# MST210/Unit 1/Book A/Initial Conditions and Boundary Conditions

Much like for first-order differential equations, we are often interested in particular solutions to differential equations - solutions that don’t contain arbitrary constants and satisfy some given conditions. For second-order differential equations, there are two classes of conditions: initial conditions and boundary conditions.

# Initial-value Problems

To specify initial conditions for a second-order differential, we are given $$y = y_0$$ and $$\frac{dy}{dx} = z_0$$ when $$x = x_0$$. These numbers are known as initial-values. Combined with a second-order differential equation, they make up an initial value problem.

To make use of $$z_0$$, it is necessary to differentiate the general solution.

Exercise 37
1. In Exercise 20b, we saw that the general solution was

$u = C \cos 3t + D \sin 3t$ where $$C$$ and $$D$$ are arbitrary constants.

Differentiating, we have

$u' = 3D \cos 3t - 3C \sin 3t$

From the initial conditions, we have

$\begin{cases} C \cos (3 \frac{\pi}{2}) + D \sin (3 \frac{\pi}{2}) = -D = 0 \\ 3D \cos (3 \frac{\pi}{2}) - 3 C \sin (3 \frac{\pi}{2}) = 3C = 1 \end{cases}$

Therefore $$C = \frac{1}{3}, D = 0$$, and so the particular solution is

$u(t) = \frac{1}{3} \cos 3t$

2. In Exercise 32a, we saw that the general solution was

$y = Ce^x + De^{2x} - 4xe^x$ where $$C$$ and $$D$$ are arbitrary constants.

Differentiating, we have

$y' = Ce^x + 2De^{2x} - 4(1 + x)e^x$

In order to satisfy the initial conditions, we must have

$\begin{cases} Ce^0 + De^{2\cdot0} - 4\cdot0\cdot e^0 = C + D = 4 \\ Ce^0 + 2De^{2\cdot 0} - 4(1 + 0)e^0 = C + 2D - 4 = 2 \end{cases}$

That is,

$\begin{cases} C + D = 4 \\ C + 2D = 6 \end{cases}$

Hence $$C = 2, D = 2$$, and so the particular solution is

$y = 2e^x + 2e^{2x} - 4xe^x = (2 - 4x)e^x + 2e^{2x}$

3. Using information from Exercise 21d and Exercise 35a, we see that the general solution is

$y = (C + Dx) e^x + 2 x^2 e^x - 3e^{2x}$

Differentiating, we have

$\begin{array}{ll} y' &= Ce^x + D(1 + x) e^x + 2x(x+2)e^x - 6e^{2x} \\ &= (2x^2 + (D + 4)x + (C + D))e^x - 6e^{2x} \end{array}$

In order to satisfy the initial conditions, we must have

$\begin{cases} (C + D \cdot 0) e^0 + 2 \cdot 0^2 e^0 - 3e^{2\cdot0} = C - 3 = 4 \\ (2\cdot0^2 + (D + 4)0 + (C + D))e^0 - 6e^{2\cdot0} = C + D - 6 = -1 \end{cases}$

That is

$\begin{cases} C = 7 \\ C + D = 5 \end{cases}$

Thus, $$C = 7, D = -2$$, and therefore the particular solution is

$\begin{array}{ll} y &= (7 - 2x) e^x + 2 x^2 e^x - 3e^{2x} \\ &= (2x^2 - 2x + 7)e^x - 3e^{2x} \end{array}$

# Boundary-Value Problems

Boundary conditions associated with a second-order differential equation with dependent variable $$y$$ and independent variable $$x$$ specify that $$y$$ or $$\frac{dy}{dx}$$ (or some combination) takes values $$y_0$$ and $$y_1$$ at two different values $$x_0$$ and $$x_1$$. These numbers are referred to as boundary values. The combination of a second-order differential equation and boundary conditions is known as a boundary-value problem.

Exercise 38

In Exercise 32a, we saw that the general solution was

$y = Ce^x + De^{2x} -4xe^x$ where $$C$$ and $$D$$ are arbitrary constants.

Differentiating, we have

$y' = Ce^x + 2De^{2x} - 4(1 + x)e^x$

In order to satisfy the boundary conditions, we must have

$\begin{cases} Ce + De^2 -4e = 0 \\ C + 2D - 4 = 2 \end{cases}$

That is

$\begin{cases} C + De = 4 \\ C + 2D = 6 \end{cases}$

Hence $$C = \frac{8 - 6e}{2 - e}, D = \frac{2}{2 - e}$$, and the required particular solution is

$y = \frac{8 - 6e}{2 - e}e^x + \frac{2}{2 - e}e^{2x} - 4xe^x$

Exercise 39

In Exercise 31, we saw that the general solution was

$y = Ce^{-x} + De^{x} -\frac{1}{2} x ^ 2 + \frac{l}{2} x - 1$

Setting $$l = 2$$, we have

$y = Ce^{-x} + De^{x} -\frac{1}{2} x ^ 2 + x - 1$

In order to satisfy the initial conditions, we must have

$\begin{cases} C + D = 1 \\ Ce^{-2} + De^{2} = 1 \end{cases}$

As shown in the solutions, this yields $$C = \frac{1}{e^2 + 1}, D = \frac{e^2}{e^2 + 1}$$, and thus the required particular solution is

$\begin{array}{ll} y &= \frac{1}{e^2 + 1} e^{x} + \frac{e^2}{e^2 + 1} e^{-x} - \frac{1}{2} x ^ 2 + x - 1 \\ &= \frac{1}{e^2 + 1} (e^{x} + e^{2-x}) - \frac{1}{2} x ^ 2 + x - 1 \end{array}$