1. Basic Concepts and Definitions:
An ordinary differential equation (or ODE) is an equation that involves an unknown function of one variable and some of its derivatives.
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Order of ODEs: The order of the highest order derivative appearing in the equation is called the order of the ODE. For example: a) $$y''(x) + 2y'(x) = e^x$$ (2nd order) b) $$y''(x) + 2x^4y'''(x) + y^5\cos x = 3$$ (3rd order) c) $$y^{(n)} + 2y^{(n-1)} + \dots + 5y''(x) + xy = \ln x$$ (nth order)
An nth order ODE can be written in the form $$F(x, y, y', y'', \dots, y^{(n)}) = 0$$, where the unknown function is denoted by $$y$$ and its independent variable by $$x$$.
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Degree of ODEs: The degree of the highest order derivative is called the degree of the equation. For example:
- $$xy' + y = \sin x$$ is an ODE of order 1 and degree 1.
- $$y''^3 + xy'^5 - e^xy = 2$$ is an ODE of order 2 and degree 3.
The general form of a first order and first degree equation is $$F(x, y, y') = 0$$.
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Solution of ODE, General Solution and Particular Solution: A function $$y(x)$$ is a solution of an ODE if, when substituted into the equation, it reduces the equation to an identity. The general solution of an ODE is a solution that contains arbitrary constants. A particular solution is a solution that does not contain any arbitrary constants.
- The general solution of the equation $$y' = 2x$$ is $$y = x^2 + C$$, where $$C$$ is an arbitrary constant.
- A particular solution of the equation $$y'' + 4y = 0$$ is $$y = 2\cos 2x + 5\sin 2x$$.
2. Initial-Value Problem (I.V.P):
An initial-value problem is a problem of solving an nth order ODE subject to n conditions of the form $$y(x_0) = y_0, y'(x_0) = y'_0, \dots, y^{(n-1)}(x_0) = y^{(n-1)}_0$$. The side conditions are called initial conditions.
Example (1):
Verify that $$y = Cx$$ is the general solution of the equation $$xy' = y$$, and hence find the solution for which $$y(1) = 3$$. Sketch the graphs of the considered solutions.
Solution We have $$y = Cx$$ and $$y' = C$$. Substituting into the given ODE, we get $$x \cdot C = Cx \Rightarrow Cx \equiv Cx$$.
Hence $$y = Cx$$ satisfies the given ODE. Since $$y = Cx$$ contains one arbitrary constant, it is the general solution of the given ODE. Substituting $$x = 1$$ and $$y = 3$$ in $$y = Cx$$ gives $$3 = C \cdot 1 \Rightarrow C = 3$$.
Thus, the function $$y = 3x$$ is a solution of the initial-value problem $$xy' = y, y(1) = 3$$.
Note that $$y = Cx$$ represents a family of straight lines passing through the origin, while $$y = 3x$$ is the line passing through the point $$(1,3)$$.
Example (2):
Find a solution of the ODE $$y'' + 4y = 0$$ that satisfies the initial conditions $$y(0) = 2$$ and $$y'(0) = -4$$.
Solution The general solution of $$y'' + 4y = 0$$ is $$y = C_1 \cos(2x) + C_2 \sin(2x)$$, where $$C_1$$ and $$C_2$$ are arbitrary constants. We have $$y' = -2C_1 \sin(2x) + 2C_2 \cos(2x)$$.
Example (2) (continued):
Substituting the initial conditions $$y(0) = 2$$ and $$y'(0) = -4$$ into the general solution, we get the following system of equations:
$$ \begin{align*} C_1 \cos(0) + C_2 \sin(0) &= 2 \ -2C_1 \sin(0) + 2C_2 \cos(0) &= -4 \ \end{align*} $$
Simplifying, we have:
$$ \begin{align*} C_1 &= 2 \ 2C_2 &= -4 \ \end{align*} $$
Solving for $$C_1$$ and $$C_2$$, we find $$C_1 = 2$$ and $$C_2 = -2$$.
Therefore, the solution of the initial-value problem $$y'' + 4y = 0, y(0) = 2, y'(0) = -4$$ is:
$$y = 2\cos(2x) - 2\sin(2x)$$
Example (3):
Solve the initial-value problem $$y' + y = 3e^{-x}, y(0) = 1$$.
Solution The given ODE is a first-order linear ODE. The integrating factor is given by $$\mu(x) = e^{\int 1 dx} = e^x$$.
Multiplying both sides of the equation by the integrating factor, we have:
$$e^xy' + e^xy = 3$$
Applying the product rule on the left-hand side, we get:
$$(e^xy)' = 3$$
Integrating both sides with respect to $$x$$, we have:
$$e^xy = 3x + C$$
Solving for $$y$$, we get:
$$y = 3xe^{-x} + Ce^{-x}$$
Substituting the initial condition $$y(0) = 1$$, we have:
$$1 = 3(0)e^{-(0)} + Ce^{-(0)}$$
Simplifying, we find $$C = 1$$.
Therefore, the solution of the initial-value problem $$y' + y = 3e^{-x}, y(0) = 1$$ is:
$$y = 3xe^{-x} + e^{-x}$$
Example (4):
Solve the initial-value problem $$y'' - 4y' + 4y = 0, y(0) = 1, y'(0) = 0$$.
