{12}$

Now the LHS is the derivative w/ respect to x for the product y*(x+8)$^{4}$ and by integrating both sides I get:

y(x+8)$^{4}$ = $\frac{(x+8)^{13}}{13}$ + C

So now I solve for C given the initial conditions and I get a big *** negative number

C = $\frac{104-8^{9}}{13}$ = -$\frac{134217624}{13}$

My final solution looks like this (gotten it 3 times now, so I must be doing the same mistake over and over...)

y = $\frac{(x+8)^{9} - 134217624}{13}$

I feel an almost instinctual compulsion to reject any answer that is ginormous or just crazy looking when it comes to homework problems.. but I submitted anyway and sure enough I'm incorrect. Anyone got a bone to throw for me?

 

[muzak]

y(x+8)$^{4}$ = $\frac{(x+8)^{13}}{13}$ + C

Solve for y here. Don't plug in the initial condition right away.

 

[muzak]

You integrated too soon, write out the dy/dx.

 

[Omega_Prime]

y(x+8)$^{4}$ = $\frac{(x+8)^{13}}{13}$ + C

Solve for y here. Don't plug in the initial condition right away.

I didn't think it'd matter... I get the same thing.

y = $\frac{(x+8)^{9}}{13}$ + C$_{1}$ | where C$_{1}$ is just another arbitrary constant that will satisfy the equation given the initial conditions. I get the same constant regardless...

8 = $\frac{8^{9}}{13}$ + C$_{1}$
C$_{1}$ = 8 - $\frac{8^{9}}{13}$ = -$\frac{134217624}{13}$

 

[Omega_Prime]

You integrated too soon, write out the dy/dx.

Ok let me look at that, thanks.

 

[Omega_Prime]

No luck. Oh well, thanks for your time I guess.

 

[muzak]

That's such a ridiculous question, who came up with it?

I keep getting something different than you, a really big number. I think you're supposed to divide the C by the (x+8)^4. You can't change that to another constant because (x+8)^4 isn't a constant. I keep getting -4.22889088e10. Kinda of a ridiculous C and problem.

 

[dynamicsolo]

Now the LHS is the derivative w/ respect to x for the product y*(x+8)$^{4}$ and by integrating both sides I get:

y(x+8)$^{4}$ = $\frac{(x+8)^{13}}{13}$ + C

You're fine up to here. But for y(0) = 8 , shouldn't this be

$$8 \cdot 8^{4} = \frac{8^{13}}{13} + C$$

$$\Rightarrow 8^{5} = \frac{8^{13}}{13} + C ,$$

and so

$$C = 8^{5} - \frac{8^{13}}{13} ?$$

[Note: you are not obligated to simplify this -- in some situations, it isn't even desirable to do so.]

Your solution function is then

$$y = \frac{(x+8)^{9}}{13} + [ 8^{5} - \frac{8^{13}}{13} ] \cdot \frac{1}{(x+8)^{4}} .$$

[In your entry above, don't forget that the "arbitrary constant" also getting divided by (x+8)⁴.]

It is pretty easy to see, in this form, that your initial condition is satisfied.


As far as how you'd come up with such a strange differential equation and solution function, this can arise from the sort of "mixing problems" involving finding the mass function of a substance where solutions of different concentrations are being added and drained from a tank (in which they are "instantaneously mixed"), with the inflow rate being greater than the drainage rate.

 

[Omega_Prime]

Thank you so much guys, I was banging my head on my desk with that. The solution

y = $\frac{(x+8)^{9}}{13}$ + (8$^{5}$ - $\frac{8^{13}}{13}$)(x+8)$^{-4}$

checked out.