Question

Using the method of undetermined coefficients, write the trial solution of the equation
y"+2y'+5y=x(e)1/x cos 2x
and hence solve it .

Answer 1

Answer on Question #40492 – Math - Differential Calculus

Using the method of undetermined coefficients, write the trial solution of the equation $y'' + 2y' + 5y = x(e)1/x \cos 2x$ and hence solve it.

Solution.

We have the differential equation

$$y'' + 2y' + 5y = xe^{-x} \cos 2x$$

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

Find the complementary solution by solving

$$y'' + 2y' + 5y = 0$$

Assume a solution will be proportional to $e^{\lambda x}$ for some constant $\lambda$. Substitute $y = e^{\lambda x}$ into the differential equation:

$$\left(e^{\lambda x}\right)'' + 2\left(e^{\lambda x}\right)' + 5e^{\lambda x} = 0$$ $$\lambda^2 e^{\lambda x} + 2\lambda e^{\lambda x} + 5e^{\lambda x} = 0$$

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

$$(\lambda^2 + 2\lambda + 5)e^{\lambda x} = 0$$

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

$$\lambda^2 + 2\lambda + 5 = 0$$

Solve for $\lambda$:

$$\lambda = -1 + 2i \text{ or } \lambda = -1 - 2i$$

The roots $\lambda = -1 \pm 2i$ give $y_1 = c_1 e^{(-1 + 2i)x}$, $y_2 = c_2 e^{(-1 - 2i)x}$ as solutions, where $c_1$ and $c_2$ are arbitrary constants.

The general solution is the sum of the above solutions:

$$y = y_1 + y_2 = \frac{c_1}{e^{(1 - 2i)x}} + \frac{c_2}{e^{(1 + 2i)x}}$$

Apply Euler's identity $e^{\alpha + i\beta} = e^{\alpha} \cos \beta + ie^{\alpha} \sin \beta$:

$$y = c_1 \left(\frac{\cos 2x}{e^x} + i \frac{\sin 2x}{e^x}\right) + c_2 \left(\frac{\cos 2x}{e^x} - i \frac{\sin 2x}{e^x}\right)$$

Regroup:

$$y = \frac{(c_1 + c_2) \cos 2x}{e^x} + \frac{i(c_1 - c_2) \sin 2x}{e^x}$$

Redefine $c_1 + c_2$ as $c_1$ and $i(c_1 - c_2)$ as $c_2$, since these are arbitrary constants:

$$y = \frac {c _ {1} \cos 2 x}{e ^ {x}} + \frac {c _ {2} \sin 2 x}{e ^ {x}}$$

Determine the particular solution to

$$y ^ {\prime \prime} + 2 y ^ {\prime} + 5 y = x e ^ {- x} \cos 2 x$$

by the method of undetermined coefficients. The particular solution is of the form:

$$y _ {p} = x \left(\frac {A \cos 2 x}{e ^ {x}} + \frac {B x \cos 2 x}{e ^ {x}} + \frac {C \sin 2 x}{e ^ {x}} + \frac {D x \sin 2 x}{e ^ {x}}\right),$$

where $\frac{A\cos 2x}{e^x} +\frac{Bx\cos 2x}{e^x} +\frac{C\sin 2x}{e^x} +\frac{Dx\sin 2x}{e^x}$ was multiplied by $x$ to account for $\frac{\cos 2x}{e^x}$ in the complementary solution.

Solve for unknown constants $A, B, C, D$ .

Compute $y_{p}^{\prime}$ :

$$\begin{array}{l} y _ {p} ^ {\prime} = \left(\frac {A x \cos 2 x}{e ^ {x}} + \frac {B x ^ {2} \cos 2 x}{e ^ {x}} + \frac {C x \sin 2 x}{e ^ {x}} + \frac {D x ^ {2} \sin 2 x}{e ^ {x}}\right) ^ {\prime} = \frac {A \cos 2 x}{e ^ {x}} - \frac {A x \cos 2 x}{e ^ {x}} - \\ - \frac {2 A x \sin 2 x}{e ^ {x}} - \frac {B x ^ {2} \cos 2 x}{e ^ {x}} + \frac {2 B x \cos 2 x}{e ^ {x}} - \frac {2 B x ^ {2} \sin 2 x}{e ^ {x}} + \frac {2 C x \cos 2 x}{e ^ {x}} + \frac {C \sin 2 x}{e ^ {x}} - \frac {C x \sin 2 x}{e ^ {x}} + \\ + \frac {2 D x ^ {2} \cos 2 x}{e ^ {x}} - \frac {D x ^ {2} \sin 2 x}{e ^ {x}} + \frac {2 D x \sin 2 x}{e ^ {x}} \\ \end{array}$$

