Ordinary Differential Equations/One-dimensional first-order linear equations

Definition

One-dimensional first-order inhomogenous linear ODEs are ODEs of the form

$x^{\prime }(t)+f(t)x(t)=g(t)$

for suitable (that is, mostly, continuous) functions $f,g:ℝ\to ℝ$; note that when $g\equiv 0$, we have a homogenous equation instead.

General solution

First we note that we have the following superposition principle: If we have a solution $x_h$ ("$h$" standing for "homogenous") of the problem

${x_h}^{\prime }(t)+f(t)x_h(t)=0$

(which is nothing but the homogenous problem associated to the above ODE) and a solution to the actual problem $x_p$; that is a function $x_p$ such that

${x_p}^{\prime }(t)+f(t)x_p(t)=g(t)$

("$p$" standing for "particular solution", indicating that this is only one of the many possible solutions), then the function

$x(t):=a{x}_h(t)+x_p(t)$ ($a\in ℝ$ arbitrary)

still solves $x^{\prime }(t)+f(t)x(t)=g(t)$, just like the particular solution $x_p$ does. This is proved by computing the derivative of $x$ directly.

In order to obtain the solutions to the ODE under consideration, we first solve the related homogenous problem; that is, first we look for $x_h$ such that

${x_h}^{\prime }(t)+f(t)x_h(t)=0\iff {x_h}^{\prime }=-f(t)x_h$.

It may seem surprising, but this gives actually a very quick path to the general solution, which goes as follows. Separation of variables (and using ${\ln }^{-1}=\exp$) gives

$x_h(t)=\exp (-\underset{t_0}{\overset{t}{\int }}f(s)ds)$,

since the function

$G(t):=-\underset{t_0}{\overset{t}{\int }}f(s)ds$

is an antiderivative of $t\mapsto -f(t)$. Thus we have found the solution to the related homogenous problem.

For the determination of a solution $x_p$ to the actual equation, we now use an Ansatz: Namely we assume

$x_p(t)=c(t)x_h(t)$,

where $c:ℝ\to ℝ$ is a function. This Ansatz is called variation of the constant and is due to Leonhard Euler. If this equation holds for $x_p$, let's see what condition on $c$ we get for $x_p$ to be a solution. We want

${x_p}^{\prime }(t)+f(t)x_p(t)=g(t)$, that is (by the product rule and inserting $x_h$):

$c^{\prime }(t)\exp (-\underset{t_0}{\overset{t}{\int }}f(s)ds)=c^{\prime }(t)\exp (-\underset{t_0}{\overset{t}{\int }}f(s)ds)+c(t)(-1)\exp (-\underset{t_0}{\overset{t}{\int }}f(s)ds)f(t)+f(t)c(t)\exp (-\underset{t_0}{\overset{t}{\int }}f(s)ds)=g(t)$.

Putting the exponential on the other side, that is

$c^{\prime }(t)=g(t)\exp (\underset{t_0}{\overset{t}{\int }}f(s)ds)$

or

$c(t)=\underset{t_0}{\overset{t}{\int }}g(r)\exp (\underset{t_0}{\overset{r}{\int }}f(s)ds)dr+C_1$.

Since all the manipulations we did are reversible, all functions of the form

$C_2\exp (-\underset{t_0}{\overset{t}{\int }}f(s)ds)+(\underset{t_0}{\overset{t}{\int }}g(r)\exp (\underset{t_0}{\overset{r}{\int }}f(s)ds)dr+C_1)\exp (-\underset{t_0}{\overset{t}{\int }}f(s)ds)$ ($C_1,C_2\in ℝ$ arbitrary)

are solutions. If we set $C:=C_2+C_1$, we get the general solution form

$C\exp (-\underset{t_0}{\overset{t}{\int }}f(s)ds)+(\underset{t_0}{\overset{t}{\int }}g(r)\exp (\underset{t_0}{\overset{r}{\int }}f(s)ds)dr)\exp (-\underset{t_0}{\overset{t}{\int }}f(s)ds)$.

