Definition

An equation that involves an unknown function and its derivatives

First Order DEs

Seperable DEs

Can be written in the form f(y)dy = g(x)dx, with the y on the one side and the x on the other.

The DE can be solved by integrating on both sides \int f(y)dy = \int g(x)dx, and solving for y taking note of any initial conditions.

Linear DEs

Can be written in the form y' + P(x)y = Q(x), with all the x terms not involving y on the one side.

Find the integrating factor: I(x) = e^{\int P(x)dx}

By multiplying both sides of the equation by the integrating factor we get I(x)y' + I(x)P(x)y = I(x)Q(x), which implies (by the product rule) (I(x)y)' = I(x)Q(x), and y = \frac{1}{I(x)}\int I(x)Q(x)dx.

Exact Equations

Differential

If z = f(x,y) has continuous first partial derivatives, then its differential is dz = \frac{\partial f}{\partial x} dx + \frac{\partial f}{\partial y} dy.

In this section, we are interested in DEs involving the differential of f(x,y) = c. That is equations of the form, \frac{\partial f}{\partial x} dx + \frac{\partial f}{\partial y} dy = 0.

Exact Equations

An equation of the form M(x,y)dx + N(x,y)dy = 0 is called an exact equations if M(x,y)dx + N(x,y)dy is the differential of some function f(x,y). That is, if \partial f/\partial x = M(x,y) \partial f/\partial y = N(x,y)

Theorem

M(x,y)dx + N(x,y)dy (M, N continuous) is an exact differential iff \frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}

Method of Solution

Suppose you are asked to solve a DE of the form M(x,y)dx + N(x,y)dy = 0.

If \frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}, then you know that this is an exact eqution, and so \partial f/\partial x = M(x,y) \quad \partial f/\partial y = N(x,y).

Integrating M(x,y) wrt x gives f(x,y) = \int M(x,y) dx+ g(y).

Differentiating wrt y gives \frac{f(x,y)}{\partial y} = \frac{\partial}{\partial y}\int M(x,y) dx+ g'(y) = N(x,y)

Solve for g'(y) in the above equation, integrate it to get g(y). After substituting, we get the solution which is of the form f(x,y) = c.

Second Order DEs

Linear Homogenous DEs

ay'' + by' + cy = 0

By supposing that all solutions are of the form y(x) = e^{rx}, we can get the characteristic equation ar^2 + br + c = 0. This equation has solutions r_{1,2} = \frac{-b\pm \sqrt{b^2 - 4ac}}{2a} = , which imply solutions to the DE of the form y=c_1 e^{r_1 x} + c_2 e^{r_2 x}.

There are three classes of solutions to the quadratic, with corresponding DE solutions:

• distinct real roots: if r_1, r_2 are real, they can be written as \alpha \pm \beta y=c_1 e^{r_1 x} + c_2 e^{r_2 x} = e^{\alpha x} (c_3 \cosh\beta x + c_4 \sinh\beta x)
• complex roots: if r_1, r_2 are complex, they can be written as \alpha \pm i\beta y=c_1 e^{r_1 x} + c_2 e^{r_2 x} = e^{\alpha x} (c_1 \cos\beta x + c_2 \sin\beta x)
• repeated real roots y=c_1 e^{r_1 x} + c_2 xe^{r_2 x}.

Solutions by Substitution

Given a relatively complicated DE, we try substitute different functions, to try convert it to a simpler, recognizable form.

Homogeneous Equations

Bernoulli’s Equation

Linear Substitution

Higher order DEs

Theory of Linear Equations

Initial-Value Problem (IVP)

For a linear DE, an nth order IVP asks to solve g(x) = a_{n}(x)y^{(n)} + a_{n-1}(x)y^{(n-1)} + \ldots + a_{1}(x)y' + a_{0}(x)y subject to the intial values y(x_0) = y_0, y'(x_0) = y_1, \ldots, y^{(n-1)}(x_0) = y_{n-1}.

Theorem (existence and uniqueness)

Let a_n(x), a_{n-1}(x), \ldots, a_1(x), a_0(x), and g(x) be continuous on an interval, and let a_n(x) \neq 0 for every x in this interval. If x = x_0 is any point in this interval, then a solution y(x) of the IVP exists on the interval and is unique.

Boundary-Value Problem (BVP)

For a linear DE, an nth order (n\geq 2) BVP asks to solve g(x) = a_{n}(x)y^{(n)} + a_{n-1}(x)y^{(n-1)} + \ldots + a_{1}(x)y' + a_{0}(x)y subject to the boundary conditions y(x_0) = y_0, y(x_1) = y_1

A BVP can have many solutions, a unique solution, or no solutions.

Homogeneous Equations

Consider a linear nth-order DE of the form g(x) = a_{n}(x)y^{(n)} + a_{n-1}(x)y^{(n-1)} + \ldots + a_{1}(x)y' + a_{0}(x)y. This equation is homogenous if g(x)=0, else it is said to be nonhomogeneous.

To solve a nonhomogenous DE, it must be possible to solve the associated homogenous equation (i.e. where the g(x) is set to 0).

Superposition Principle

If y_1, y_2, \ldots, y_k are solutions of a homogenous nth-order DE, then any linear combination of the solutions is also a solution.

• Linear Dependence/Independence
• Wronskian
• Fundamental Set of Solutions
• Existence of a Fundamental Set
• General Solution-Homogeneous Equations
• General Solution-Nonhomogeneous Equations
• Complementary Function
• Superposition Principle—Nonhomogeneous Equations

Reduction or Order

In this section, we look at homogenous linear 2nd-order DEs of the form a_2(x)y'' + a_1(x)y' + a_0(x)y = 0. This DE has the general solution y = c_1 y_1 + c_2 y_2, where y_1, y_2 are linearly independent.

Suppose now that we know y_1 is a solution but we need y_2. Now, there must exist some function u(x) such that y_2 = u(x)y_1. We substitute the RHS into the DE, and then solve for u(x).

• General case

Homogeneous Linear Equations with Constant Coefficients

Appendix

Linear Combination

The variable y is a linear combination of x_1, x_2, \ldots, x_n if y = c_1 x_1 + c_2 x_2 + \ldots + c_n x_n, where the c_i are arbitrary constants.

Power Series Solution method

Say you want to solve an ODE for f(x). Substitutes the Taylor expansion f(x) around x into the equation and solve for the Taylor coefficients.

This method can be used to solve Legendre’s ODE: (1-x^2) \frac{d^2y}{dx^2} - 2x \frac{dy}{dx} + k(k+1)y = 0.

The polynomials obtained from solving Legendre’s ODE for different integer values of k are the Legendre polynomials.

Rodrigues formula is a quicker way to calculate Legendre polynomials: P_k(x) = \frac{1}{2^kk!} \frac{d^k}{dx^k}(x^2 - 1)^k. It is derived from the recursion relation arising from the power series solution method.

Frobenius method