Differential Equation

Differential Equation - Fanyu Zhao

Key Facts

1. Properties

Shorts:

D.E. Differential Equation
O.D.E. Ordinary Differential Equation
G.S. General Solution
I.V.P. Initial Value Probelms
B.V.P. Boundary Value Problems

1.1 Order

Order $n$ below.

y^{(n)} = \frac{d^{n} y}{d x^{n}}

1.2 Degree

The Power of the derivative is called Degree.

(y^{(n)})^{r} = (\frac{d^{n} y}{d x^{n}})^{r}

$y^{(0)} - zero^{th}\ derivative\ of\ y$ $y$ itself.

1.3 Linear - ODE

$y$ $r=1$ .

1.4 IVP & BVP

By solving the D.E., we normally get the general solution, G.S, the general solution is with constant term,C. To specify the solution, we need to plug in specific value into D.E., and calculate "C".

Initial Value Problems

In this case, the specific values are given at a certain point of "x". For example,

y (x_{0}) = 1, y^{'} (x_{0}) = 2, y^{(n)} = 10

$y$ $x=x_0$ .

Boundary Value Problems

In this case, values are given at different points of "x". For example,

y (0) = 1, y (1) = 3, y (10) = 4

2. First Order Ordinary Differential Equations ODE

General form:

y^{'} = f (x, y)

2.1 Missing One Variable

Case 1. y is missing

\begin{matrix} y^{'} = f (x) \\ \Leftrightarrow \frac{d y}{d x} = f (x) \\ d y = f (x) d x \\ \int d y = \int f (x) d x \\ s o, y = \int f (x) d x + C \end{matrix}

Case 2. x is missing

\begin{matrix} y^{'} = f (y) \\ \Leftrightarrow \frac{d y}{d x} = f (y) \\ \frac{1}{f (y)} d y = d x \Leftrightarrow \int \frac{1}{f (y)} d y = \int d x \\ s o, x = \int \frac{1}{f (y)} d y + c \end{matrix}

2.2 Variable Separable

y^{'} = g (x) \cdot h (y)

"x" and "y" are separable, we can follow the same steps .

\begin{matrix} \frac{d y}{d x} = g (x) \cdot h (y) \\ \frac{1}{h (y)} d y = g (x) d x \\ \int \frac{1}{h (y)} d y = \int g (x) d x \\ \int \frac{1}{h (y)} d y = \int g (x) d x + C \end{matrix}

Rationale: If "x" and "y" are separable, we move all "y" and "dy" into one side, and all "x" and "dx" into the other side.

2.3 Linear Equations

y^{'} + P (x) \cdot y = Q (x)

Integrating Factor $R(x)$ .

\underset{= \frac{d}{d x} (R (x) y)}{\underset{⏟}{R (x) \cdot [y^{'} + P (x) \cdot y]}} = R (x) \cdot Q (x)

I.F. $R(x)$ needs to satisfy,

\begin{matrix} \frac{d}{d x} (R (x) y) = R^{'} (x) y + R (x) y^{'} \\ = R (x) P (x) y + R (x) y^{'} \end{matrix}

So, we need

R^{'} (x) = R (x) P (x)

Therefore, our problem becomes,

\begin{matrix} \frac{d}{d x} (R (x) y) = R (x) Q (x) \\ \int \frac{d}{d x} (R (x) y) d x = \int R (x) Q (x) d x \\ R (x) y = \int R (x) Q (x) d x + C \end{matrix}

The satisfied I.F., is,

R (x) = e^{\int P (x) d x} = e x p (\int P (x) d x)

$R(x)=exp(\int P(x)\ dx)$ From both side of the equation, we could then easily solve the D.E., and the answer is the following,

\begin{matrix} R (x) y = \int R (x) Q (x) d x + C \\ y = \frac{\int R (x) Q (x) d x}{R (x)} + \frac{C}{R (x)} \end{matrix}

3 Second Order ODE

General form:

y^{″} = f (x, y, y^{'})

P.S. we may find later that the number of arbitrary constant "C" in the solution is equal to the highest order of derivatives.

3.1 Simplest Cases

$y', y$ are missing

y^{″} = f (x)

Easiest one, integrate it twice.

Case 2. Only "y" is missing

y^{″} = f (x, y^{'})

All "y"s in this equation are in derivatives of none zero.

