Question

1. a) Solve the following ordinary differential equations:

i) dy/dx + 4xy = x

ii) d²y/dx² + 4dy/dx - 12y = cos2x .

Answer 1

Answer on Question #70742 – Math – Differential Equations

Question

1. a) Solve the following ordinary differential equations:

i) $\frac{dy}{dx} + 4xy = x$

ii) $\frac{d^2y}{dx^2} + \frac{4dy}{dx} - 12y = \cos 2x$.

Solution

i)

$$\begin{array}{l} y' + 4xy = x \\ y' = x(1 - 4y) \\ \frac{dy}{1 - 4y} = xdx \\ \int \frac{dy}{1 - 4y} = \int xdx \\ -\frac{1}{4} \ln(1 - 4y) = \frac{x^2}{2} + C_1 \\ \ln(1 - 4y) = -2x^2 - 4C_1 \\ 1 - 4y = C_2 e^{-2x^2} \\ y(x) = C e^{-2x^2} + \frac{1}{4'} \end{array}$$

where $C_1, C_2, C$ are real constants.

ii)

$$y'' + 4y' - 12y = \cos 2x$$

First, solve homogeneous equation:

$$y'' + 4y' - 12y = 0$$ $$\lambda^2 + 4\lambda - 12 = 0 \rightarrow \lambda_1 = -6, \lambda_1 = 2$$

Thus, the general solution of the homogeneous equation is

$$y_{homogeneous} = C_1 e^{-6x} + C_2 e^{2x},$$

where $C_1, C_2$ are arbitrary real constants.

Now, let's find any solution for non-homogeneous equation in the form

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

If we substitute it in equation, we shall get

$$-4A \cos 2x - 4B \sin 2x - 8A \sin 2x + 8B \cos 2x - 12A \cos 2x - 12B \sin 2x = \cos 2x$$

As $\cos 2x$ and $\sin 2x$ are linearly independent,

$$\begin{array}{l} -4A + 8B - 12A = 1 \\ -4B - 8A - 12B = 0 \\ \end{array}$$

Solving this system

$$A = -\frac{1}{20}, B = \frac{1}{40}$$

Thus, the general solution of the non-homogeneous equation is

$$y(x) = y_{homogeneous} + \tilde{y} = C_1 e^{-6x} + C_2 e^{2x} - \frac{1}{20} \cos 2x + \frac{1}{40} \sin 2x$$

Answer: i) $y(x) = C e^{-2x^2} + \frac{1}{4}$; ii) $y(x) = C_1 e^{-6x} + C_2 e^{2x} - \frac{1}{20} \cos 2x + \frac{1}{40} \sin 2x$.

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