Question

solve the second order differential equation

y" - 3y' + 2y = 3e^(-x) - 10cos3x y(0) = 1 y'(0) =2

Answer 1

Answer on Question #49494 – Math– Differential Calculus | Equations

Question:

Solve the second order differential equation

$$y'' - 3y' + 2y = 3e^{\wedge}(-x) - 10\cos(3x)y(0) = 1 \quad y'(0) = 2$$

Solution:

Solve $-3 \frac{dy(x)}{dx} + \frac{d^2y(x)}{dx^2} + 2y(x) = 3e^{-x} - 10\cos(3x)$, such that $y(0) = 2$ and $y'(0) = 2$:

The general solution will be the sum of the complementary solution and particular solution.

Find the complementary solution by solving $\frac{d^2y(x)}{dx^2} - 3\frac{dy(x)}{dx} + 2y(x) = 0$:

Assume a solution will be proportional to $e^{\lambda x}$ for some constant $\lambda$.

Substitute $y(x) = e^{\lambda x}$ into the differential equation:

$$\frac{d^2}{dx^2}(e^{\lambda x}) - 3 \frac{d}{dx}(e^{\lambda x}) + 2 e^{\lambda x} = 0$$

Substitute $\frac{d^2}{dx^2}(e^{\lambda x}) = \lambda^2 e^{\lambda x}$ and $\frac{d}{dx}(e^{\lambda x}) = \lambda e^{\lambda x}$:

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

Factor out $e^{\lambda x}$:

$$(\lambda^2 - 3\lambda + 2) e^{\lambda x} = 0$$

Since $\pmb{e}^{\lambda x} \neq 0$ for any finite $\lambda$, the zeros must come from the polynomial:

$$\lambda^2 - 3\lambda + 2 = 0$$

Factor:

$$(\lambda - 2)(\lambda - 1) = 0$$

Solve for $\lambda$:

$$\lambda = 1 \text{ or } \lambda = 2$$

The root $\lambda = 1$ gives $y_1(x) = c_1 e^x$ as a solution, where $c_1$ is an arbitrary constant.

The root $\lambda = 2$ gives $y_2(x) = c_2 e^{2x}$ as a solution, where $c_2$ is an arbitrary constant.

The general solution is the sum of the above solutions:

$$y(x) = y_1(x) + y_2(x) = c_1 e^x + c_2 e^{2x}$$

Determine the particular solution to $\frac{d^2 y(x)}{dx^2} + 2y(x) - 3\frac{dy(x)}{dx} = 3e^{-x} - 10\cos(3x)$ by the method of undetermined coefficients:

The particular solution will be the sum of the particular solutions to

$$\frac{d^2 y(x)}{dx^2} + 2y(x) - 3\frac{dy(x)}{dx} = 3e^{-x} \text{ and } \frac{d^2 y(x)}{dx^2} + 2y(x) - 3\frac{dy(x)}{dx} = -10\cos(3x).$$

The particular solution to $\frac{d^2 y(x)}{dx^2} + 2y(x) - 3\frac{dy(x)}{dx} = 3e^{-x}$ is of the form:

$$y_{p_1}(x) = \frac{a_1}{e^x}$$

The particular solution to $\frac{d^2 y(x)}{dx^2} + 2y(x) - 3\frac{dy(x)}{dx} = -10\cos(3x)$ is of the form:

$$y_{p_2}(x) = a_2 \cos(3x) + a_3 \sin(3x)$$

Sum $y_{p_1}(x)$ and $y_{p_2}(x)$ to obtain $y_p(x)$:

$$y_p(x) = y_{p_1}(x) + y_{p_2}(x) = \frac{a_1}{e^x} + a_2 \cos(3x) + a_3 \sin(3x)$$

Solve for the unknown constants $a_1, a_2$, and $a_3$:

Compute $\frac{dy_p(x)}{dx}$:

$$\begin{aligned} \frac{dy_p(x)}{dx} &= \frac{d}{dx} \left( \frac{a_1}{e^x} + a_2 \cos(3x) + a_3 \sin(3x) \right) \\ &= -\frac{a_1}{e^x} - 3a_2 \sin(3x) + 3a_3 \cos(3x) \end{aligned}$$

Compute $\frac{d^2y_p(x)}{dx^2}$ :

$$\begin{array}{l} \frac{d^2 y_p(x)}{d x^2} = \frac{d^2}{d x^2} \left(\frac{a_1}{e^x} + a_2 \cos(3x) + a_3 \sin(3x)\right) \\ = \frac{a_1}{e^x} - 9 a_2 \cos(3x) - 9 a_3 \sin(3x) \end{array}$$

