- #1
Omega_Prime
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Homework Statement
Solve the initial value problem:
y'+[itex]\frac{4y}{x+8}[/itex]=(x+8)[itex]^{8}[/itex] , 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[itex]^{\int P(x)dx}[/itex] = e[itex]^{\int \frac{4}{x+8}dx}[/itex] = e[itex]^{4 ln |x+8|}[/itex] = (x+8)[itex]^{4}[/itex] (I took the positive expression since any number raised to the 4th power is positive...)
Then multiply the equation by I:
(x+8)[itex]^{4}[/itex]*y' + (x+8)[itex]^{4}[/itex]*4(x+8)[itex]^{-1}[/itex]*y = (x+8)[itex]^{4}[/itex]*(x+8)[itex]^{8}[/itex]
Simplify
y'(x+8)[itex]^{4}[/itex]+4y(x+8)[itex]^{3}[/itex] = (x+8)[itex]^{12}[/itex]
Now the LHS is the derivative w/ respect to x for the product y*(x+8)[itex]^{4}[/itex] and by integrating both sides I get:
y(x+8)[itex]^{4}[/itex] = [itex]\frac{(x+8)^{13}}{13}[/itex] + C
So now I solve for C given the initial conditions and I get a big *** negative number
C = [itex]\frac{104-8^{9}}{13}[/itex] = -[itex]\frac{134217624}{13}[/itex]
My final solution looks like this (gotten it 3 times now, so I must be doing the same mistake over and over...)
y = [itex]\frac{(x+8)^{9} - 134217624}{13}[/itex]
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?