May 12th, 2020

# Abstract

When I got first introduced to differential equations, I had a love-hate relationship with it. Mainly due to some back-of-the-book problems we were given and never-ending projects we were assigned to. After some time, differential equations is a way to truly understand physics and the foundations of gravity, fields, and everything. This articles is merely an intro on manually solving common forms of differential equations. Hope you enjoy

## Quick notes

• $$f_x \iff \partial_x f$$

• $$A,B,C$$ are usually constants

• $$c_k$$ is usually solution’s constant that is defined with initial conditions

• Most of the functions are $$\mathbb{R}^k \to \mathbb{R}$$, $$k \in \mathbb{N}^+$$

• If you found a typo or want to comment, feel free to email me. Email on top of the page.

## First-order, linear

Those equations have the form \begin{align*} y' + p(t) y = q(t) \end{align*} Find \begin{align*} \mu(t) = e^{\int p(t) dt} \end{align*} Then \begin{align*} \frac{d}{dt}(\mu(t)y) & = q(t) \mu(t) \\ \implies y & = \frac{\int q(t) \mu(t) dt}{\mu(t)} \end{align*}

## First-order, separable

Those equations have the form:

\begin{align*} \frac{dy}{dx} = f(x)g(y) \end{align*}

Find the solution by solving

\begin{align*} \int \frac{dy}{g(y)} = \int f(x) dx \end{align*}

Solve for exact (explicit) values of $$y$$

## Exact equations

They have the form

\begin{align*} M(x,y) + N(x,y) \frac{dy}{dx} = 0 \end{align*}

• $$(\xi)$$ If $$M_y = N_x$$

$$\implies$$ Find such $$F(x,y)=C$$, where

\begin{align*} F_x = M, \quad F_y = N \end{align*}

• otherwise, make it exact, such that

\begin{align*} \frac{M_y-N_x}{N} \text{ only depends on } x \text{ or } \frac{N_x-M_y}{M} \text{ only depends on } y \end{align*}

Find

\begin{align*} \mu(x) = e^{\int \frac{M_y-N_x}{N} dx} \quad\text{or}\quad \mu(y) = e^{\int \frac{N_x-M_y}{M} dy} \end{align*}

multiply both by $$M$$ and $$N$$, so the condition $$M_y = N_y$$ is satisfied. Go to $$(\xi)$$ and proceed with finding $$F(x,y)$$

## Second-order, linear, constant-coefficient equations

They have the form

\begin{align*} y” + p y' + q y = f(t) \end{align*}

• First, solve for the homogeneous case, where $$y” + p y' + q y = 0$$

Make a characteristic polynomial, let $$y = e^{rt}$$:

$$r^2+pr+q=0$$

Find roots, solution (general) will be

\begin{align*} y = c_1 e^{r_1 t} + c_2 e^{r_2 t} \end{align*}

• if repeated root $$\implies y = c_1 e^{rt} + c_2 t e^{rt}$$

• if $$r = \alpha \pm i \beta$$ $$\implies y=c_1 \cos(\beta t)e^{\alpha t} + c_2 \sin(\beta t) e^{\alpha t}$$

### Solving for particular solution $$y_p(t)$$

#### Undetermined coefficients (superpositioned) for $$f(t)$$

Whatever is in $$f(t)$$, start adding up the corresponding coefficients into $$y_p(t)$$

• $$e^{nt} \to Ae^{nt}$$

• $$t^m \to A_m t^m + \ldots + A_1 t + A_0$$

• $$\cos(\beta t)$$ or $$\sin(\beta t) \to Acos(\beta t) + B\sin(\beta t)$$

 NOTE should not be linearly dependent with the general solution. If it is, multiply by $$t$$ until it is linearly independent.

#### Variation of parameters

Seek $$y_p(t) = v_1(t)y_1(t)+v_2(t)y_2(t)$$, where

\begin{align*} \begin{cases}v_1'y_1+v_2'y_2=0\\v_1'y_1'+v_2'y_2'=f(t)\end{cases} \end{align*}

So the final solution is $$y(t)=c_1 y_1(t) + c_2 y_2(t) + y_p(t)$$

## Second-order, linear, variable-coefficient equations

Equations have the form

• $$(1)$$: $$a(t)y” + b(t)y'+c(t)y = f(t)$$

• $$(2)$$: $$y” + p(t)y'+q(t)y = g(t)$$

In general case, guess the first homogeneous solution (try $$y_1=e^t$$) and use reduction of order to find the second homogeneous solution, so that \begin{align*} & y_2(t) = v(t)y_1(t) \\ & \implies y_2” + p(t)y_2' + q(t)y_2 = 0 \\ & \implies (v(t)y_1(t))”+p(t)(v(t)y_1(t))'+q(t)(v(t)y_1(t)) = 0 \\ \end{align*}

 NOTE Also applicable with form $$(1)$$

You will probably have another differential equation emerge from above. It should have lower order than our current equation, so just refer to one of the techniques above to find $$v(t)$$ and then you can find $$y_2(t)=v(t)y_1(t)$$

Use variation of parameters to find a particular solution. It’s that system with $$v$$

 NOTE What you if you have a Cauchy-Euler equation?

