Question

1. a) Solve the following ordinary differential equations:

i) 1/x sin ydx + [(ln x)(cos y) + y)] dy = 0

ii) d^2 y / dx^2 + dy/dx - 12y = 4e^2x

b) Solve the initial value problem:

d^2 y / dx^2 + 2(dy/dx) + 2y=0, y(0)=2, y'(0)=1

Answer 1

Answer on Question #58204 – Math – Differential Equations

Question

1. a) Solve the following ordinary differential equation:

i) $1 / x \sin y \, dx + [(\ln x)(\cos y) + y)] \, dy = 0$

Solution

$$\frac{\sin y}{x} \, dx + (\ln x \cos y + y) \, dy = 0$$

Let us split the left-hand side of equation into two parts

$$\left(\frac{\sin y}{x} \, dx + \ln x \cos y \, dy\right) + y \, dy = 0$$

In the first part

$$\frac{\sin y}{x} \, dx = \sin y \, d(\ln x)$$

and

$$\ln x \cos y \, dy = \ln x \, d(\sin y)$$ $$\begin{array}{l} \left(\frac{\sin y}{x} \, dx + \ln x \cos y \, dy\right) + y \, dy = (\sin y \cdot d(\ln x) + \ln x \cdot d(\sin y)) + y \, dy \\ = d(\sin y \cdot \ln x) + y \, dy = 0 \end{array}$$

We integrate the last expression and get

$$\ln x \cdot \sin y + \frac{y^2}{2} = C,$$

where $C$ is an arbitrary real constant.

Answer: $\ln x \cdot \sin y + \frac{y^2}{2} = C$.

Question

1. a) Solve the following ordinary differential equation:

ii) $\mathrm{d}^2 y / \mathrm{d}x^2 + \mathrm{d}y / \mathrm{d}x - 12y = 4e^2x$

Solution

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

Let us find a solution in the form of

$$y = e^{\mu x}$$

First we solve the homogeneous equation

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

which has the characteristic equation

$$\mu^2 + \mu - 12 = 0$$

Applying Vieta's formula we get $\mu = 3$ or $\mu = -4$.

Solution to the homogeneous equation is $y_h = Ce^{3x} + De^{-4x}$.

Now, find a particular solution to non-homogeneous equation in the form of

$$y_p = \alpha e^{2x}$$ $$\frac{d^2 y_p}{dx^2} + \frac{dy_p}{dx} - 12y_p = 4e^{2x}$$ $$\frac {d ^ {2}}{d x ^ {2}} (\alpha e ^ {2 x}) + \frac {d}{d x} (\alpha e ^ {2 x}) - 1 2 (\alpha e ^ {2 x}) = 4 e ^ {2 x}$$ $$4 \alpha e ^ {2 x} + 2 \alpha e ^ {2 x} - 1 2 e ^ {2 x} = 4 e ^ {2 x}$$ $$4 \alpha + 2 \alpha - 1 2 = 4 \rightarrow 6 \alpha = 1 6 \rightarrow \alpha = \frac {1 6}{6} = \frac {8}{3},$$

hence

$$y _ {p} = \frac {8}{3} e ^ {2 x}$$

and the general solution is

$$y = y _ {h} + y _ {p} = C e ^ {3 x} + D e ^ {- 4 x} + \frac {8}{3} e ^ {2 x}$$

Answer: $y = Ce^{3x} + De^{-4x} + \frac{8}{3}e^{2x}$ .

Question

b) Solve the initial value problem:

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

Solution

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

Let us find a solution in the form of

$$y = e ^ {\mu x}$$

An homogeneous equation

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

has a characteristic equation

$$\mu^ {2} + 2 \mu + 2 = 0$$

Applying Vieta's formula we get $\mu = -1 + i$ or $\mu = -1 - i$ .

Solution to homogeneous equation is $y = Ce^{-x}\sin x + De^{-x}\cos x$ To meet the initial conditions

$$y (0) = 2,$$ $$y ^ {\prime} (0) = 1,$$

consider

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

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

www.AssignmentExpert.com