$P=y'$ $\frac{dP}{dx}=y''$ .

P^{'} = f (x, P)

$P$ later to get the G.S. for this D.E.

$y'$ $x$ are missing

y^{″} = f (y)

$p=y'$ .

y^{″} = \frac{d^{2} y}{d x^{2}} = \frac{d p}{d x} = \frac{d p}{d y} \frac{d y}{d x} = \frac{d p}{d y} \cdot p = f (y)

Case 4. "x" is missing

y^{″} = f (y^{'}, y)

$p=y'$ ,

\frac{d^{2} y}{d x^{2}} = P \frac{d P}{d y} = f (P, y)

The problem would then become the first order O.D.E.

4. Linear ODE

General form,

a_{n} (x) \cdot y^{(n)} + a_{n - 1} (x) \cdot y^{(n - 1)} + . . . + a_{1} (x) y^{'} + a_{0} (x) y = g (x)

We then transform the equation,

L - Linear Differential Operator of Order "n"

We apply the notation:

D \equiv \frac{d}{d x}, D^{r} \equiv \frac{d^{r}}{d x^{r}}, s o D^{r} y \equiv \frac{d^{r} y}{d x^{r}}

\begin{matrix} L = a_{n} D^{n} + a_{n - 1} D^{n - 1} + . . . + a_{1} D + a_{0} \\ L y = g \end{matrix}

$Ly=g$ is given by

y = y_{c} + y_{p}

$y_c$ $y_p$ - Particular Integral (or Particular Solution).

\begin{matrix} \begin{array}{r} y_{c} \leftarrow L y = 0 \\ y_{p} \leftarrow L y = g \end{array}} \Rightarrow G . S . y = y_{c} + y_{p} \end{matrix}

$\exists \vec{\lambda}\quad s.t.\quad \lambda_1 y_1+\lambda_2 y_2+...+\lambda_n y_n=0$ , or in other words, variables could be represented by the linear combination of others.
$\lambda_1 y_1+\lambda_2 y_2+...+\lambda_n y_n=0$ $\lambda_1=\lambda_2=...=\lambda_n=0$ .

4.1 Homogenous - Auxiliary Equation A.E.

L y = 0 = g

$g(x)=0, \forall x, L\ y=0$ is said to be homogeneous.

\begin{matrix} H o m o g e n e i t y : {\begin{array}{r} L (y_{1} + y_{2}) = L (y_{1}) + L (y_{2}) \\ y_{p} \leftarrow L (c \cdot y) = c \cdot L (y) \end{array} \end{matrix}

For example, for a second order D.E.

L y = a \frac{d^{2} y}{d x^{2}} + b \frac{d y}{d x} + c y = 0

$y=e^{\lambda x}$ , and plug it back to our D.E.

L \cdot e^{λ x} = (a D^{2} + b D + c) \cdot e^{λ x}

Hence,

λ : a λ^{2} + b λ + c = 0

$\lambda s$ $\lambda=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$

$b^2-4ac$ would affects the form of solutions.

$b^2-4ac>0$ $y=C_1e^{\lambda_1 x}+C_2 e^{\lambda_2 x}$
$b^2-4ac=0$ $\lambda_1=\lambda_2=\lambda, \quad y=C_1e^{\lambda x}+C_2 xe^{\lambda x}$
$b^2-4ac<0$ $\lambda \in \mathbb{C}, \lambda=p\pm iq$ $p=-\frac{b}{2a},\quad q=\frac{1}{2a}\sqrt{|b^2-4ac|}$ , and the G.E. is
$y = C_{1} e^{λ_{1} x} + C_{2} e^{λ_{2} x} = C_{1} e^{(p + i q) x} + C_{2} e^{(p - i q) x}$
$e^{\pm i\theta}=cos\ \theta\pm i\cdot sin \ \theta$ .
$y = e^{p x} (A c o s q x + B s i n q x)$

Therefore, in a "n" order case, we would apply the similar reasoning.

Case 1: "n" distinct roots, then G.E. is
$y = β_{1} e^{λ_{1} x} + β_{2} e^{λ_{2} x} + . . . + β_{n} e^{λ_{n} x}$

Case 2: r - fold roots, then G.E. is

y = e^{λ x} (a_{1} + a_{2} x + a_{3} x^{2} + . . . + a_{r} x^{r - 1})

Case 3: Complex Number.

