Question

Solve the initial value problem y"+3y'=0; y(0)=3, y'(0)=6

Answer 1

Solve the initial value problem y"+3y'=0; y(0)=3, y'(0)=6.

The characteristic equation is

$k^2+3k=0 \\ k(k+3)=0 \implies k_1=0, k_2=-3.$

We have two real distinct roots, then the general solution is

$y(t)=C_1e^{0t}+C_2e^{-3t}=C_1+C_2e^{-3t}.$

Note that $y'(t)=-3C_2e^{-3t}.$

Applying the initial conditions gives the following system of equations for the coefficients $C_1, C_2$ :

$3=y(0)=C_1+C_2, \quad (1)\\ 6=y'(0)=-3C_2 \implies C_2=-2.$

From (1): $C_1=3-C_2=3-(-2)=5$.

We get the solution $y(t)=5-2e^{-3t}.$