Question

Find the general solution the differential equation
y''-3y'=8*exp(3x)+4sinx

Answer 1

Answer on Question #58666 - Math - Differential Equations

Question

$$y'' - 3y' = 8e^{3x} + 4\sin x.$$

Solution

The general solution will be the sum of the complementary solution $y_0$ and a particular solution $y^*$:

$$y = y_0 + y^*.$$

Solving the characteristic equation

$$k^2 - 3k = 0$$

gives

$$k_1 = 0, \, k_2 = 3.$$

Then the solution $y_0$ of the differential equation $y'' - 3y' = 0$ is

$$y_0 = C_1 + C_2 e^{3x}.$$

Find a particular solution $y^*$ of the differential equation

$$y'' - 3y' = 8e^{3x} + 4\sin x$$

by the method of undetermined coefficients:

$$y^* = x A e^{3x} + B \cos x + C \sin x.$$

Solve for unknown constants $A, B, C$:

$$y^{*'} = A e^{3x} + 3A x e^{3x} - B \sin x + C \cos x;$$ $$y^{'''} = 3A e^{3x} + 3A e^{3x} + 9x A e^{3x} - B \cos x - C \sin x = 6A e^{3x} + 9x A e^{3x} - B \cos x - C \sin x.$$

Plug a particular solution into the differential equation:

$$6A e^{3x} + 9x A e^{3x} - B \cos x - C \sin x - 3(A e^{3x} + 3A x e^{3x} - B \sin x + C \cos x) = 8e^{3x} + 4 \sin x.$$ $$3A e^{3x} + (-B - 3C) \cos x + (3B - C) \sin x = 8e^{3x} + 4 \sin x.$$

Equating like terms gives the system of equations:

$$\left\{ \begin{array}{c} 3 A = 8, \\ - B - 3 C = 0, \\ 3 B - C = 4. \end{array} \right. \Rightarrow \left\{ \begin{array}{c} A = \frac {8}{3} \\ B = - 3 C \\ - 9 C - C = 4 \end{array} \right. \Rightarrow \left\{ \begin{array}{c} A = \frac {8}{3} \\ B = - 3 \cdot \left(- \frac {2}{5}\right) = \frac {6}{5} \\ C = - \frac {4}{10} = - \frac {2}{5} \end{array} \right.$$

hence

$$y ^ {*} = \frac {8}{3} e ^ {3 x} + \frac {6}{5} \cos x - \frac {2}{5} \sin x.$$

Then the general solution of $y'' - 3y' = 8e^{3x} + 4\sin x$ is

$$y = y _ {0} + y ^ {*} = C _ {1} + C _ {2} e ^ {3 x} + \frac {8}{3} e ^ {3 x} + \frac {6}{5} \cos x - \frac {2}{5} \sin x.$$

Answer: $y = C_{1} + C_{2}e^{3x} + \frac{8}{3} e^{3x} + \frac{6}{5}\cos x - \frac{2}{5}\sin x.$

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