Question

d^2x/dt^2+2 dx/dt+2x = 4cost +2sint x(0) =0; x(0)=3

Answer 1

Answer on Question #83181– Math – Differential Equations

Question

$$\frac {d ^ {2} x}{d t ^ {2}} + 2 \frac {d x}{d t} + 2 x = 4 \cos t + 2 \sin t, x (0) = 0; x (0) = 3.$$

Solution

Solve second order linear nonhomogeneous differential equation with constant coefficients.

$$\frac {d ^ {2} x}{d t ^ {2}} + 2 \frac {d x}{d t} + 2 x = 4 \cos t + 2 \sin t.$$

The general solution of a nonhomogeneous equation is the sum of the general solution $x_0(t)$ of the related homogeneous equation and a particular solution $x_1(t)$ of the nonhomogeneous equation:

$$x (t) = x _ {0} (t) + x _ {1} (t).$$

First we solve the related homogeneous equation

$$\frac {d ^ {2} x}{d t ^ {2}} + 2 \frac {d x}{d t} + 2 x = 0.$$

Find the roots of the corresponding characteristic equation:

$$k ^ {2} + 2 k + 2 = 0,$$ $$D = 2 ^ {2} - 4 \cdot 2 = - 4 = 4 i ^ {2}, \text{ so } k _ {1, 2} = \frac {- 2 \pm \sqrt {4 i}}{2} = - 1 \pm i.$$

If the roots of the characteristic equation are the complex numbers $k_{1,2} = \alpha \pm i\beta$, then the general solution of the homogeneous equation $x(t) = e^{\alpha t}(C_1\cos \beta t + C_2\sin \beta t)$. In our case we have:

$$x _ {0} (t) = e ^ {- t} \left(C _ {1} \cos t + C _ {2} \sin t\right).$$

Now find the particular solution $x_{1}(t)$ of the nonhomogeneous equation. The right side of a nonhomogeneous differential equation is the combination of trigonometric functions. In this case, it's more convenient to look for a solution using the method of undetermined coefficients. Write the particular solution $x_{1}(t)$ in the form:

$$x _ {1} (t) = A \cos t + B \sin t, \text{ where } A, B \text{ are the undetermined coefficients}.$$

Find $\frac{dx_1}{dt}$, $\frac{d^2x_1}{dt^2}$:

$$\frac {d x _ {1}}{d t} = - A \sin t + B \cos t,$$ $$\frac {d ^ {2} x _ {1}}{d t ^ {2}} = - A \cos t - B \sin t.$$

Substitute them back into the original differential equation:

$$- A \cos t - B \sin t - 2 A \sin t + 2 B \cos t + 2 A \cos t + 2 B \sin t = 4 \cos t + 2 \sin t,$$ $$(A + 2 B) \cos t + (B - 2 A) \sin t = 4 \cos t + 2 \sin t.$$

Therefore we can write the following system of equations to determine the coefficients $A, B$:

$$\left\{ \begin{array}{l} A + 2 B = 4, \\ B - 2 A = 2, \end{array} \right. \left\{ \begin{array}{l} A + 4 A + 4 = 4, \\ B = 2 A + 2, \end{array} \right. \left\{ \begin{array}{l} A = 0, \\ B = 2. \end{array} \right.$$

Thus, the particular solution has the form:

$$x _ {1} (t) = 2 \sin t.$$

Respectively, the general solution of the original nonhomogeneous equation is written as:

$$x (t) = e ^ {- t} \left(C _ {1} \cos t + C _ {2} \sin t\right) + 2 \sin t.$$

Find the solution with the initial conditions $x(0) = 0; x(0) = 3$.

$$x (0) = e ^ {0} \left(C _ {1} \cos 0 + C _ {2} \sin 0\right) + 2 \sin 0 = C _ {1}.$$

So we have that $C_1 = 0$ and $C_1 = 3$. But it's impossible and the solution with the initial conditions $x(0) = 0; x(0) = 3$ doesn't exist.

But if the initial conditions $x(0) = 0; x'(0) = 3$ should be considered we can find the solution.

First find $x'(t)$:

$$x ^ {\prime} (t) = - e ^ {- t} \left(C _ {1} \cos t + C _ {2} \sin t\right) + e ^ {- t} \left(- C _ {1} \sin t + C _ {2} \cos t\right) + 2 \cos t,$$ $$x ^ {\prime} (0) = - e ^ {0} \left(C _ {1} \cos 0 + C _ {2} \sin 0\right) + e ^ {0} \left(- C _ {1} \sin 0 + C _ {2} \cos 0\right) + 2 \cos 0 = - C _ {1} + C _ {2} + 2,$$

So

$$\left\{ \begin{array}{c} C _ {1} = 0, \\ - C _ {1} + C _ {2} + 2 = 3, \end{array} \right. \left\{ \begin{array}{c} C _ {1} = 0, \\ C _ {2} = 1. \end{array} \right.$$

Therefore, the solution of the initial-value problem is:

$$x (t) = e ^ {- t} \left(0 \cdot \cos t + 1 \cdot \sin t\right) + 2 \sin t = e ^ {- t} \sin t + 2 \sin t.$$

Answer: The solution doesn't exist for the initial conditions $x(0) = 0; x(0) = 3$.

For the initial conditions $x(0) = 0; x'(0) = 3$ the solution of the initial-value problem is

$$x(t) = e^{-t} \sin t + 2 \sin t.$$

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