Characteristic equation with real distinct roots

Characteristic equation with real distinct roots

Lessons

In the previous section we came up with a method to solve linear homogeneous constant coefficient second order differential equations:

Ay+By+Cy=0Ay''+By'+Cy=0

By using the characteristic equation:By using the characteristic equation:

Ar2+Br+C=0Ar^2+Br+C=0

And solving this quadratic will yield two roots, r1,r2r_1,r_2. Let’s suppose that both r1r_1 and r2r_2 are distinct and real.

So the solution will be:

y1(x)=er1xy_1 (x)=e^{r_1 x}
y2(x)=er2xy_2 (x)=e^{r_2 x}

Or in full generality:

y(x)=c1er1x+c2er2xy(x)=c_1 e^{r_1 x}+c_2 e^{r_2 x}

This is the general solution. We can find a particular solution with initial parameters

y(x0)=y0,y(x1)=y1y(x_0 )=y_0, y' (x_1 )=y_1.
  • Introduction
    Using the Characteristic Equation with Real Distinct Roots

  • 1.
    Using the Characteristic Equation with Real Distinct Roots
    Find the particular solution to the following differential equation:

    y9y=0y''-9y=0

    With initial values y(0)=2,y(0)=1y(0)=2, y' (0)=1

  • 2.
    Find the particular solution to the following differential equation:

    6y+8y8y=06y''+8y'-8y=0

    With initial values y(2)=1,y(2)=4y(2)=1, y' (2)=4

  • 3.
    Find the particular solution to the following differential equation:

    y5y+3y=0y''-5y'+3y=0

    With initial values y(0)=0,y(0)=1y(0)=0, y' (0)=1