Repeated Roots: 4y'' - 12y' + 9y = 0, y(1) = 4, y'(1) = 2

[jjm5119]

Ok I keep trying this problem and I can't see where I'm going wrong.

4y''-12y'+9y=0 with initial conditions y(1)=4, y'(1)=2

So I use the characteristic equation (2r-3)(2r-3)=0, and find my r to be 3/2

this follows that the solution will be y=c1e^(3/2)t + c2te^(3/2)t

so then i sub in my initial condition y(1) and find that c1e^(3/2)+c2e^(3/2)t=4

then I take the derivative and use it's initial condition to find that y'(1)=(3/2)c1e^(3/2)+(3/2)c2e^(3/2)+c2e^(3/2)

solving for c1 from c1e^(3/2)+c2e^(3/2)t=4 gives me c1=4-c2e^(3/2)t

Plugging c1 back into y' and solving for c2 I get that c2=-4/e^(3/2)

Then I plug my c2 into c1=4/e^(3/2)-c2 to get c1=4/e^(3/2)-(-4/e^(3/2))

So i get c1=8e^/(3/2) and c2=-4/e^(3/2)

Therefore my solution is 8e^/(3/2)*e^((3/2)t)-4/e^((3/2)t)*t*e^((3/2)t)

This answer is not coming out to be correct, I believe I am screwing up my constants but I am not sure. Any help would be appreciated, thanks.

 

[skeeter]

Re: Repeated Roots Problem

I checked your work ... I also get

\(\displaystyle c_1 = 8e^{-\frac{3}{2}}\) and \(\displaystyle c_2 = -4e^{-\frac{3}{2}}\)

\(\displaystyle y = 8e^{-\frac{3}{2}} \cdot e^{\frac{3}{2}t} - 4e^{-\frac{3}{2}}t \cdot e^{\frac{3}{2}t}\)

\(\displaystyle y = 8e^{\frac{3}{2}(t-1)} - 4te^{\frac{3}{2}(t-1)}\)

\(\displaystyle y' = 8e^{\frac{3}{2}(t-1)} - 6te^{\frac{3}{2}(t-1)}\)

\(\displaystyle y'' = 6e^{\frac{3}{2}(t-1)} - 9te^{\frac{3}{2}(t-1)}\)

check me, but won't those expressions for y, y', and y'' satisfy the original differential equation?

 

[jjm5119]

Ah thanks man. I kept putting the second term to be -4e^((-3/2)t)*e^((3/2)t) instead of -4e^(-3/2)t*e^((3/2)t). In other words I was raising e to the (3/2)t instead of just multiplying it by t.