# Examples of differential equations

Differential equations arise in many problems in physics, engineering, and other sciences. The following examples show how to solve differential equations in a few simple cases when an exact solution exists.

## Separable first-order ordinary differential equations

Equations in the form ${\displaystyle {\frac {dy}{dx}}=f(x)g(y)}$ are called separable and solved by ${\displaystyle {\frac {dy}{g(y)}}=f(x)dx}$ and thus ${\displaystyle \int {\frac {dy}{g(y)}}=\int f(x)dx}$. Prior to dividing by ${\displaystyle g(y)}$, one needs to check if there are stationary (also called equilibrium) solutions ${\displaystyle y=const}$ satisfying ${\displaystyle g(y)=0}$.

## Separable (homogeneous) first-order linear ordinary differential equations

A separable linear ordinary differential equation of the first order must be homogeneous and has the general form

${\displaystyle {\frac {dy}{dt}}+f(t)y=0}$

where ${\displaystyle f(t)}$ is some known function. We may solve this by separation of variables (moving the y terms to one side and the t terms to the other side),

${\displaystyle {\frac {dy}{y}}=-f(t)\,dt}$

Since the separation of variables in this case involves dividing by y, we must check if the constant function y=0 is a solution of the original equation. Trivially, if y=0 then y'=0, so y=0 is actually a solution of the original equation. We note that y=0 is not allowed in the transformed equation.

We solve the transformed equation with the variables already separated by Integrating,

${\displaystyle \ln |y|=\left(-\int f(t)\,dt\right)+C}$

where C is an arbitrary constant. Then, by exponentiation, we obtain

${\displaystyle y=\pm e^{\left(-\int f(t)\,dt\right)+C}=\pm e^{C}e^{-\int f(t)\,dt}}$.

Here, ${\displaystyle e^{C}>0}$, so ${\displaystyle \pm e^{C}\neq 0}$. But we have independently checked that y=0 is also a solution of the original equation, thus

${\displaystyle y=Ae^{-\int f(t)\,dt}}$.

with an arbitrary constant A, which covers all the cases. It is easy to confirm that this is a solution by plugging it into the original differential equation:

${\displaystyle {\frac {dy}{dt}}+f(t)y=-f(t)\cdot Ae^{-\int f(t)\,dt}+f(t)\cdot Ae^{-\int f(t)\,dt}=0}$

Some elaboration is needed because ƒ(t) might not even be integrable. One must also assume something about the domains of the functions involved before the equation is fully defined; the solution above assumes the real case.

If ${\displaystyle f(t)=\alpha }$ is a constant, the solution is particularly simple, ${\displaystyle y=Ae^{-\alpha t}}$ and describes, e.g., if ${\displaystyle \alpha >0}$, the exponential decay of radioactive material at the macroscopic level. If the value of ${\displaystyle \alpha }$ is not known a priori, it can be determined from two measurements of the solution. For example,

${\displaystyle {\frac {dy}{dt}}+\alpha y=0,y(1)=2,y(2)=1}$

gives ${\displaystyle \alpha =\ln(2)}$ and ${\displaystyle y=4e^{-\ln(2)t}=2^{2-t}}$.

## Non-separable (non-homogeneous) first-order linear ordinary differential equations

First-order linear non-homogeneous ODEs (ordinary differential equations) are not separable, they can be solved by the following approach, known as an integrating factor method. Consider first-order linear ODEs of the general form:

${\displaystyle {\frac {dy}{dx}}+p(x)y=q(x)}$

The method for solving this equation relies on a special integrating factor, μ:

${\displaystyle \mu =e^{\int _{x_{0}}^{x}p(t)\,dt}}$

We choose this integrating factor because it has the special property that its derivative is itself times the function we are integrating, that is:

${\displaystyle {\frac {d{\mu }}{dx}}=e^{\int _{x_{0}}^{x}p(t)\,dt}\cdot p(x)=\mu p(x)}$

Multiply both sides of the original differential equation by μ to get:

${\displaystyle \mu {\frac {dy}{dx}}+\mu {p(x)y}=\mu {q(x)}}$

Because of the special μ we picked, we may substitute /dx for μ p(x), simplifying the equation to:

${\displaystyle \mu {\frac {dy}{dx}}+y{\frac {d{\mu }}{dx}}=\mu {q(x)}}$

Using the product rule in reverse, we get:

${\displaystyle {\frac {d}{dx}}{(\mu {y})}=\mu {q(x)}}$

Integrating both sides:

${\displaystyle \mu {y}=\left(\int \mu q(x)\,dx\right)+C}$

Finally, to solve for y we divide both sides by ${\displaystyle \mu }$:

${\displaystyle y={\frac {\left(\int \mu q(x)\,dx\right)+C}{\mu }}}$

Since μ is a function of x, we cannot simplify any further directly.

