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How are the homogeneous roots r=−2r=-2 and r=−1r=-1 obtained for the ODE x¨+3x˙+2x=2e−t\ddot{x}+3\dot{x}+2x=2e^{-t}?

The homogeneous roots are obtained by solving the associated homogeneous equation x¨+3x˙+2x=0\ddot{x}+3\dot{x}+2x=0. By substituting the exponential trial solution x=ertx=e^{rt}, the differential equation converts into the characteristic polynomial r2+3r+2=0r^2+3r+2=0, which factors to (r+2)(r+1)=0(r+2)(r+1)=0, yielding the roots r=−2r=-2 and r=−1r=-1.

Conditions

  • The ODE has constant coefficients.
  • The trial solution is of the form x=ertx=e^{rt}.
  • The characteristic equation is a quadratic polynomial.

Reasoning, step by step

  1. Set the right-hand side of the ODE to zero to form the homogeneous equation: x¨+3x˙+2x=0\ddot{x}+3\dot{x}+2x=0.
  2. Substitute the trial solution x=ertx=e^{rt}, which gives x˙=rert\dot{x}=re^{rt} and x¨=r2ert\ddot{x}=r^2e^{rt}.
  3. Divide by erte^{rt} (which is never zero) to obtain the characteristic equation: r2+3r+2=0r^2+3r+2=0.
  4. Factor the quadratic equation: (r+2)(r+1)=0(r+2)(r+1)=0.
  5. Solve for rr to find the roots r=−2r=-2 and r=−1r=-1.

Example

The speaker says, 'for the homogeneous equation if we just try x equals e to the rt, we get the characteristic equation r squared plus 3r plus 2 equals 0'. The board writes 'r2+3r+2=0r^2 + 3r + 2 = 0 => (r+2r+2)(r+1r+1)=0'.

Common misconceptions

  • Forgetting to set the right-hand side to zero before finding the homogeneous roots.
  • Assuming the roots are r=2r=2 and r=1r=1 by ignoring the signs in the factored equation.

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