Question

Solve the following IVP
(d^2y)/(dx^2) + (dy)/(dx) - 2y = -6 sin2x -18 cos2x
y(0)=2, y'(0)=2

Answer 1

Answer on Question # 66662 – Math – Differential Equations Question

Solve the following IVP

$$\begin{array}{l} (d ^ {2} y) / (d x ^ {2}) + (d y) / (d x) - 2 y = - 6 \sin 2 x - 18 \cos 2 x \\ y (0) = 2, y ^ {\prime} (0) = 2 \end{array}$$

Solution

We have IVP

$$\frac {d ^ {2} y}{d x ^ {2}} + \frac {d y}{d x} - 2 y = - 6 \sin 2 x - 18 \cos 2 x, \quad y (0) = 2, y ^ {\prime} (0) = 2$$

Find first the general solution of this equation which can be written as

$$y = y _ {c} + y _ {p}$$

where $y_{p}$ is a particular solution of original equation and $y_{c}$ is the general solution of the related homogeneous equation

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

To find $y_{c}$ we substitute solution as exponential function $y = e^{mx}$ into the equation. We get

$$m ^ {2} e ^ {m x} + m e ^ {m x} - 2 e ^ {m x} = 0$$

or

$$(m ^ {2} + m - 2) e ^ {m x} = 0$$

Then we get the auxiliary equation

$$m ^ {2} + m - 2 = 0$$

with roots $m = 1$ and $m = -2$. So the solution of the complementary equation is

$$y _ {c} = c _ {1} e ^ {x} + c _ {2} e ^ {- 2 x}$$

where $c_{1}$ and $c_{2}$ are some constants. Next we find a particular solution in the form

$$y _ {p} = A \sin 2 x + B \cos 2 x$$

Then

$$y _ {p} ^ {\prime} = 2 A \cos 2 x - 2 B \sin 2 x$$ $$y _ {p} ^ {\prime \prime} = - 4 A \sin 2 x - 4 B \cos 2 x$$

Substituting $y_{p}, y_{p}^{\prime}$ and $y_{p}^{\prime \prime}$ into the given differential equation gives

$$- 4 A \sin 2 x - 4 B \cos 2 x + 2 A \cos 2 x - 2 B \sin 2 x - 2 A \sin 2 x - 2 B \cos 2 x = - 6 \sin 2 x - 18 \cos 2 x$$

or

$$- (6 A + 2 B) \sin 2 x - (6 B - 2 A) \cos 2 x = - 6 \sin 2 x - 18 \cos 2 x$$

This is true if

$$6 A + 2 B = 6 \quad \text{and} \quad 6 B - 2 A = 18$$

The solution of this system is $A = 0$ and $B = 3$. So, the particular solution is

$$y _ {p} = 3 \cos 2 x$$

The general solution of the given equation is

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

Now we find $c_{1}$ and $c_{2}$ using the initial conditions $y(0) = 2, y'(0) = 2$:

$$y (0) = c _ {1} e ^ {0} + c _ {2} e ^ {0} + 3 \cos 0 = c _ {1} + c _ {2} + 3 = 2$$ $$y ^ {\prime} = c _ {1} e ^ {x} - 2 c _ {2} e ^ {- 2 x} - 6 \sin 2 x$$ $$y'(0) = c_1 e^0 - 2c_2 e^0 - 6\sin0 = c_1 - 2c_2 = 2$$

We have system of equation

$$\left\{ \begin{array}{l} c_1 + c_2 = -1 \\ c_1 - 2c_2 = 2 \end{array} \right.$$

The solution of this system is $c_1 = 0$, $c_2 = -1$, so the solution of IVP is

$$y = -e^{-2x} + 3\cos2x$$

Answer: The solution of IVP is

$$y = -e^{-2x} + 3\cos2x$$

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