Solution The given ODE is a second-order linear ODE with constant coefficients. The characteristic equation is:
$$r^2 - 4r + 4 = 0$$
Factoring, we have:
$$(r - 2)^2 = 0$$
This gives us a repeated root $$r = 2$$.
Therefore, the complementary function is:
$$y_c(x) = (C_1 + C_2x)e^{2x}$$
To find the particular solution, we assume a solution of the form $$y_p(x) = Ax^2e^{2x}$$, where $$A$$ is a constant to be determined.
Differentiating twice, we have:
$$y_p'(x) = 2Axe^{2x} + 2Ax^2e^{2x}$$
$$y_p''(x) = 4Ae^{2x} + 4Axe^{2x} + 4Ax^2e^{2x}$$
Substituting into the ODE, we get:
$$(4Ae^{2x} + 4Axe^{2x} + 4Ax^2e^{2x}) - 4(2Axe^{2x} + 2Ax^2e^{2x}) + 4(Ax^2e^{2x}) = 0$$
Simplifying, we have:
$$4Ae^{2x} = 0$$
Since $$e^{2x}$$ is never zero, we must have $$A = 0$$.
Therefore, the particular solution is $$y_p(x) = 0$$.
The general solution of the ODE is the sum of the complementary function and the particular solution:
$$y(x) = y_c(x) + y_p(x) = (C_1 + C_2x)e^{2x}$$
Substituting the initial conditions $$y(0) = 1$$ and $$y'(0) = 0$$, we have:
$$C_1 = 1$$
$$C_2 + 2C_1 = 0$$
Solving for $$C_2$$, we find $$C_2 = -2$$.
Therefore, the solution of the initial-value problem $$y'' - 4y' + 4y = 0, y(0) = 1, y'(0) = 0$$ is:
$$y(x) = (1 - 2x)e^{2x}$$
Part 1:
The Particular Integral of the Nonhomogeneous Equations
I. The method of variation of parameters:
Consider the ODE of the second order $$L(D)y = (a_0D^2 + a_1D + a_2)y = f(x)$$
or $$a_0y'' + a_1y' + a_2y = f(x)$$
(16) Let $$y_c.f. = c_1 y_1 + c_2 y_2$$, where $$c_1, c_2$$ are constants, we shall seek the particular integral of the nonhomogeneous equation in the form: $$y_p.i. = c_1(x) y_1 + c_2(x) y_2$$
where $$c_1(x), c_2(x)$$ are functions of $$x$$ to be determined. We have $$y'_{p.i.} = c_1' y_1 + c_1 y_1' + c_2' y_2 + c_2 y_2'$$
We choose $$c_1, c_2$$ to satisfy the condition $$c_1' y_1 + c_2' y_2 = 0$$
Thus, $$y'{p.i.} = c_1 y_1' + c_2 y_2'$$, and $$y''{p.i.} = c_1' y_1' + c_1 y_1'' + c_2' y_2' + c_2 y_2''$$, Substituting in (16) we get $$c_1(a_0y_1'' + a_1y_1' + a_2y_1) + c_2(a_0y_2'' + a_1y_2' + a_2y_2) + a_0(c_1' y_1' + c_2' y_2') = f(x)$$
Since $$y_1, y_2$$ are solutions of the homogeneous equation, $$a_0y_1'' + a_1y_1' + a_2y_1 = 0, a_0y_2'' + a_1y_2' + a_2y_2 = 0$$
Then $$c_1' y_1' + c_2' y_2' = f(x) a_0$$ (17)
Equations (17), (18) give unique values for $$c_1', c_2'$$. Then by integration we may find $$c_1, c_2$$.
Example (31):
Find the general solution of $$y'' + y = \cos x$$
Solution
The complementary function is $$y_c.f. = c_1 \cos x + c_2 \sin x$$ Thus, $$y_1 = \cos x$$, $$y_2 = \sin x$$. We seek the particular integral of the given nonhomogeneous equation in the form $$y_p.i. = c_1(x) \cos x + c_2(x) \sin x$$
The system of equations (17), (18) of this case are $$c_1' \cos x + c_2' \sin x = 0$$
$$-c_1' \sin x + c_2' \cos x = \cos x$$
Solving for $$c_1', c_2'$$ we find that $$c_1' = -1$$, $$c_2' = \cos x$$
Integrating gives $$c_1 = -x$$, $$c_2 = \ln(\sin x)$$
Thus, $$y_p.i. = -x \cos x + \ln(\sin x) \sin x$$
Finally, $$y_g.s. = y_c.f. + y_p.i.$$
$$= (c_1 - x) \cos x + (c_2 + \ln(\sin x))\sin x$$
Generally, we seek a particular integral of the nonhomogeneous equation of order $$n$$ in the form: $$y_p.i. = c_1(x) y_1 + c_2(x) y_2 + \ldots + c_n(x) y_n$$
where, $$y_c.f. = c_1 y_1 + c_2 y_2 + \ldots + c_n y_n$$
The unknown function $$c_1(x), \ldots , c_n(x)$$ satisfy the system of equations: $$c_1' y_1 + c_2' y_2 + \ldots + c_n' y_n = 0$$
$$c_1' y_1' + c_2' y_2' + \ldots + c_n' y_n' = 0$$
$$\ldots$$
$$c_1' y_1^{(n-1)} + c_2' y_2^{(n-1)} + \ldots + c_n' y_n^{(n-1)} = f(x) a_0$$
