homogeneous equation solution help

dimon

New member
Joined
Jul 9, 2006
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15
Hello

I'm having trouble finding solution for the following equation

4d^2y/dt^2+6dy/dt = 0

with y(0) = 4 and dy(0)/dt = 3

please give me some Ideas
 
\(\displaystyle \L\\4y"+6y'=0\,\ y(0)=4\,\ y'(0)=3\)

auxiliary equation:

\(\displaystyle \L\\4m^{2}+6m=0\)

\(\displaystyle \L\\m=(0,\frac{-3}{2})\)are the roots of the auxiliary equation.

general equation:

\(\displaystyle \L\\C_{1}e^{\frac{-3t}{2}}+C_{2}\)

Using y(0)=4 and solving for C1 in terms of C2:

\(\displaystyle \L\\4=C_{1}e^{\frac{-3(0)}{2}}+C_{2}\)

We find that \(\displaystyle \L\\C_{1}=4-C_{2}\)....[1]

Using the derivative of the general equation, y'(0)=3 and subbing in [1], we find that \(\displaystyle \L\\C_{2}=6\)

Therefore, \(\displaystyle \L\\C_{1}=-2\)

So, we have:

\(\displaystyle \H\\-2e^{\frac{-3t}{2}}+6\)
 
Thanks alot.

Actually I was able to find the roots of the auxiliary equation, however, I'm not
completely sure how you found C1 and C2
 
I had a small typo in my post. I fixed it.

Anyway, C1 is easy to solve for. If you set up your equation using the

initial condition, \(\displaystyle y(0)=4\,\ e^{0}=1\), so you can see you're left with

\(\displaystyle C_{1}=4-C_{2}\)

Now, use the other initial condition, \(\displaystyle y'(0)=3\) to find \(\displaystyle C_{2}\)

Sub \(\displaystyle C_{1}=4-C_{2}\)into the derivative of your general equation:

\(\displaystyle \L\\\frac{d}{dt}[(4-C_{2})e^{\frac{-3t}{2}}+C_{2}]\)

=\(\displaystyle \L\\\frac{3(C_{2}-4)e^{\frac{-3(0)}{2}}}{2}=3\)

Solve for \(\displaystyle C_{2}\). You should get \(\displaystyle C_{2}=6\).
 
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