How are the homogeneous roots and obtained for the ODE ?
Conditions
- The ODE has constant coefficients.
- The trial solution is of the form .
- The characteristic equation is a quadratic polynomial.
Reasoning, step by step
- Set the right-hand side of the ODE to zero to form the homogeneous equation: .
- Substitute the trial solution , which gives and .
- Divide by (which is never zero) to obtain the characteristic equation: .
- Factor the quadratic equation: .
- Solve for to find the roots and .
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 ' => ()()=0'.
Common misconceptions
- Forgetting to set the right-hand side to zero before finding the homogeneous roots.
- Assuming the roots are and by ignoring the signs in the factored equation.
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