# 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