Characteristic equation with repeated roots

Characteristic equation with repeated 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:

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

Using the quadratic formula:

r=B±B24AC2Ar=\frac{-B\pm\sqrt{B^2-4AC}}{2A}

But what if B2=4ACB^2=4AC?

r=B±B24AC2A=B2Ar=\frac{-B\pm\sqrt{B^2-4AC}}{2A}=\frac{-B}{2A}

Throughout the videos it will be shown that our solutions are:

y1(x)=eB2Axy_1(x)=e^{\frac{-B}{2A}x}
y2(x)=xeB2Axy_2(x)=xe^{\frac{-B}{2A}x}

Or in full generality:
y(x)=c1er1x+c2xer2xy(x)=c_1 e^{r_1 x}+c_2 xe^{r_2 x}

Where r1=r2=B2Ar_1=r_2=-\frac{B}{2A}
  • 1.
    What is the solution to the Characteristic Equation with Repeated Roots?

  • 2.
    Determining the Characteristic Equation with Repeated Roots
    Find the particular solution to the following differential equation:

    y6y+9y=0y''-6y'+9y=0

    With initial values y(0)=3,y(0)=2y(0)=3, y'(0)=2

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

    4y12y+9y=04y''-12y'+9y=0

    With initial values y(2)=e3,y(2)=12e3y(2)=e^3, y'(2)=\frac{1}{2}e^3