Section 1.4 Linear equations and the integrating factor
Objectives
After this section, you will be able to:
-
Identify a linear first-order differential equation and write a first-order linear equation in standard form,
-
Solve initial value problems for first-order linear differential equations by integrating factors, and
-
Write solutions to first-order linear initial value problems in integral form if needed.
One of the most important types of equations we will learn how to solve are the so-called
linear equations. In fact, the majority of the course is about linear equations. In this section we focus on the
first order linear equations.
Definition 1.4.1.
A first order equation is linear if we can put it into the form:
\begin{equation}
y' + p(x) y = f(x).\tag{1.4.1}
\end{equation}
The word "linear" means linear in
\(y\) and
\(y'\text{;}\) no higher powers nor functions of
\(y\) or
\(y'\) appear. The dependence on
\(x\) can be more complicated.
Solutions of linear equations have nice properties. For example, the solution exists wherever
\(p(x)\) and
\(f(x)\) are defined, and has the same regularity (read: it is just as nice). We’ll see this in detail in
Section 1.5. But most importantly for us right now, there is a method for solving linear first order equations. In
Section 1.1, we saw that we could easily solve equations of the form
\begin{equation*}
\frac{dy}{dx} = f(x)
\end{equation*}
because we could directly integrate both sides of the equation, since the left hand side was the derivative of something (in this case, \(y\)) and the right side was only a function of \(x\text{.}\) We want to do the same here, but the something on the left will not be the derivative of just \(y\text{.}\)
The trick is to rewrite the left-hand side of
(1.4.1) as a derivative of a product of
\(y\) with another function. Let
\(r(x)\) be this other function, and we can compute, by the product rule, that
\begin{equation*}
\frac{d}{dx}\Bigl[r(x)y\Bigr] = r(x)y' + r'(x) y.
\end{equation*}
Now, if we multiply
(1.4.1) by the function
\(r(x)\) on both sides, we get
\begin{equation*}
r(x)y' + p(x)r(x)y = f(x)r(x)
\end{equation*}
and the first term on the left here matches the first term from our product rule derivative. To make the second terms match up as well, we need that
\begin{equation*}
r'(x) = p(x)r(x).
\end{equation*}
This equation is separable! We can solve for the \(r(x)\) here by separating variables to get that
\begin{equation*}
\frac{dr}{r} = p(x)\, dx
\end{equation*}
so that
\begin{equation*}
\ln{|r|} = \int p(x)\, dx
\end{equation*}
or
\begin{equation*}
r(x) = e^{\int p(x) \,dx} .
\end{equation*}
With this choice of \(r(x)\text{,}\) we get that
\begin{equation*}
r(x) y' + r(x) p(x) y = \frac{d}{dx}\Bigl[ r(x) y \Bigr],
\end{equation*}
so that if we multiply
(1.4.1) by
\(r(x)\text{,}\) we obtain
\(r(x) y' + r(x) p(x) y\) on the left-hand side, which we can simplify using our product rule derivative above to obtain
\begin{equation*}
\frac{d}{dx}\Bigl[ r(x) y \Bigr] = r(x)f(x) .
\end{equation*}
Now we integrate both sides. The right-hand side does not depend on \(y\) and the left-hand side is written as a derivative of a function. Afterwards, we solve for \(y\text{.}\) The function \(r(x)\) is called the integrating factor and the method is called the integrating factor method.
This method works for any first order linear equation, no matter what \(p(x)\) and \(f(x)\) are. In general, we can compute:
\begin{align*}
y' + p(x) y \amp = f(x) \\
e^{\int p(x) \,dx} y' + e^{\int p(x) \,dx} p(x) y \amp = e^{\int p(x) \,dx} f(x)\\
\frac{d}{dx}\left[ e^{\int p(x) \,dx} y \right] \amp = e^{\int p(x) \,dx} f(x) \\
e^{\int p(x) \,dx} y \amp = \int e^{\int p(x) \,dx} f(x) \,dx + C \\
y \amp = e^{-\int p(x) \,dx} \left( \int e^{\int p(x) \,dx} f(x) \,dx + C \right) \text{.}
\end{align*}
Advice: Do not try to remember the formula itself, that is way too hard. It is easier to remember the process and repeat it.
Of course, to get a closed form formula for
\(y\text{,}\) we need to be able to find a closed form formula for the integrals appearing above.
Example 1.4.2.
Solve
\begin{equation*}
y' + 2xy = e^{x-x^2}, \qquad y(0) = -1 .
\end{equation*}
Solution.
