# Solving Second Order Differential Equations with Real Distinct Roots

fact
Remember that the "characteristic equation" for a linear second order differential equation $$ay'' + by' + cy = 0$$ is given by: $$ar^2 + br + c = 0$$
fact
If the roots of a second order differential equation's characteristic equation are real and different ($$\lambda_1, \lambda_2$$) then the solutions to the equation are: $$y_1 = e^{\lambda_1t}$$ and $$y_2 = e^{\lambda_2t}$$ So the general solution for the differential equation is: $$y = c_1e^{\lambda_1t} + c_2e^{\lambda_2t}$$
example

Solve the IVP $$2y'' + 4y' = 0$$, $$\quad y(0) = -6$$, $$\quad y'(0) = 6$$

Our characteristic equation is $$2r^2 + 4r = 0 \implies r^2 + 2r = 0$$ $$r(r+2) = 0 \implies r=0, r=-2$$ So the general solution is: $$y = c_1e^{0t} + c_2e^{-2t} = c_1 + c_2e^{-2t}$$ Now we use the initial conditions to find $$c_1$$ and $$c_2$$. $$y(0) = -6 \implies c_1 + c_2 = -6$$ $$y'(0) = 6 \implies -2c_2 = 6$$ So $$c_2 = -3$$ and $$c_1 = -6 -c_2 = -6 +3 = -3$$ Which finally gives us our solution $$y(t) = -3 -3e^{-2t}$$
practice problems