NotesAtHKU

First Order Differential Equations

Differential equations are equations that involve a function and its derivatives.

Order of differential equations

A n ordered differential equation is an equation of the form:

F(x,y,y'\dots y'_n)=0

Where $y'_n$ is the nth derivative of $y$ with respect to $x$ . The highest degree of the derivative is n for a n-ordered differential equation. Note that $y$ is really just $y(x)$ (A function of x)

Solving linear 1st-ODEs

Linear differential equations

Linear differential equations does not contain non-linear functions. (e.g. $\sin y$ ) Otherwise, it's a non-linear ODE.

Solving by integrating factors

We can solve a linear 1st-ODE as followed, given a particular solution of $y(x)$ :

y'+p(x)y=q(x)\quad : \quad\times\ e^{\int p(x)}

The multiplied integration factor $e^{p(x)}$ will give us a product of the product rule, then we simply integrate both sides to solve for $y$ . Make sure that the coefficient of $y'$ is 1.

g(y)}^{-1}dy & =f(x)dx \\

We can solve a separable equation as followed, given a particular solution of $y(x)$ :

\begin{aligned} \frac{dy}{dx} & =f(x)g(y) \\ \int{g(y)}^{-1}\:dy & =\int f(x)\:dx \Leftarrow [(x,y)\to c] \end{aligned}

Solving non-linear 1st-ODEs

Bernoulli's equation

A non-linear 1st-ODE of the form can be solved by:

y'+p(x)y=q(x)y^n\quad n\in\mathbb{R}\quad : \quad \text{sub }u=y^{1-n}\to y,y'

The subsitution $u=y^{1-n}$ will turn the equation into a linear ODE, then simply solve using integrating factors.

Riccati's equation

A non-linear 1st-ODE of the form can be solved by the following, given a particular solution of $y(x)$ :

y'=p(x)y^2+q(x)y+r(x)\quad : \quad \text{sub }y=y(x) + u^{-1}

Homogeneous equations

A homogeneous equation has it's $x$ and $y$ terms in the same degree. (e.g. $x^2+xy+y^2=0$ )

A homogeneous 1st-ODE of the form can be solved by the following, given a particular solution of $y(x)$ :

y'=f(\frac{y}{x})\quad : \quad \text{sub }u=\frac{y}{x} \to y'=u + xu'

We can divide the formula by $x^n$ or $y^n$ to get the equation in the desired form (every term is the ratio $\frac{y}{x}$ ). Otherwise, we can shift the origin using $X=x-n$ and $Y=y-m$ .

After substitution, we will get a separable equation after the substitution, and the particular solution is used.

A partial derivative is a derivative of a function with respect to one of its variables, with the others held constant. The following notation expresses the partial derivative of $f$ with respect to $x$ :

\frac{\partial f}{\partial x}

\begin{aligned} F & =2x+y \\ \frac{\partial F}{\partial x} & =2 \\ \end{aligned}

Exact equations

An exact equation is simply a 1st-ODE where $dF=0$ .

The expressed equation $dF$ is exact if:

dF=Mdx+Ndy\quad:\quad\frac{\partial M}{\partial y}=\frac{\partial N}{\partial x}

Solving exact equations

To find the solution of an exact equation:

\begin{aligned} M=\frac{\partial F}{\partial x} & ,\quad N=\frac{\partial F}{\partial y} \\ \partial F & = M\partial x \\ F & = \int M dx + g(y) \\ \frac{\partial F}{\partial y} & = N \\ \frac{\partial \int M dx + g(y)}{\partial y} & = N \to g'(y) \\ \int g'(y)\:dy & = g(y) \to F \end{aligned}

$g(y)$ is present as we are integrating partially with respect to $x$ , and $g(y)$ is the constant of integration.

Hence, the solution would be:

\int M dx + g(y)=c

Second Order Differential Equations

Solving homogeneous linear 2nd-ODEs

Constant coefficient Homogeneous 2nd-ODEs

The term homogeneous is used differently from the previous section.