For example,

\begin{matrix} λ^{6} - 5 λ^{4} = 0 \\ λ^{4} (λ^{2} - 5) = 0 \end{matrix}

$\pm \sqrt{5}$ For the second part. The G.S. is then,

y = A e^{\sqrt{5} x} + B e^{- \sqrt{5} x} + (C + D x + E x^{2} + F x^{3}) e^{0 x}

4.2 Non-homogeneous Case

G.S. is

y = y_{c} + y_{p}

$g(x)=0, so\ Ly=0$ .

The second term of RHS is the specific solution of our D.E. The way of calculating it is called "Guesswork".

There are three cases we may encounter:

$g(x)=p_0+p_1x+...+p_m x^m$
$g(x)=Ce^{kx}$ $g(x)=Cxe^{kx}$ .
$g(x)$ $sin\ ax$ $cos\ ax$ .

For example, g(x)=e^x

\begin{matrix} y^{″} + 3^{'} + 2 y = 3 e^{5 x} \\ y_{c} = A e^{- x} + B e^{- 2 x} . \end{matrix}

$y_p=Ce^{5x}$ $e^{5x}$ $y_p$ back to our D.E. and find "C".

C (5^{2} + 15 + 2) e^{5 x} = 3 e^{5 x} \to C = \frac{1}{14}

Our G.E. of this problem would be,

\begin{matrix} y = y_{c} + y_{p} \\ = A e^{- x} + B e^{- 2 x} + \frac{1}{14} e^{5 x} \end{matrix}

For example, g(x)=x^2

y^{″} + 3 y^{'} + 2 y = x^{2}

$y_c=Ae^{-x}+Be^{-2x}$

$y_p=p_0+p_1x+p_2 x^2$ $p_0=7/4,\quad p_1=-3/2, \quad p_2=1/2$

So our G.S. would be,

y = A e^{- x} + B e^{- 2 x} + \frac{7}{4} - \frac{3}{2} x + \frac{1}{2} x^{2}

Failure Case Example,

y^{″} - 5 y^{'} + 6 y = e^{2 x}

$y_c=Ae^{2x}+Be^{3x}$ $y_p$ $y_p=Ce^{2x}$ Would be impossible, because answers are folded. Let's see,

C e^{2 x} (4 - 10 + 6) = e^{2 x}

$y_p=Cxe^{kx}$ Instead.

5. Linear ODE with Variable Coefficients - Euler Equation

Cauchy-Euler Equation form:

L y = a x^{2} \frac{d^{2} y}{d x^{2}} + β x \frac{d y}{d x} + c y = g (x)

Each derivative terms is consisted with the derivatives multiplied by the same power "x", where the order of derivatives is same as the power of "x".

$y=y_c+y_p$ as well.

5.1 Homogeneous Part

$y_c$ :

$y=x^\lambda$ . Plug it back to the homogeneous D.E.

\begin{matrix} a x^{2 λ} λ (λ - 1) x^{λ - 2} + β λ x^{λ} + c x^{λ} = 0 \\ (a λ^{2} + (β - a) λ + c) x^{λ} = 0 \end{matrix}

$x^\lambda\neq0$ $a\lambda^2+(\beta-a)\lambda+c=0$ , to be the root.

$b^2-4ac >, =, < 0$ are in discussion.

5.2 Reduce to Constant Coefficient

Simplify the problem,

x^{2} y^{″} - x y^{'} + y = l n x

$x=e^t$ $t=ln\ x$ , by the chain rule,

\begin{matrix} \frac{d y}{d x} = \frac{d y}{d t} \frac{d t}{d x} = \frac{1}{x} \frac{d y}{d t} \\ \frac{d^{2}}{d x^{2}} = \frac{d}{d x} (\frac{d y}{d x}) = \frac{d}{d x} (\frac{1}{x} \frac{d y}{d t}) \\ = \frac{1}{x} \frac{d}{d x} \frac{d y}{d t} - \frac{1}{x^{2}} \frac{d y}{d t} = \frac{1}{x^{2}} \frac{d^{2} y}{d t^{2}} - \frac{1}{x^{2}} \frac{d y}{d t} \end{matrix}

The Euler Equation would then become,

y^{″} - 2 y^{'} + y = t

By Fanyu Zhao