Substitute the particular solution $y_{p}(x)$ into the differential equation:

$$\begin{array}{l} \frac{d^2 y_p(x)}{d x^2} - 3 \frac{d y_p(x)}{d x} + 2 y_p(x) = \frac{3}{e^x} - 10 \cos(3x) \\ \left(\frac{a_1}{e^x} - 9 a_2 \cos(3x) - 9 a_3 \sin(3x)\right) - 3 \left(-\frac{a_1}{e^x} - 3 a_2 \sin(3x) + 3 a_3 \cos(3x)\right) + \\ 2 \left(\frac{a_1}{e^x} + a_2 \cos(3x) + a_3 \sin(3x)\right) = \frac{3}{e^x} - 10 \cos(3x) \end{array}$$

Simplify:

$$\frac{6 a_1}{e^x} + (-7 a_2 - 9 a_3) \cos(3x) + (9 a_2 - 7 a_3) \sin(3x) = \frac{3}{e^x} - 10 \cos(3x)$$

Equate the coefficients of $e^{-x}$ on both sides of the equation:

$$6 a_1 = 3$$

Equate the coefficients of $\cos(3x)$ on both sides of the equation:

$$-7 a_2 - 9 a_3 = -10$$

Equate the coefficients of $\sin(3x)$ on both sides of the equation:

$$9 a_2 - 7 a_3 = 0$$

Solve the system:

$$a_1 = \frac{1}{2}$$ $$a_2 = \frac{7}{13}$$ $$a_3 = \frac{9}{13}$$

Substitute $a_1, a_2$ , and $a_3$ into $y_p(x) = a_1 e^{-x} + a_3 \sin(3x) + a_2 \cos(3x)$ :

$$y_p(x) = \frac{1}{2 e^x} + \frac{7}{13} \cos(3x) + \frac{9}{13} \sin(3x)$$

The general solution is:

$$y(x) = y_c(x) + y_p(x) = \frac{1}{2 e^x} + \frac{7}{13} \cos(3x) + \frac{9}{13} \sin(3x) + c_1 e^x + c_2 e^{2x}$$

Solve for the unknown constants using the initial conditions:

Compute $\frac{dy(x)}{dx}$:

$$\begin{aligned} \frac{dy(x)}{dx} &= \frac{d}{dx} \left( \frac{1}{2e^x} + \frac{7}{13} \cos(3x) + \frac{9}{13} \sin(3x) + c_1 e^x + c_2 e^{2x} \right) \\ &= -\frac{1}{2e^2} + \frac{27}{13} \cos(3x) - \frac{21}{13} \sin(3x) + c_1 e^x + 2c_2 e^{2x} \end{aligned}$$

Substitute $y(0) = 2$ into $y(x) = c_1 e^x + c_2 e^{2x} + \frac{e^{-x}}{2} + \frac{9}{13} \sin(3x) + \frac{7}{13} \cos(3x)$:

$$c_1 + c_2 + \frac{27}{26} = 2$$

Substitute $y'(0) = 2$ into $\frac{dy(x)}{dx} = c_1 e^x + 2c_2 e^{2x} - \frac{e^{-x}}{2} - \frac{21}{13} \sin(3x) + \frac{27}{13} \cos(3x)$:

$$c_1 + 2c_2 + \frac{41}{26} = 2$$

Solve the system:

$$\begin{aligned} c_1 &= \frac{3}{2} \\ c_2 &= -\frac{7}{13} \end{aligned}$$

Substitute $y(0) = 2$ into $y(x) = c_{1} e^{x} + c_{2} e^{2x} + \frac{e^{-x}}{2} + \frac{9}{13} \sin(3x) + \frac{7}{13} \cos(3x)$:

$$c_{1} + c_{2} + \frac{27}{26} = 2$$

Substitute $y'(0) = 2$ into $\frac{dy(x)}{dx} = c_{1} e^{x} + 2 c_{2} e^{2x} - \frac{e^{-x}}{2} - \frac{21}{13} \sin(3x) + \frac{27}{13} \cos(3x)$:

$$c_{1} + 2 c_{2} + \frac{41}{26} = 2$$

Solve the system:

$$\begin{array}{l} c_{1} = \frac{3}{2} \\ c_{2} = -\frac{7}{13} \\ \end{array}$$

Substitute $c_{1} = \frac{3}{2}$ and $c_{2} = -\frac{7}{13}$ into $y(x) =$

$$c_{1} e^{x} + c_{2} e^{2x} + \frac{e^{-x}}{2} + \frac{9}{13} \sin(3x) + \frac{7}{13} \cos(3x);$$

Answer:

$$y(x) = -\frac{7}{13} e^{2x} + \frac{3}{2} e^{x} + \frac{e^{-x}}{2} + \frac{9}{13} \sin(3x) + \frac{7}{13} \cos(3x)$$

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