A homogeneous 2nd-ODE is of the form:

ay''+by'+cy=\mathbf{0}

Where $a,b,c$ are constants. (Coefficients are constants)

We first use the following substitution:

ay''+by'+cy=0:\quad y=e^{\lambda x}\implies a\lambda^2+b\lambda+c=0\to \lambda

To find the general solution, we put the $\lambda$ roots into the quadratic characteristic equation:

$\lambda_1\ne\lambda_2:\quad y=c_1\mathbf{e}^{\lambda_1x}+c_2\mathbf{e}^{\lambda_2x}$
$\lambda_1=\lambda_2:\quad y=c_1\mathbf{e}^{\lambda x}+c_2x\mathbf{e}^{\lambda x}$
$\{\lambda_{1,2}=\alpha\pm\beta i\}\in\mathbb{C}:\quad y=c_{1}\mathbf{e}^{\alpha x}\cos(\beta x)+c_{2}\mathbf{e}^{\alpha x}\sin(\beta x)$ If given particular solutions of $y$ and $y'$ , we can solve for $c_1$ and $c_2$ by finding $y'$ with our general solution and substituting.

Cauchy-Euler equations

A Cauchy-Euler equation is a slight variation of homogenous 2nd-ODEs, which is of the following form and can be solved by:

ax^2y''+bxy'+cy=0\quad:\quad y=x^\lambda\implies a(\lambda^2-\lambda)+b\lambda+c=0

The general solutions is similar to that of the homogeneous 2nd-ODEs, but with all terms of $x\to \ln x$ :

$e^{\lambda x}\to x^\lambda$
$e^{\lambda x}\to x^\lambda,\quad x\to \ln x$
$e^{\alpha x}\to x^\alpha,\quad \beta x \to \beta \ln x$

Solving non-homogeneous linear 2nd-ODEs

Constant coefficient Non-homogeneous 2nd-ODEs

A non-homogeneous 2nd-ODE is of the form:

F:\ ay''+by'+cy=g(x)

Where $a,b,c$ are constants. (Coefficients are constants)

We first solve for $\lambda_{1,2}$ for the complementary homogenous function $F_c$ to get $Y_c$ :

F_c:\ ay''+by'+cy=0\ \to Y_c

The general solution $y$ for the non-homogeneous 2nd-ODE $F$ is:

y=Y_c+Y_p

Where $Y_p$ is a particular solution of $y$ . To solve for $Y_p$ , we can use the following methods:

Method of undetermined coefficients

To solve for $Y_p$ for a non-homogeneous 2nd-ODE, let $Y_p$ as the following if $g(x)$ consists of:

$e^{ax}\to Y_p=Ae^{ax}$
$\sin x\text{ and / or }\cos x\to Y_p=A\sin x+B\cos x$
$x^n\to Y_p=Ax^n=A_nx^n+A_{n-1}x^{n-1}+\dots+A_0$ (polynomial of degree $n$ )

Important things to note:

$\blacktriangleright$ If $g(x)$ is a product of multiple components, $Y_p$ is the product of the different results.

$\blacktriangleright$ If $Y_p$ consists of a non-constant term that exists in $Y_c$ , we must multiply $Y_p$ by $x^i$ and repeat the process.

We then substitute $y=Y_p\to F$ and solve for $A$ and $B$ .

For Cachy-Eular equations, we instead multiply $Y_p$ by $(\ln x)$ if $Y_p$ consists of a non-constant term that exists in $Y_c$ .

Variation of parameters

We can use this method when we are unable to see a particular solution for $Y_p$ in the above method.

Note that for $Y_c$ is in the form of $c_1y_1+c_2y_2$ . To solve for $Y_p$ for a non-homogeneous 2nd-ODE:

Y_{P}=-y_1\int\frac{y_2g(x)}{W}dx+y_2\int\frac{y_1g(x)}{W}dx,\quad W=y_1y_2'-y_2y_1'

Note that for Cachy-Eular equations, $g(x)$ is defined as the function with the coefficient of $y''$ as 1, hence, $g(x)\to \frac{g(x)}{ax^2}$ .