Compute $y_{p}^{\prime \prime}$

$$\begin{array}{l} y _ {p} ^ {\prime \prime} = \left(\frac {A x \cos 2 x}{e ^ {x}} + \frac {B x ^ {2} \cos 2 x}{e ^ {x}} + \frac {C x \sin 2 x}{e ^ {x}} + \frac {D x ^ {2} \sin 2 x}{e ^ {x}}\right) ^ {\prime \prime} = - \frac {4 A x \cos 2 x}{e ^ {x}} + \\ + A \cos 2 x \left(- \frac {2}{e ^ {x}} + \frac {x}{e ^ {x}}\right) - 4 A \left(e ^ {- x} - \frac {x}{e ^ {x}}\right) \sin 2 x - \frac {4 B x ^ {2} \cos 2 x}{e ^ {x}} + B \cos 2 x \left(\frac {2}{e ^ {x}} + \frac {x ^ {2}}{e ^ {x}} - \frac {4 x}{e ^ {x}}\right) - \\ - 4 B \left(- \frac {x ^ {2}}{e ^ {x}} + \frac {2 x}{e ^ {x}}\right) \sin 2 x + 4 C \cos 2 x \left(e ^ {- x} - \frac {x}{e ^ {x}}\right) - \frac {4 C x \sin 2 x}{e ^ {x}} + C \left(- \frac {2}{e ^ {x}} + \frac {x}{e ^ {x}}\right) \sin 2 x + \\ + 4 D \cos 2 x \left(- \frac {x ^ {2}}{e ^ {x}} + \frac {2 x}{e ^ {x}}\right) - \frac {4 D x ^ {2} \sin 2 x}{e ^ {x}} + D \left(\frac {2}{e ^ {x}} + \frac {x ^ {2}}{e ^ {x}} - \frac {4 x}{e ^ {x}}\right) \sin 2 x \\ \end{array}$$

Substitute the particular solution $y_{p}$ into the differential equation and simplify:

$$\frac {(2 B + 4 C) \cos 2 x}{e ^ {x}} + \frac {8 D x \cos 2 x}{e ^ {x}} + \frac {(- 4 A + 2 D) \sin 2 x}{e ^ {x}} - \frac {8 B x \sin 2 x}{e ^ {x}} = \frac {x \cos 2 x}{e ^ {x}}$$

Equate the coefficients of $e^{-x} \cos 2x$ on both sides of the equation:

$$2 B + 4 C = 0$$

And similarly:

$$\begin{array}{l} 8 D = 1 \\ - 4 A + 2 D = 0 \\ \end{array}$$ $$- 8 B = 0$$

Solve the system:

$$A = \frac {1}{16}$$ $$B = 0$$ $$C = 0$$ $$D = \frac {1}{8}$$

Substitute $A, B, C, D$ into $y_{p}$:

$$y _ {p} = \frac {x \cos 2 x}{16 e ^ {x}} + \frac {x ^ {2} \sin 2 x}{8 e ^ {x}}$$

So the general solution is:

$$y = \frac {c _ {1} \cos 2 x}{e ^ {x}} + \frac {c _ {2} \sin 2 x}{e ^ {x}} + \frac {x \cos 2 x}{16 e ^ {x}} + \frac {x ^ {2} \sin 2 x}{8 e ^ {x}}$$

Answer:

$$y (x) = \frac {c _ {1} \cos 2 x}{e ^ {x}} + \frac {c _ {2} \sin 2 x}{e ^ {x}} + \frac {x \cos 2 x}{16 e ^ {x}} + \frac {x ^ {2} \sin 2 x}{8 e ^ {x}}$$