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Exercise

$y'''-y''-2y'=0$

Step-by-step Solution

1

Obtain the characteristic equation

$r^{3}-r^{2}-2r=0$
2

Find the solutions to the cubic equation $r^{3}-r^{2}-2r=0$

$r=0,\:r=-1,\:r=2$
3

Use a formula to find the general solution to the differential equation. Substituting each solution to the characteristic equation ($r$ values) into the formula $y=e^{rx}$ gives us a linearly independent solution. Then the general solution to the differential equation is the sum of all linearly independent solutions obtained

$y=C_0e^{0x}+C_1e^{-x}+C_2e^{2x}$
4

Simplifying

$y=C_0+C_1e^{-x}+C_2e^{2x}$

Final answer to the exercise

$y=C_0+C_1e^{-x}+C_2e^{2x}$

Try other ways to solve this exercise

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  • Exact Differential Equation
  • Linear Differential Equation
  • Separable Differential Equations
  • Homogeneous Differential Equation
  • Integrate by partial fractions
  • Product of Binomials with Common Term
  • FOIL Method
  • Integrate by substitution
  • Integrate by parts
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