Question

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Answer 1

Answer on Question #60154 – Math – Differential Equations

Question

Find a homogeneous linear differential equation with constant coefficients whose general solution is given by

$$y(x) = c_1 + c_2 e^{2x} \cos 5x + c_3 e^{2x} \sin 5x.$$

Solution

General solution of a homogeneous linear differential equation with constant coefficients can be written as

$$y(x) = c_1 e^{\lambda_1 x} + c_2 e^{\lambda_2 x} + c_3 e^{\lambda_3 x},$$

where $\lambda_1, \lambda_2, \lambda_3$ are the complex numbers.

The solution $y(x) = c_1 + c_2 e^{2x} \cos 5x + c_3 e^{2x} \sin 5x$ gives us the hint, that

$$\lambda_1 = 0, \lambda_2 = 2 + 5i, \lambda_3 = 2 - 5i.$$

Let's write the characteristic equation:

$$(\lambda - \lambda_1)(\lambda - \lambda_2)(\lambda - \lambda_3) = 0.$$ $$\begin{array}{l} (\lambda - 0) \cdot (\lambda - (2 + 5i)) \cdot (\lambda - (2 - 5i)) = \\ = \lambda \cdot ((\lambda - 2) - 5i) \cdot ((\lambda - 2) + 5i) = \\ = \lambda \cdot ((\lambda - 2)^2 + 25) = \lambda \cdot (\lambda^2 - 4\lambda + 4 + 25) = \\ = \lambda \cdot (\lambda^2 - 4\lambda + 4 + 25) = \lambda \cdot (\lambda^2 - 4\lambda + 29) = \\ = \lambda^3 - 4\lambda^2 + 29\lambda = 0. \end{array}$$

As a solution of any linear differential equation with constant coefficients can be found in the form of $y = e^{\lambda x}$, $y'(x) = \lambda e^{\lambda x}$, $y''(x) = \lambda^2 e^{\lambda x}$, $y'''(x) = \lambda^3 e^{\lambda x}$, it's obvious, that

$$e^{\lambda x} \lambda^3 - 4 e^{\lambda x} \lambda^2 + 29 e^{\lambda x} \lambda = 0$$

and the differential equation is $y'''(x) - 4y''(x) + 29y'(x) = 0$.

Answer: $y'''(x) - 4y''(x) + 29y'(x) = 0$.

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