Question

Q. solve (D4+8D2−9)y=9x3+5cos2x

Answer 1

Answer on Question #64774 – Math – Differential Equations

Question

Solve $(D4+8D2-9)y=9x3+5\cos2x$

Solution

Rewrite the equation $\left(D^{4} + 8D^{2} - 9\right)y = 9x^{3} + 5\cos 2x$ as

$$y ^ {(4)} + 8 y ^ {\prime \prime} - 9 y = 9 x ^ {3} + 5 \cos 2 x.$$

The general solution will be the sum of the complementary and a particular solutions. Find the complementary solution by solving

$$y ^ {(4)} + 8 y ^ {\prime \prime} - 9 y = 0.$$

Assume a solution will be $e^{\lambda x}$, where $\lambda$ is a constant.

Substituting $y = e^{\lambda x}$ into the differential equation

$$\begin{array}{l} (e ^ {\lambda x}) ^ {(4)} + 8 (e ^ {\lambda x}) ^ {\prime \prime} - 9 e ^ {\lambda x} = 0, \\ \lambda^ {4} e ^ {\lambda x} + 8 \lambda^ {2} e ^ {\lambda x} - 9 e ^ {\lambda x} = 0, \\ e ^ {\lambda x} (\lambda^ {4} + 8 \lambda^ {2} - 9) = 0, \\ e ^ {\lambda x} \neq 0, \lambda^ {4} + 8 \lambda^ {2} - 9 = 0, \\ \left(\lambda^ {2} + 9\right) \left(\lambda^ {2} - 1\right) = 0, \\ \left(\lambda + 3 i\right) \left(\lambda - 3 i\right) \left(\lambda - 1\right) \left(\lambda + 1\right) = 0. \end{array}$$

Solving for $\lambda$:

$$\lambda = 1 \text{ or } \lambda = - 1 \text{ or } \lambda = 3 i \text{ or } \lambda = - 3 i.$$

The complementary solution is

$$y _ {c} = C _ {1} e ^ {x} + C _ {2} e ^ {- x} + C _ {3} \cos 3 x + C _ {4} \sin 3 x.$$

Determine a particular solution of $y^{(4)} + 8y^{\prime \prime} - 9y = 9x^{3} + 5\cos 2x$ via the method of undetermined coefficients.

A particular solution will be the sum of the particular solutions to $y^{(4)} + 8y^{\prime \prime} - 9y = 9x^{3}$ and $y^{(4)} + 8y^{\prime \prime} - 9y = 5\cos 2x$.

A particular solution to $y^{(4)} + 8y^{\prime \prime} - 9y = 9x^{3}$ is given by

$$y _ {1} = a _ {1} + a _ {2} x + a _ {3} x ^ {2} + a _ {4} x ^ {3}.$$

A particular solution to $y^{(4)} + 8y^{\prime \prime} - 9y = 5\cos 2x$ is given by

$$y _ {2} = a _ {5} \cos 2 x + a _ {6} \sin 2 x.$$

Adding $y_{1}$ and $y_{2}$ one obtains $y_{p}$:

$$y _ {p} = y _ {1} + y _ {2} = a _ {1} + a _ {2} x + a _ {3} x ^ {2} + a _ {4} x ^ {3} + a _ {5} \cos 2 x + a _ {6} \sin 2 x.$$

To find the unknown coefficients $a_1, a_2, a_3, a_4, a_5$ and $a_6$ first we need to compute

$$\begin{array}{l} (y _ {p}) ^ {(4)} = \left(a _ {1} + a _ {2} x + a _ {3} x ^ {2} + a _ {4} x ^ {3}\right) ^ {(4)} + \left(a _ {5} \cos 2 x + a _ {6} \sin 2 x\right) ^ {(4)} = \\ = a _ {5} \cdot 2 ^ {4} \cos \left(2 x + \frac {\pi}{2} \cdot 4\right) + a _ {6} \cdot 2 ^ {4} \sin \left(2 x + \frac {\pi}{2} \cdot 4\right) = 16 a _ {5} \cos 2 x + 16 a _ {6} \sin 2 x \\ \end{array}$$ $$(y _ {p}) ^ {\prime \prime} = \left(a _ {1} + a _ {2} x + a _ {3} x ^ {2} + a _ {4} x ^ {3}\right) ^ {\prime \prime} + \left(a _ {5} \cos 2 x + a _ {6} \sin 2 x\right) ^ {\prime \prime} = 2 a _ {3} + 6 a _ {4} x - 4 a _ {5} \cos 2 x - 4 a _ {6} \sin 2 x$$

Substituting the particular solution $y_{p}$ and $\left(y_{p}\right)^{(4)}, \left(y_{p}\right)^{\prime \prime}$ into the differential equation

$$(y_p)^{(4)} + 8(y_p)'' - 9y = 9x^3 + 5\cos 2x,$$

one gets

$$\begin{array}{l} 16a_5 \cos 2x + 16a_6 \sin 2x + 8(2a_3 + 6a_4x - 4a_5 \cos 2x - 4a_6 \sin 2x) - \\ - 9(a_1 + a_2x + a_3x^2 + a_4x^3 + a_5 \cos 2x + a_6 \sin 2x) = 9x^3 + 5 \cos 2x. \end{array}$$

Simplifying

$$(-9a_1 + 16a_3) + (-9a_2 + 48a_4) - 9a_3x^2 - 9a_4x^3 - 25a_5 \cos 2x - 25a_6 \sin 2x = 9x^3 + 5 \cos 2x.$$

Then

$$\left\{ \begin{array}{l} -9a_1 + 16a_3 = 0 \\ -9a_2 + 48a_4 = 0 \\ -9a_3 = 0 \\ -9a_4 = 9 \\ -25a_5 = 5 \\ -25a_6 = 0 \end{array} \right.$$

The third equation $(-9a_3 = 0)$ of the system gives $a_3 = 0$ and substituting for $a_3$ into the first equation $(-9a_1 + 16a_3 = 0)$ one gets $-9a_1 = 0$, hence $a_1 = 0$.

The fourth equation $(-9a_4 = 9)$ of the system gives $a_4 = -1$ and substituting for $a_4$ into the second equation $(-9a_2 + 48a_4 = 0)$ one gets $-9a_2 - 48 = 0$, hence $a_2 = -\frac{48}{9} = -\frac{16}{3}$.

It follows from the fifth equation $(-25a_5 = 5)$ of the system that $a_5 = -\frac{5}{25} = -\frac{1}{5}$.

It follows from the sixth equation $(-25a_6 = 0)$ of the system that $a_6 = 0$.

Solution of the system is

$$\left\{ \begin{array}{l} a_1 = 0 \\ a_2 = -\frac{16}{3} \\ a_3 = 0 \\ a_4 = -1 \\ a_5 = -\frac{1}{5} \\ a_6 = 0 \end{array} \right.$$

Then

$$y_p = -\frac{16}{3}x - x^3 - \frac{1}{5}\cos 2x \text{ and}$$ $$y = y_c + y_p = C_1 e^x + C_2 e^{-x} + C_3 \cos 3x + C_4 \sin 3x - x^3 - \frac{16}{3}x - \frac{1}{5} \cos 2x.$$

Answer:

$$y = y_c + y_p = C_1 e^x + C_2 e^{-x} + C_3 \cos 3x + C_4 \sin 3x - x^3 - \frac{16}{3}x - \frac{1}{5} \cos 2x.$$

Answer provided by https://www.AssignmentExpert.com