First Order ODEs

First order ODE is equations that has the highest derivative of order one. It takes the form

The general solution to this equation is some function $y\left(x\right)$ that satisfies the equation.

Separable Equation

A separable equation is of the form

Linear Equation

A linear equation is of the form

For now, we assume $a_1\left(x\right)\ne0$ for all $x\in I$, then the equation above is simplified into

• If $q\left(x\right)=0$ for all $x\in I$, it is called a homogeneous equation;

• Otherwise, it is called an inhomogeneous equation.

Homogeneous

A homogeneous linear equation is of the form

which is a separable equation. The general solution to it is

If given an initial condition $y\left(\xi\right)=\eta$, then the solution is

Inhomogeneous

An inhomogeneous linear equation is of the form

Let

and multiply $u\left(x\right)$, which is called the integrating factor, to both sides of the equation, we get $\left(uy\right)'=uq$. Therefore the solution is

If given initial condition $y\left(\xi\right)=\eta$, then we can find a specific constant $C$.

Method of substitution

In addition to separable equations and linear equations for which we have a direct solving techniques, some other first order equations can be transformed into separable equations or linear equations by method of substitution. The following are some types of those transformable equations.

Equation: $y'=f\left(ax+by+c\right)$

Consider a equation of the form

where $a,b,c\in \mathbb{R}$, $b\ne 0$.

Let $u=ax+by+c$, then,

which is a separable equation of $u$.

Homogeneous Equation

A homogeneous equation has the form

This type of equation has an important characteristic that its invariant under zoom, i.e., if make change $x\to ax$, $y\to ay$, the equation does not change.

Tip:

We solve this kind of equations by transforming them into a separable equation.

First, let $\displaystyle z=\frac{y}{x}$, and therefore, $y=zx$, and $y'=z'x+z$. Substitute that in the equation and,

which is a separable equation. Here are some examples,

• Example

  Find the general solution to the following differential equation

  $$y'=\frac{y+x}{x-y}$$

  Solution:

  First, make the left hand side life a homogeneous equation,

  $$y'=\frac{y/x+1}{1-y/x}$$

  and let $z=y/x$. Then, the equation becomes,

  $$xz'=\frac{z^2+1}{1-z}$$

  integrate both side

  $$\int \frac{1}{x}\mathrm{d}x=\int\frac{1-z}{z^2+1}\mathrm{d}z$$

  which is

  $$\ln x +C=\arctan z-\frac{1}{2}\ln\left(z^2+1\right)$$

  Change back into $y$, and get

  $$\arctan \left(\frac{y}{x}\right)=\ln\sqrt{x^2+y^2}+C$$

  Now, suppose we are in the polar coordinate, where $\displaystyle \theta=\arctan y/x$, and $r=x^2+y^2$,

  $$\theta=\ln r+C$$

  or equivalently,

  $$r=C_1e^{\theta}$$

  which is known as the exponential spiral.

Bernoulli’s Equation

A Bernoulli’s equation has the form

where $n\ne 0,1$. We solve this kind of equations by transforming them into a linear equation.

First, multiply the equation by $\left(1-n\right)y^{-n}$, and get

Next, let $\displaystyle v=y^{1-n}$, and observe that $\displaystyle v'=\left(1-n\right)\frac{y'}{y^n}$, thus the equation becomes,

which is a linear equation. Here is an example.

• Example

  Find the general solution of the following equation

  $$y'+\frac{y}{1+x}+\left(1+x\right)y^4=0$$

  Solution:

  First, divide the equation by $y^4$, which gives

  $$\frac{y'}{y^4}+\frac{1}{1+x}\cdot\frac{1}{y^3}+\left(1+x\right)=0$$

  Then, the substitution should be $v=y^{-3}$, and $v'=-3y^{-4}y'$. Thus, the equation becomes

  $$v'-\frac{3}{1+x}v-3\left(1+x\right)=0$$

  which is a linear equation.

  Next, the integrating factor is

  $$u\left(x\right)=\exp\left(-\int\frac{3}{1+x}\mathrm{d}x\right)=\frac{1}{\left(1+x\right)^3}$$

  Multiply the last equation by $u$ gives

  $$\frac{v'}{\left(1+x\right)^3}-\frac{3}{\left(1+x\right)^4}v=3\left(1+x\right)$$

  which is in fact,

  $$\left(\frac{1}{\left(1+x\right)^3}v\right)'=3\left(1+x\right)$$

  Therefore,

  $$v=\left(1+x\right)^2\left(C+Cx-3\right)$$

  and substitute back to get

  $$y=\frac{\mathrm{sgn}\left(C+Cx-3\right)}{\sqrt[3]{\left(1+x\right)^2\left|C+Cx-3\right|}}$$

Integral Curves

Even with the previous cases, there are not very much equations that we can solve. Now, we generalize the solution of a differential by not requiring it to be a explicit function. That is, we can solve (1) if we can found a function $\phi\left(x,y\right)$, such that (1) can be transformed into the form

Then, we integrate it with respect to $x$ and get a general solution

for some constant $c$. This kind of solution is know as the integral curve.

Now, we discuss when can we write an equation into (2). By chain rule of partial differentiation, we have

compare this with equation (1), we know that for such a nice function $\phi$ to exist, we require

Recall from the theory of vector field, this means that $\phi$ is indeed a potential function. This means that the condition above is equivalent to

Those differential equations with the above condition satisfied is said to be exact.