If $Y_p$ consists of a non-constant term that exists in $Y_c$ , we simply discard it (merging constants).

Solving ODEs with Laplace Transforms

Laplace transform

The Laplace transform is a technique used to solve linear ODEs with constant coefficients. The Laplace transform of a function $f(t)$ is defined as:

\mathcal{L}\{f(t)\}=\int_{0}^{\infty}e^{-st}f(t)dt=F(t)

Where $s$ is a complex number.

Properties of Laplace transforms

The following are some properties of Laplace transforms:

$\mathcal{L}\{f+g\} = \mathcal{L}\{f\}+\mathcal{L}\{g\}$
$\mathcal{L}\{kf\} = k\mathcal{L}\{f\}$

Laplace transform of derivatives

The Laplace transform of the derivative of a function $y(t)$ is:

\begin{aligned} \mathcal{L}\{y'\} & =sY(s)-y(0) \\ \mathcal{L}\{y''\} & =s^2Y(s)-sy(0)-y'(0) \end{aligned}

Where $f(0)$ is the initial condition of $f$ .

	$f$	$\mathcal{L}\{f\}$
0.	$a$	$\frac{a}{s}$
0.	$e^{at}$	$\frac{1}{s-a}$
0.	$f*g$	$F(s)G(s)$
1.	$t^n$	$\frac{n!}{s^{n+1}}$
2.	$t^nf(t)$	$(-1)^n\frac{d^n}{ds^n}\mathcal{L}\{f(t)\}$
3.	$\sin at$	$\frac{a}{s^2+a^2}$
4.	$\cos at$	$\frac{s}{s^2+a^2}$
5.	$e^{at}f(t)$	$F(s-a)$
6.	$f(t-a)H(t-a)$	$e^{-as}\mathcal{L}\{f(t)\}$

Convolution operator:\ $f*g(t)=\int_0^tf(\tau)g(t-\tau)\:d\tau$

Inverse Laplace transform

The inverse Laplace transform is the reverse operation of the Laplace transform.

\mathcal{L}^{-1}\{F(t)\}=f(t)

You basically think backwards like how you'd do intergration sometimes. Remember to use the rules!

Partial fractions

A fraction can be decomposed into partial fractions if the degree of the numerator is less than the degree of the denominator. If not, then perform long division first.

$\frac{f(x)}{g(x)h(x)} = \frac{A}{g(x)} + \frac{B}{h(x)}$
$\frac{f(x)}{g^2(x)h(x)} = \frac{A}{g(x)} + \frac{B}{g^2(x)} + \frac{C}{h(x)}$
$\frac{f(x)}{(x^2+1)} = \frac{Ax+B}{x^2+1}$

We can solve for $A$ , $B$ , and $C$ by multiplying the denominator (to make left side $f(x)$ only) and solving for the numerator.

Solving ODEs with Laplace transforms

To solve a linear ODE with constant coefficients using Laplace transforms:

Take the Laplace transform of both sides of ODE ( $y(t) \to Y(s)$ )
Solve for the Laplace transform of the function $Y(s)$
Convert $Y(s)$ into partial fractions
Find the inverse Laplace transform of the function ( $Y(s)\to y(t)$ )

Orinary Differential Equations

First Order Differential Equations

Order of differential equations

Solving linear 1st-ODEs

Linear differential equations

Solving by integrating factors

g(y)}^{-1}dy & =f(x)dx \\

Solving non-linear 1st-ODEs

Bernoulli's equation

Riccati's equation

Homogeneous equations

Exact equations

Partial derivatives

Exact equations

Solving exact equations

Second Order Differential Equations

Solving homogeneous linear 2nd-ODEs

Constant coefficient Homogeneous 2nd-ODEs

Cauchy-Euler equations

Solving non-homogeneous linear 2nd-ODEs

Constant coefficient Non-homogeneous 2nd-ODEs

Method of undetermined coefficients

Variation of parameters

Solving ODEs with Laplace Transforms

Laplace transform

Properties of Laplace transforms

Laplace transform of derivatives

Inverse Laplace transform

Partial fractions

Solving ODEs with Laplace transforms

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