We want now to prove that these constitute all the solutions to the equation under consideration. Thus, set

$x_C(t):=C\exp (-\underset{t_0}{\overset{t}{\int }}f(s)ds)+(\underset{t_0}{\overset{t}{\int }}g(r)\exp (\underset{t_0}{\overset{r}{\int }}f(s)ds)dr)\exp (-\underset{t_0}{\overset{t}{\int }}f(s)ds)$

and let $x_2(t)$ be any other solution to the inhomogenous problem under consideration. Then $x_C-x_2$ solves the homogenous problem, for

${x_C}^{\prime }(t)-{x_2}^{\prime }(t)-f(t)(x_C(t)-x_2(t))={x_C}^{\prime }(t)-f(t)x_C(t)-({x_2}^{\prime }(t)-f(t)x_2(t))=g(t)-g(t)=0$.

Thus, if we prove that all the homogenous solutions (and in particular the difference $x_C-x_2$) are of the form

$C\exp (-\underset{t_0}{\overset{t}{\int }}f(s)ds)$,

then we may subtract

$D\exp (-\underset{t_0}{\overset{t}{\int }}f(s)ds)$

from $x_C-x_2$ for an appropriate $D\in ℝ$ to obtain zero, which is why $x_2$ is then of the desired form.

Thus, let $x_h$ be any solution to the homogenous problem. Consider the function

$t\mapsto x_h(t)\cdot \exp (\underset{t_0}{\overset{t}{\int }}f(s)ds)$.

We differentiate this function and obtain by the product rule

${x_h}^{\prime }(t)\exp (\underset{t_0}{\overset{t}{\int }}f(s)ds)+f(t)\exp (\underset{t_0}{\overset{t}{\int }}f(s)ds)x_h(t)=-f(t)x_h(t)\exp (\underset{t_0}{\overset{t}{\int }}f(s)ds)+f(t)\exp (\underset{t_0}{\overset{t}{\int }}f(s)ds)x_h(t)=0$

since $x_h$ is a solution to the homogenous problem. Hence, the function is constant (that is, equal to a constant $C\in ℝ$), and solving

$x_h(t)\cdot \exp (\underset{t_0}{\overset{t}{\int }}f(s)ds)=C$

for $x_h$ gives the claim.

We have thus arrived at:

Note that imposing a condition $x(t_0)=x_0$ for some $x_0\in ℝ$ enforces $C=x_0$, whence we got a unique solution for each initial condition.

Exercises

• Exercise 3.2.1: First prove that $\frac{d}{dt}\ln (t^2)=\frac{2}{t}$. Then solve the ODE $x^{\prime }(t)+\frac{2}{t}x(t)=\frac{1}{t^2}$ for a function existent on $[1,\mathrm{\infty })$ such that $x(1)=c$ for $c\in ℝ$ arbitrary. Use that a similar version of theorem 3.1 holds when $f,g$ are only defined on a proper part of $ℝ$; this is because the proof carries over.

Clever Ansatz for polynomial RHS

First note that RHS means "Right Hand Side". Let's consider the special case of a 1-dim. first-order linear ODE

$x^{\prime }(t)+cx(t)=a_i{t}^i$ ($c\in ℝ$ arbitrary),

where we used Einstein summation convention; that is, $a_i{x}^i$ stands for $\sum _{i=0}^m{a}_i{t}^i$ for some $m\in \mathbb{N}$. In the notation of above, we have $f(t)\equiv c$ and $g(t)=a_i{t}^i$.

Using separation of variables, the solution to the corresponding homogenous problem $g\equiv 0$ is easily seen to equal $x_h(t)=C\exp (-ct)$ for some capital $C\in ℝ$.

To find a particular solution $x_p$, we proceed as follows. We pick the Ansatz to assume that $x_p$ is simply a polynomial; that is

$x_p(t)=b_i{t}^i$

for certain coefficients $b_i$.

Exercises

• Exercise 3.3.1: Find all solutions to the ODE $x^{\prime }(t)+2x(t)=2t^2+4t+3$. (Hint: What does theorem 3.1 say about the number of solutions to that problem with a given fixed initial condition?)