Question

Solve the initial value problem:
2 2 ,0 )0( ,2 )0( 1
2
2
+ + y = y = y′ =
dx
dy
dx
d y

Answer 1

Answer on Question #62062 – Math – Differential Equations

Question

Solve the initial value problem

$$\frac {d ^ {2} y}{d x ^ {2}} + 2 \frac {d y}{d x} + 2 y = 0, y (0) = 2, y ^ {\prime} (0) = 1$$

Solution

In this question

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

is a second order homogeneous equation with constant coefficients.

Characteristic equation is

$$\begin{array}{l} r ^ {2} + 2 r + 2 = 0 \\ D = 2 ^ {2} - 4 \cdot 2 = - 4 < 0. \\ \end{array}$$

Its roots are complex:

$$r _ {1, 2} = - 1 \pm i.$$

The general solution of (2) is given by

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

In order to find a particular solution, we use the initial conditions (1) to determine $C_1$ and $C_2$ .

First, we have

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

Since

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

we get

$$y ^ {\prime} (0) = - C _ {1} e ^ {0} \cos 0 - C _ {1} e ^ {0} \sin 0 - C _ {2} e ^ {0} \sin 0 + C _ {2} e ^ {0} \cos 0 = - 2 + C _ {2} = 1;$$

hence

$$C _ {2} = 3.$$

Thus, the solution of the initial value problem (1) is

$$y = 2 e ^ {- x} \cos x + 3 e ^ {- x} \sin x.$$

Answer: $y = 2e^{-x}\cos x + 3e^{-x}\sin x.$

www.AssignmentExpert.com