Question

Solve the following ordinary differential equations: i) dy/dx + 4xy = x, (2) d^2y/dx^2 + 4 dy/dx - 12 y =cos 2x

Answer 1

Answer on Question #70069-Math-Differential Equations

Solve the following ordinary differential equations:

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

Solution

$$\frac{dy}{dx} = x - 4xy = x(1 - 4y)$$ $$\frac{dy}{1 - 4y} = x\,dx$$

Integrating both sides,

$$\left(\frac{1}{4}\right)\ln(1 - 4y) = -\left(\frac{x^2}{2}\right) + c$$ $$\ln(1 - 4y) = -2x^2 + 4c$$ $$C = 4c$$ $$y = \frac{1 + e^{-2x^2 + C}}{4}$$

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

Solution

1) $\frac{d^2y}{dx^2} + \frac{4dy}{dx} - 12y = 0$

$$y = e^{\lambda x}$$ $$\lambda^2 + 4\lambda - 12 = 0$$ $$\lambda_1 = -6, \lambda_2 = 2.$$ $$y = c_1 e^{-6x} + c_2 e^{2x}.$$

2)

$$y^* = A \cos 2x + B \sin 2x$$ $$y^{*'} = -2A \sin 2x + 2B \cos 2x$$ $$y^{*''} = -4A \cos 2x - 4B \sin 2x$$

So,

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

We have

$$-4A + 8B - 12A = 1$$ $$-4B - 8A - 12B = 0$$ $$8A = -16B$$ $$A = -2B.$$ $$-4(-2B) + 8B - 12(-2B) = 1$$ $$8B + 8B + 24B = 1$$ $$B = \frac{1}{40}$$ $$A = -2\frac{1}{40} = -\frac{1}{20}$$

The solution is

$$y = c_1 e^{-6x} + c_2 e^{2x} - \frac{1}{20} \cos 2x + \frac{1}{40} \sin 2x.$$

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