Solution to Differential Equation: y'-4y=9e^(7t) with Initial Condition y(0)=5

[glid02]

Here's the problem:
Solve y'-4y=9e^(7t) with y(0)=5

p(t)=-4
mu(t)=e^(int[-4dt])=e^(-4t)

multiplying both sides by mu(t)
e^(-4t)(y'-4y)=9e^(3t)
Dt(e^(-4t)y)=9e^(3t)
e^(-4t)y=3e^(3t)+c
y=(3e^(3t))/(e^(-4t))+c/e^(-4t)

now when y(0)=5
5=3+c
c=2

so y=(3e^(3t))/(e^(-4t))+2

That's not right, I also tried y=(3e^(3t))/(e^(-4t))(5/3)
and that's not right either. Any idea what I'm doing wrong?

Thanks.

Never mind, figured it out, I didn't divide c by the left side of the equation.

 

[joob]

your on the right track. You have the correct integration factor.
[tex] \mu(t)= e^{-4t} [tex]

 

[joob]

hmmm, still haven't gotten the hang of how to type formulas here. Ok let's if I get it right this time...
After you set up the problem you actually need to do a full integration on the RHS, and you need to evaluate your LFS at the endpoints of integration...
[tex] \frac{d}{dt} y(t)e^{-4t}= 9e^{3t} \rightarrow y(t)e^(-4t') \Big^t_0= \int^t_0 9e^{3t'}dt' [\tex]

 

[joob]

ok sorry , see if i can get it now...
$$\frac{d}{dt} y(t)e^{-4t}= 9e^{3t} \rightarrow y(t)e^(-4t') \Big^t_0= \int^t_0 9e^{3t'}dt'$$

 

[joob]

last try...
$$\frac{d}{dt} \left[y(t)e^{-4t} \right]= 9e^{3t} \rightarrow y(t)e^{-4t'} \Big| _0^t= \int^t_0 9e^{3t'}dt'$$

 

[HallsofIvy]

Why do you have primes on the ts? On the right hand side, yes, the t inside the integral is a "dummy" variable and so should be distinguished from the t outside. I would be inclined to use a completely different letter, say s. There certainly shouldn't be a prime on the t in the exponential on the left.
To finish, what is
$$\int_0^t 9e^{3s}ds$$