2.1: Linear Differential Equations

Our first solution method comes from a convenient observation in the next example.

Example \(\PageIndex{1}\)

Determine the general solution of \(t^3y' +3t^2y = 4\).

Solution

Let's try to directly integrate the equation:

\[ \notag
\int (t^3y' + 3t^2y) dt = 4 \int dt.
\]

The integrand on the left is just the product rule

\[ \notag
(t^3y)'  = t^3y' +3t^2y,
\]

and so we can rewrite this differential equation as \((t^3 \cdot y)' = 4\). We then have

\[ \notag
\begin{align*}
\int (t^3y)' dt &= \int 4 dt, \\
t^3y + C_1 &= 4t+C_2, \\
t^3y &= 4t + C, \quad (\text{relabeling} \,\, C = C_2 - C_1) \\
y &= 4t^{-2} + Ct^{-3}.
\end{align*}
\]

(Or does it look better as below?)

\[ \notag
\begin{align*}
(t^3y)' = 4 \quad &\Longleftrightarrow \quad \int (t^3y)' dt = \int 4 dt, \\
&\Longleftrightarrow \quad t^3y + C_1 = 4t+C_2, \\
&\Longleftrightarrow \quad t^3y = 4t + C, \quad (\text{relabeling} \,\, C = C_2 - C_1) \\
&\Longleftrightarrow \quad y = 4t^{-2} + Ct^{-3}.
\end{align*}
\]

Note that we combined the arbitrary constants of integration in the final expression. Without extra information, this expression is called the general solution to the differential equation. To pinpoint a specific solution and the value of \(C\), we need additional information such as the value of \(y(0)\).

 Linear First-Order Differential Equations

Part of what enabled the solution technique in the previous example is that only \(y\) and \(y'\) appeared, and not functional expressions like \((y')^2\) or \(\cos(y)\). This motivates the following definition.

Definition: First Order Linear differential equation

A first-order differential equation is linear if it can be written in the form
\[ \notag
y' + p(t) y = g(t)
\]
for functions \(p(t)\) and \(q(t)\). An equation is said to be written in standard form if it is expressed this way.

Example:

1) \(y^{\prime} + 3y = 0\)
2) \(y^{\prime} - 2y = \cos(t); \quad y(0) = 1\) 
3) \(t^3y' + 3t^2y = 4\) (This is the equation from the first example)

The above examples can be written in this form, and so are linear equations. The following two examples cannot be written in this form, and so are non-linear equations.

4) \(y' = \cos(y) + 1\)
5) \(\frac{dv}{dt} = -kv^2 \quad \) (This is the Bernoulli Drag Equation)

The product rule technique from Example 1 is almost usable for a linear equation written in standard form. We explore this in the next example.

Example \(\PageIndex{2}\)

Determine the unique solution of \(\frac{dy}{dt} + \frac{1}{2}y = \frac{1}{2}e^{\frac{t}{3}}\).

Solution

We can't immediately use the product rule as before, but we can eventually use it by multiplying through the entire equation by some mystery function as in

\[ \notag
\frac{dy}{dt} + \frac{1}{2}y = \frac{1}{2}e^{\frac{t}{3}} \quad \Longleftrightarrow \quad \mu(t)\frac{dy}{dt} + \frac{1}{2}\mu(t)y = \frac{1}{2}\mu(t)e^{\frac{t}{3}},
\]

and ask which function(s) \(\mu(t)\) satisfy the constraint \(\frac{d}{dt}\mu(t) = \frac{1}{2}\mu(t)\). This new equation is similar to a first-order differential equation that we solved in the previous chapter, and its general solution is \(\mu(t) = Ce^{\frac{t}{2}}\). We can choose \(C=1\) so that

\[ \notag
\mu(t) = e^{\frac{t}{2}}.
\]

Now the left-hand side of the new differential equation satisfies the product-rule:

\[ \notag
(\mu(t)y)' = \mu(t)\frac{dy}{dt} + \frac{1}{2}\mu(t)y \quad \Longleftrightarrow \quad (e^{\frac{t}{2}}y)' = e^{\frac{t}{2}}\frac{dy}{dt} + \frac{1}{2}e^{\frac{t}{2}}y.
\]