They have the form $$at^2y”+bty'+cy=0$$

then $$y=t^r \implies ar^2+(b-a)r+c=0$$

• if $$r$$ is repeated, $$y_1=t^r$$, $$y_2=ln|t|t^r$$

• if $$r=\alpha\pm i\beta$$, $$y_1=t^{\alpha}\cos(\beta ln|t|)$$ and $$y_2=t^{\alpha}\sin(\beta ln|t|)$$

Generally, solution has the form $$y=c_1t^{r_1}+c_2t^{r_2}$$

## Higher-order, linear equations

\begin{align*} a_n(t)y^{(n)}+a_{n-1}(t)y^{(n-1)}+\ldots+a_1(t)y'+a_0(t)y=g(t) \end{align*} All second-order methods above extend to $$n^{th}$$ order.

## Laplace transform

Laplace is a holy grail of solving differential equations with initial values defined. Laplace is the same kind of Bible to engineers like Taylor Series is. \begin{align*} \mathcal{L}\{f\}(s) = \int_0^{\infty} e^{-st} f(t) dt \end{align*} assuming $$f$$ is piecewise continuous and of exponential order.

Table of common Laplace transformations
$$f(t)$$ $$\mathcal{L}\{f\}(s)$$
$$1$$ $$\frac{1}{s}$$
$$e^{at}$$ $$\frac{1}{s-a}$$
$$\sin(bt)$$ $$\frac{b}{s^2+b^2}$$
$$\cos(bt)$$ $$\frac{s}{s^2+b^2}$$
$$u(t-a)$$ $$\frac{e^{-as}}{s}$$
$$\delta(t-a)$$ $$e^{-as}$$

Where $$u(t)$$ is the Heaviside step function and $$\delta(t)$$ is the Dirac delta function.

Some Laplace transform properties:

• $$\mathcal{L}\{e^{at}f(t)\}(s) = \mathcal{L}\{f(t)\}(s-a)$$

• $$\mathcal{L}\{t^nf(t)\}(s) = s^n\mathcal{L}\{f\}(s)-s^{n-1}f(0)-\ldots-sf^{(n-2)}(0)-f^{(n-1)}(0)$$

• $$\mathcal{L}\{t^nf(t)\}(s) = (-1)^n \frac{d^n}{ds^n} \mathcal{L}\{f(t)\}(s)$$

If $$f$$ is a T-periodic function, \begin{align*} \mathcal{L}\{f(t)\}(s) = \frac{\int_0^T e^{-sT} f(t) dt}{1-e^{-sT}} \end{align*} where $$\int_0^T e^{-sT} f(t) dt = \mathcal{L}\{f_T(t)\}(s)$$, the sum of integrals of different parts of the piecewise function.

Convolutions:

• $$(f*g)(t) = \int_0^t f(t-v)g(v)dv$$

• $$\mathcal{L}\{(f*g)(t)\} = \mathcal{L}\{f(t)\}(s)\cdot \mathcal{L}\{g(t)\}(s)$$

• $$(f*g)(t) = \mathcal{L}^{-1}\{F\cdot G\}(t)$$, where $$F=\mathcal{L}\{f\}(s)$$ and $$G=\mathcal{L}\{g\}(s)$$

Heaviside/unit step function:

• $$\mathcal{L}\{u(t-a)f(t)\}(s) = e^{-as}\mathcal{L}\{f(t+a)\}(s)$$

• $$\mathcal{L}^{-1}\{e^{-as}F(s)\}(t)=u(t-a)\mathcal{L}^{-1}\{F(s)\}(t-a)$$

If IVP is not at 0, define some new function like $$w(t)=y(t+\alpha)$$, and solve for $$w$$. Finally, you can offset to find $$y$$

Step (block) function: \begin{align*} \Pi_{a,b}(t) = u(t-a)-u(t-b) \end{align*} so \begin{align*} \mathcal{L}\{\Pi_{a,b}(t)\}(s)=\frac{e^{-sa}-e^{-sb}}{s} \end{align*}