For some first order equations that are not exact, we can somehow turn it exact by multiplying with a function $\mu\left(x,y\right)$, which is called the integrating factor or Euler’s Multiplier.

After doing so, condition (3) then becomes

or

This turns out to be a partial differential equation for $\mu$, which is hard to solve generally. However, if we assume that $\mu$ only depend on $x$ or $y$, it then becomes an ordinary differential equation that hopefully we can solve.

For example, assume $\mu$ only depend on $x$, then

Second Order ODEs

A second-order linear ODE takes the form of

Note, we will leave the case where the coefficient of $y''$ is not 1 till later. When all of $p,q,r$ are not dependent on $t$, we say that it is an equation with constant coefficients. In addition, if $g=0$ for all possible $t$, then we say that the equation is homogeneous, otherwise inhomogeneous.

It can be proved that the general solution to a linear, second-order homogeneous ODE has the form

where $c_1,c_2\in\mathbb{R}$ are constants, and $y_1,y_2$ are linearly independent. See proof below.

In addition to the differential equation, we will often impose initial conditions

Then the constants $c_1,c_2$ are specified.

Constant Coefficient Homogeneous Equations

We begin by studying equations that have constant coefficients and are homogeneous.

where $A,B\in\mathbb{R}$. Our goal is to find to independent solutions. This can be done basically by guessing with exponential function $y=e^{rt}$. Substituting this back to equation (1) gives

Since $e^{rt}$ never vanishes, it can be cancelled, what is left is the following

which is a simple quadratic equation, known as the characteristic equation of (1). Based on the number of solution of (2), there are the following three different cases.

Two Real Solutions, $A^2-4B>0$

When there are two different real solutions $r_1,r_2$ to the characteristic solution, we immediately get two independent solutions, $y_1=e^{r_1t},y_2=e^{r_2t}$. Then, the general solution is

Two Complex solutions, $A^2-4B<0$

In this case, we get two complex conjugate solutions $a\pm bi$ for equation (2).

Now, if we plug in the formula given above, we still get $y_1,y_2$ which are independent solution. However, they are complex functions. We can of course just say that we don’t care about this, but sometimes when we use this ODE in some physical models, we can’t interpret them if the solution is complex. For example, in a damped oscillation model, $y$ should represent a displacement from the equilibrium position, complex solution is not tolerable. Therefore, we still want to figure out a general form for the real solution of this equation.

We do this by first noticing an important property: if $u+vi$ is a complex solution, where $u,v$ are real functions of $t$, then both $u$ and $v$ are a solution.

Therefore, the real part and imaginary part of the solution $e^{\left(a+bi\right)t}$ are two independent solutions, and the general solution is

Remark:

There is another way to write the general real solutions, but with real coefficients $c,d$.

One Single Solution, $A^2-4B=0$

In this case, we only have one real solution $r$ for the characteristic equation. One obvious solution is $y_1=e^{rt}$, but the problem is what is the other solution?

There is a theorem related to this condition saying if $y_1$ is a solution to the equation

then, there exist some function $u\left(t\right)$, such that $y=uy_1$ is also a solution. The proof of this theorem has something to do with Jordan Normal Form of the fundamental system, which is omitted here.

Therefore, what we need to do is to plug in $y=uy_1$ and get the function $u$,and then we have the second independent solution.

Let $r=-a$, and suppose $y$ is a solution to $y''+2ay'+a^2y=0$, then

Add those three equations up to get

Just one choice of $u$ is enough, and therefore we set $u=t$, the second independent solution is

and the general solution to this case is

Inhomogeneous Equation

Next, we want to consider a more general case, which is to find the solution to

We claim that the general solution to this equation is a superposition of a particular solution $y_{p}$ and the general solution $y_{c}$ to the associated homogeneous equation

Proof:

1. All $y_p+y_c$ are solutions. This can be done by plugging those solutions into the equation.

2. No other solutions. Suppose $u$ is any solution, then $L\left(u\right)=g\left(t\right)$. Since $y_p$ is a particular solution, $L\left(y_p\right)=g\left(t\right)$. Subtract those two equations and get $L\left(u-y_p\right)=0$. Therefore, $u-y_p=c_1y_1+c_2y_2$ for some constant $c_1,c_2$.

We can use the method of variation of parameters once we have found two independent solutions to the associated homogeneous equation (5). (Notice that it’s also possible to find the particular solution by inspection. First, try some simple functions before diving into the calculations!)

Let $y_1,y_2$ be two solutions to equation (5) and set

Observe that equation (4) actually only imposes one condition on two unknown functions, which therefore gives us some freedom in the choice of $c_1\left(t\right)$ and $c_2\left(t\right)$. Therefore, we can determine those two functions by imposing another condition that can simplify our calculation.

Computing

and notice that the $y''$ will contain no terms of the second order derivative of $c_1$ and $c_2$ if

With this condition, plugging $y\left(t\right)$ back to (4) and get

Therefore, we only need to solve for $c_1'\left(t\right)$ and $c_2'\left(t\right)$ from conditions (6) and (7) and then integrate to get $c_1$ and $c_2$.

Homogeneous Equation

Now, we proceed to equations that are homogeneous without constant coefficients.

There is no general method for solving this kind of equation for arbitrary function $p$ and $q$. However, we can still get its general solution if we can guess one solution.

Let $y_1$ be a solution and set

where $v\left(t\right)$ is some unknown function. Plugging $y_2$ into (3) and get

from which we can solve for $v'$ and then $v$.