First note that \(p(x) = 2x\) and \(f(x) = e^{x-x^2}\text{.}\) The integrating factor is \(r(x) = e^{\int p(x)\, dx} = e^{x^2}\text{.}\) We multiply both sides of the equation by \(r(x)\) to get
\begin{align*}
e^{x^2} y' + 2xe^{x^2}y \amp = e^{x-x^2} e^{x^2} \\
\frac{d}{dx} \left[ e^{x^2} y \right] \amp = e^x \text{.}
\end{align*}
We integrate
\begin{align*}
e^{x^2} y \amp = e^x +C \\
y \amp = e^{x-x^2} + C e^{-x^2}\text{.}
\end{align*}
Next, we solve for the initial condition \(-1 = y(0) = 1 + C\text{,}\) so \(C=-2\text{.}\) The solution is
\begin{equation*}
y = e^{x-x^2} - 2 e^{-x^2}\text{.}
\end{equation*}
Note that we do not care which antiderivative we take when computing
\(e^{\int p(x) dx}\text{.}\) You can always add a constant of integration, but those constants will not matter in the end.
Checkpoint 1.4.3.
Try it! Add a constant of integration to the integral in the integrating factor and show that the solution you get in the end is the same as what we got above.
Since we cannot always evaluate the integrals in closed form, it is useful to know how to write the solution in definite integral form. A definite integral is something that you can plug into a computer or a calculator. Suppose we are given
\begin{equation*}
y' + p(x) y = f(x) , \qquad y(x_0) = y_0 .
\end{equation*}
Look at the solution and write the integrals as definite integrals.
\begin{equation}
y(x) = e^{-\int_{x_0}^x p(s)\, ds} \left( \int_{x_0}^x e^{\int_{x_0}^t p(s)\, ds}
f(t) \,dt + y_0 \right).\tag{1.4.2}
\end{equation}
You should be careful to properly use dummy variables here. If you now plug such a formula into a computer or a calculator, it will be happy to give you numerical answers.
Checkpoint 1.4.4.
Check that
\(y(x_0) = y_0\) in formula
(1.4.2).
Example 1.4.5.
Solve the initial value problem
\begin{equation*}
ty' + 4y = t^2 - 1 \qquad y(1) = 3.
\end{equation*}
Solution.
In order to solve this equation, we want to put the equation in standard form, which is
\begin{equation*}
y' + \frac{4}{t}y = t - \frac{1}{t}.
\end{equation*}
In this form, the coefficient \(p(t)\) of \(y\) is \(p(t) = \frac{4}{t}\text{,}\) so that the integrating factor is
\begin{equation*}
r(t) = e^{\int p(t)\ dt} = e^{\int \frac{4}{t}\ dt} = e^{4\ln(t)}.
\end{equation*}
Since \(4\ln(t) = \ln(t^4)\text{,}\) we have that \(r(t) = t^4\text{.}\) Multiplying both sides of the equation by \(t^4\) gives
\begin{equation*}
t^4y' + 4t^3 y = t^5 - t^3
\end{equation*}
where the left hand side is \(\frac{d}{dt}(t^4y)\text{.}\) Therefore, we can integrate both sides of the equation in \(t\) to give
\begin{equation*}
t^4 y = \frac{t^6}{6} - \frac{t^4}{4} + C
\end{equation*}
and we can solve out for \(y\) as
\begin{equation*}
y(t) = \frac{t^2}{6} - \frac{1}{4} + \frac{C}{t^4}.
\end{equation*}
To solve for \(C\) using the initial condition, we plug in \(t=1\) to get that we need
\begin{equation*}
3 = \frac{1}{6} - \frac{1}{4} + C \qquad C = \frac{37}{12}.
\end{equation*}
Therefore, the solution to the initial value problem is
\begin{equation*}
y(t) = \frac{t^2}{6} - \frac{1}{4} + \frac{37/12}{t^4}.
\end{equation*}
Example 1.4.6.
Solve the initial value problem
\begin{equation*}
y' + 2xy = 3 \qquad y(0) = 4.
\end{equation*}
Solution.
This equation is already in standard form. Since the coefficient of \(y\) is \(p(x) = 2x\text{,}\) we know that the integrating factor is
\begin{equation*}
r(x) = e^{\int p(x)\ dx} = e^{x^2}.
\end{equation*}
We can multiply both sides of the equation by this integrating factor to give
\begin{equation*}
y'e^{x^2} + 2xe^{x^2}y = 3e^{x^2}
\end{equation*}
and then want to integrate both sides. The left-hand side of the equation is \(\frac{d}{dx}[e^{x^2}y]\text{,}\) so the antiderivative of that side is just \(ye^{x^2}\text{.}\) For the right-hand side, we would need to compute
\begin{equation*}
\int 3e^{x^2}\ dx,
\end{equation*}
which does not have a closed-form expression. Therefore, we need to represent this as a definite integral. Since our initial condition gives the value of \(y\) at zero, we want to use zero as the bottom limit of the integral. Therefore, we can write the solution as
\begin{equation*}
ye^{x^2} = \int_0^x 3e^{s^2}\ ds + C
\end{equation*}
and so can solve for \(y\) as
\begin{equation*}
y(x) = e^{-x^2} \int_0^x 3e^{s^2}\ ds + Ce^{-x^2}.
\end{equation*}
Plugging in the initial condition gives that
\begin{equation*}
y(0) = 4 = e^{-0} \int_0^0 3e^{s^2}\ ds + Ce^{-0} = C.