## Constant-coefficient, homogeneous systems of ODE

\begin{align*} \vec{x}' = A \vec{x}, \quad \text{where } A\in\mathbb{R}^{n\times n},\quad x\in\mathbb{R}^n \end{align*}

If $$A$$ has n linearly independent eigenvectors $$\vec{u_i}$$ associated to n eigenvalues $$\lambda_i$$, then a general solution of the system is given by $$\vec{x}(t) = c_1 e^{\lambda_1 t}\vec{u_1}+c_2e^{\lambda_2t}\vec{u_2} + \ldots + c_ne^{\lambda_nt}\vec{u_n}$$

• If $$\lambda=\alpha \pm i \beta$$, so $$\vec{u}=\vec{a}+i\vec{b}$$, we have $$\vec{x}=c_1e^{\alpha t}(\cos(\beta t)\vec{a}-\sin(\beta t)\vec{b}) + c_2e^{\alpha t}(\cos(\beta t)\vec{b}+\sin(\beta t)\vec{a})$$

• Matrix exponential

$$e^{At} = \sum_{k=0}^{\infty} \frac{A^k t^k}{k!}$$, where $$A^0=I$$, an identity matrix.

• Find solutions for any eigenvalues

• Compute the characteristic polynomial $$p(\lambda)$$ of $$A$$

\begin{align*} p(\lambda)=det(A-\lambda I) \end{align*}

• Factor $$p(\lambda)$$ into linear factors to yield

\begin{align*} p(\lambda) = c(\lambda-\lambda_1)^{m_1} \cdot \ldots \cdot (\lambda-\lambda_k)^{m_k},\quad c=\pm 1 \end{align*}

• For each $$\lambda_j$$, find $$m_j$$ linearly independent generalized eigenvectors $$\{\vec{u_j}^{m_1},\cdots,\vec{u_j}^{m_j}\}$$ satisfying

\begin{align*} (A-\lambda_i I)^{m_j} \vec{u} = \vec{0} \end{align*}

• For each $$\vec{u_j}^i$$ computed in the previous step, compute $$e^{At}\vec{u_j}^i$$ by

\begin{align*} e^{At}\vec{u_j}^i & =e^{\lambda_jt}e^{(A-\lambda_jI)t}\vec{u_j}^i \\ & =e^{\lambda_jt}(\vec{u_j}^i+t(A-\lambda_jI)\vec{u_j}^i+\cdots+\frac{t^{m_j-1}}{(m_j-1)!}(A-\lambda_jI)^{m_j-1}\vec{u_j}^i) \end{align*}

## Linear systems of ODE

\begin{align*} \vec{x}' = A(t)\vec{x} + \vec{f}(t), \quad \text{where } A\in\mathbb{R}^{n\times n}, x\in\mathbb{R}^n, \quad f\in\mathbb{R}^n \end{align*}

If $$X(t)$$ is a matrix whose columns are made up of n linearly independent homogeneous solutions ($$X(t)$$ is the fundamental matrix), then a general solution may be written as $$\vec{x}(t_0)=\vec{x_0}$$ \begin{align*} \vec{x}(t) = X(t)X^{-1}(t_0)\vec{x_0}+X(t)\int_{t_0}^{t}X^{-1}(s)f(s)ds \end{align*} If $$A(t)$$ is constant-coefficient, then we recover Duhamel’s formula: \begin{align*} \vec{x}(t) = e^{A(t-t_0)}x_0 + \int_{t_0}^{t}e^{A(t-s)}f(s)ds \end{align*}

## Applications

There are many applications of differential equations in classical mechanics, fields, etc. Below you will find just a snippet of some very common Physics 1/2 scenarios

### Falling object

\begin{align*} m\frac{dv}{dt}=mg-bv \end{align*}

where $$b$$ is the air resistance

### Fluid mix, define $$R_{in}$$ and $$R_{out}$$

\begin{align*} \frac{dx}{dt}=R_{in}-R_{out} \end{align*}

### Mass-Spring System

• Vertical spring (direction of gravity) \begin{align*} my”=-by'-k(L+y)+mg+F_{ext}(t) \end{align*} assume $$KL=mg$$, where $$b$$ is dumping, and $$k$$ is stiffness

• Horizontal spring \begin{align*} my”=-by'-ky+F_{ext}(t) \end{align*} where $$b$$ is dumping, and $$k$$ is stiffness

## Conclusion

This is as much as I can recover from my initial experience with differential equations. This article is not as much to teach you how to solve them but provide a quick lookup cheatsheet if needed or glance at different forms that we can actually solve! There are infinitely many differential equations that we cannot find an exact solution for! ◼︎