## Second-order linear ordinary differential equations

### A simple example

Suppose a mass is attached to a spring which exerts an attractive force on the mass proportional to the extension/compression of the spring. For now, we may ignore any other forces (gravity, friction, etc.). We shall write the extension of the spring at a time t as x(t). Now, using Newton's second law we can write (using convenient units):

${\displaystyle m{\frac {d^{2}x}{dt^{2}}}+kx=0,}$

where m is the mass and k is the spring constant that represents a measure of spring stiffness. For simplicity's sake, let us take m=k as an example.

If we look for solutions that have the form ${\displaystyle Ce^{\lambda t}}$, where C is a constant, we discover the relationship ${\displaystyle \lambda ^{2}+1=0}$, and thus ${\displaystyle \lambda }$ must be one of the complex numbers ${\displaystyle i}$ or ${\displaystyle -i}$. Thus, using Euler's formula we can say that the solution must be of the form:

${\displaystyle x(t)=A\cos t+B\sin t}$

See a solution by WolframAlpha.

To determine the unknown constants A and B, we need initial conditions, i.e. equalities that specify the state of the system at a given time (usually t = 0).

For example, if we suppose at t = 0 the extension is a unit distance (x = 1), and the particle is not moving (dx/dt = 0). We have

${\displaystyle x(0)=A\cos 0+B\sin 0=A=1,}$

and so A = 1.

${\displaystyle x'(0)=-A\sin 0+B\cos 0=B=0,}$

and so B = 0.

Therefore x(t) = cos t; this is an example of simple harmonic motion.

See a solution by Wolfram Alpha.

### A more complicated model

The above model of an oscillating mass on a spring is plausible but not very realistic: in practice, friction will tend to decelerate the mass and have magnitude proportional to its velocity (i.e. dx/dt). Our new differential equation, expressing the balancing of the acceleration and the forces, is

${\displaystyle m{\frac {d^{2}x}{dt^{2}}}+c{\frac {dx}{dt}}+kx=0,}$

where ${\displaystyle c}$ is the damping coefficient representing friction. Again looking for solutions of the form ${\displaystyle Ce^{\lambda t}}$, we find that

${\displaystyle m\lambda ^{2}+c\lambda +k=0.}$

This is a quadratic equation which we can solve. If ${\displaystyle c^{2}<4km}$ there are two complex conjugate roots a ± ib, and the solution (with the above boundary conditions) will look like this:

${\displaystyle x(t)=e^{at}\left(\cos bt-{\frac {a}{b}}\sin bt\right)}$

Let us for simplicity take ${\displaystyle m=1}$, then ${\displaystyle 0 and ${\displaystyle k=a^{2}+b^{2}}$.

The equation can be also solved in MATLAB symbolic toolbox as

x = dsolve('D2x+c*Dx+k*x=0','x(0)=1','Dx(0)=0')


although the solution looks rather ugly,

x = (c + (c^2 - 4*k)^(1/2))/(2*exp(t*(c/2 - (c^2 - 4*k)^(1/2)/2))*(c^2 - 4*k)^(1/2)) -
(c - (c^2 - 4*k)^(1/2))/(2*exp(t*(c/2 + (c^2 - 4*k)^(1/2)/2))*(c^2 - 4*k)^(1/2))


This is a model of a damped oscillator; the plot of displacement against time would look like this:

which resembles how one would expect a vibrating spring to behave as friction removes energy from the system.

## Linear systems of ODEs

The following example of a first order linear systems of ODEs

${\displaystyle y_{1}'=y_{1}+2y_{2}+t}$
${\displaystyle y_{2}'=2y_{1}-2y_{2}+\sin(t)}$

can be easily solved symbolically using numerical analysis software.