\end{equation*}
Therefore, the solution to the initial value problem is
\begin{equation*}
y(x) = e^{-x^2} \int_0^x 3e^{s^2}\ ds + 4e^{-x^2}.
\end{equation*}
Checkpoint 1.4.7.
Write the solution of the following problem as a definite integral, but try to simplify as far as you can. You will not be able to find the solution in closed form.
\begin{equation*}
y' + y = e^{x^2-x}, \qquad y(0) = 10 .
\end{equation*}
Answer.
\(y = e^{-x}\int_0^x e^{-s^2}\ ds + 10e^{-x}\)
Exercises Exercises
In the exercises, feel free to leave answer as a definite integral if a closed form solution cannot be found. If you can find a closed form solution, you should give that.
1.
Solve
\(y' + xy = x\text{.}\)
2.
Solve
\(y' + 6y = e^x\text{.}\)
Answer.
\(y = \frac{1}{7}e^x + Ce^{-6x}\)
3.
Solve
\(y' + 4y = x^2e^{-4x}\text{.}\)
Answer.
\(y = \frac{1}{3}x^3e^{-4x} + Ce^{-4x}\)
4.
Solve
\(y' - 3y = xe^x\text{.}\)
Answer.
\(y = -\frac{1}{2}xe^{x} - \frac{1}{4}e^x + Ce^{3x}\)
5.
Solve
\(y' + 3y = e^{4x} - e^{-2x}\) with
\(y(0) = -3\text{.}\)
Answer.
\(y = \frac{1}{7}e^{4x} - e^{-2} + \frac{27}{7}e^{-3x}\)
6.
Solve
\(y' - 2y = x + 4\text{.}\)
Answer.
\(y = -\frac{x}{2} - \frac{9}{4} + Ce^{2x}\)
7.
Solve
\(xy' + 4y = x^2 - \frac{1}{x^2}\text{.}\)
Answer.
\(y = \frac{x^2}{6} - \frac{1}{2x^2} + Cx^{-4}\)
8.
Solve
\(xy' - 3y = x-2\) with
\(y(1) = 3\text{.}\)
Answer.
\(y = -\frac{1}{2}x - \frac{2}{3} + \frac{25}{6}x^3\)
9.
Solve
\(y' - 4y = \cos{(3t)}\text{.}\)
Answer.
\(y = \frac{3}{25}\sin(3t) - \frac{4}{25}\cos(3t) + Ce^{4t}\)
10.
Solve
\(y'+3 x^2 y = x^2\text{.}\)
Answer.
\(y = C e^{-x^3} + \nicefrac{1}{3}\)
11.
Solve
\(y' + 3x^2y = \sin(x) \, e^{-x^3}\text{,}\) with
\(y(0) = 1\text{.}\)
Answer.
\(y = (2-\cos(x))e^{-x^3}\)
12.
Solve
\(y' + \cos (x) y = \cos(x)\text{.}\)
Answer.
\(y = 1 + Ce^{-\sin(x)}\)
13.
Solve the IVP
\(4ty'+y=24\sqrt{t};\ y(10000)=100.\)
Answer.
\(y = 8\sqrt{t} - 7000t^{-1/4}\)
14.
Solve the IVP
\((t^2+1)y' - 2ty = t^2+1; \ y(1)=0\text{.}\)
Answer.
\(y = (t^2 + 1)\left(\tan^{-1}(t) - \frac{\pi}{4}\right)\)
15.
Solve
\(\frac{1}{x^2+1} \, y' + x y = 3\text{,}\) with
\(y(0) = 0\text{.}\)
Answer.
\(y = e^{-\left(\frac{x^4}{4} + \frac{x^2}{2}\right)}\int_0^x 3e^{\frac{s^4}{4} + \frac{s^2}{2}}(s^2 + 1)\ ds\)
16.
Solve
\(y'+ 2\sin(2x) y = 2\sin(2x)\text{,}\) \(y(\nicefrac{\pi}{2}) = 3\text{.}\)
Answer.
\(y = 2 e^{\cos(2x)+1} + 1\)
17.
Consider the initial value problem
\begin{equation*}
5y' - 3y = e^{-2t} \qquad y(0) = a
\end{equation*}
for an undetermined value \(a\text{.}\) Solve the problem and determine the dependence on the value of \(a\text{.}\) How does the value of the solution as \(t \rightarrow \infty\) depend on the value of \(a\text{?}\)
Answer.
\(y = -\frac{1}{13}e^{-2t} + \left(a + \frac{1}{13}\right)e^{3/5}t\text{.}\) If
\(a > -\frac{1}{13}\text{,}\) the solution goes to
\(\infty\text{,}\) if
\(a = -\frac{1}{13}\text{,}\) the solution goes to 0, if
\(a < -\frac{1}{13}\text{,}\) the solution goes to
\(-\infty\text{.}\)
18.
Find an expression for the general solution to
\(y' + 3y = \sin(t^2)\) with
\(y(0) = 2\text{.}\) Simplfy your answer as much as possible.
Answer.
\(y = 2e^{-3t} + e^{-3t}\int_0^t e^{3s}\sin(s^2)\ ds\)
