\(\displaystyle \L\\4y"+6y'=0\,\ y(0)=4\,\ y'(0)=3\)
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:
\(\displaystyle \H\\-2e^{\frac{-3t}{2}}+6\)