Now we can solve:

\[ \notag
\begin{align*}
\frac{dy}{dt} + \frac{1}{2}y = \frac{1}{2}e^{\frac{t}{3}} \quad &\Longleftrightarrow \quad e^{\frac{t}{2}}\frac{dy}{dt} + \frac{1}{2}e^{\frac{t}{2}}y = \frac{1}{2}e^{\frac{t}{2}}e^{\frac{t}{3}}, \\
&\Longleftrightarrow \quad (e^{\frac{t}{2}}y)' = \frac{1}{2}e^{\frac{t}{2}+\frac{t}{3}}, \\
&\Longleftrightarrow \quad (e^{\frac{t}{2}}y)' = \frac{1}{2}e^{\frac{5t}{6}}, \\
&\Longleftrightarrow \quad \int (e^{\frac{t}{2}}y)' dt = \frac{1}{2} \int e^{\frac{5t}{6}} dt, \\
&\Longleftrightarrow \quad e^{\frac{t}{2}}y + C_1 = \frac{1}{2} \left(\frac{6}{5} e^{\frac{5t}{6}} + C_2\right), \\
&\Longleftrightarrow \quad e^{\frac{t}{2}}y = \frac{3}{5} e^{\frac{5t}{6}} + C, \quad \left(\text{relabeling} \,\, C = \left(\frac{6}{5}C_2 - C_1\right)\right)\\
&\Longleftrightarrow \quad y(t) = \frac{3}{5} e^{\frac{t}{3}} + Ce^{\frac{-t}{2}}.
\end{align*}
\]

This success motivates the following definition. 

Definition: Integrating Factor

For a Linear First-Order Differential Equation written in standard form \(y' + p(t) y = g(t)\), the function 
\[ \notag
\mu(t) = e^{\int p(t) dt}
\]
is called the integrating factor.

Caution 

When determining an integrating factor, we only need one solution that serves to enable the product rule substitution. For this reason only we can ignore the constant of integration when we calculate \(\int p(t) dt\).

Exercise \(\PageIndex{3}\)

Determine the general solution of \(t^2y' + 2ty = \sin(t)\).

Hint

We cannot use the separation of variables technique for this problem.

Answer

Even though using the integrating factor will always work for a linear equation, we don't need to use it since the left-hand side is already a product-rule expression:

\[ \notag
(t^2y)' = t^2y' + 2ty.
\]

By directly integrating, we get

\[ \notag
\begin{align*}
t^2y' + 2ty = \sin(t) \quad &\Longleftrightarrow \quad (t^2y)' = \sin(t) \\
&\Longleftrightarrow \quad \int (t^2y)' dt = \int \sin(t) dt \\
&\Longleftrightarrow \quad t^2y + C_1 = -\cos(t) + C_2 \\
&\Longleftrightarrow \quad t^2y = -\cos(t) + C \\
&\Longleftrightarrow \quad y(t) = -\frac{\cos(t)}{t^2} + Ct^{-2}.
\end{align*}
\]

Exercise \(\PageIndex{4}\)

Solve the following differential equation: \(t y^{\prime}+9 y=\frac{e^{2t}}{t^5}.\)

Answer

We are then going to create the product rule using the integrating factor of \(u(t) = e^{\int p(t) dt}.\) For the linear differential equation \( y' + p(t)y = g(t).\) We first take the original differential equation and divide both sides by \(t\) to get \(y^{\prime}+9 \frac{y}{t}=\frac{e^{2t}}{t^6}.\) Then we find the integrating factor 

\[ \notag
u(t) = e^{\int p(t) dt} = e^{\int \frac{9}{t} dt} = e^{9\int \frac{1}{t} dt} = e^{9 \ln|t|} = e^{\ln|t|^9} = t^9.
\]

Now that we have our integrating factor, we multiply it by both sides of the differential equation and get \(t^9 (y'  + \frac{9}{t}y) = \frac{e^{2t}}{t^5}\cdot t^9\) which we can rewrite as \(t^9 y'  + 9 t^8 y = e^{2t} t^4\). Then since we used the integrating factor, we have created the product rule and can rewrite the equation as \((t^9 y)' = e^{2t} t^4.\) We then integrate both sides to get \(t^9 y = \int e^{2t} t^4 dt .\) Then we can integrate the right hand side using integration by parts. We then see that our final answer is \( t^9 y =\frac{1}{4} e^{2 t}\left(2 t^4-4 t^3+6 t^2-6 t+3\right) + C\) which then simplifies to \( y =\frac{1}{4t^9} e^{2 t}\left(2 t^4-4 t^3+6 t^2-6 t+3\right) + \frac{C}{t^9}.\)

Exercise \(\PageIndex{5}\)

Solve \(\frac{d y}{d t}=a y+b \) by separating variables.

Answer

By separating the variable we have \( \frac{d y}{a y+b}=d t\) which then implies \( \int \frac{d y}{a y+b}=\int d t\) which simplifies as:

\[ \notag
\begin{align}
&\frac{\ln |a y+b|}{a}=t+C \notag\\
&\implies \ln |a y+b|=a t+C \notag \\ 
& \implies e^{\ln |a y+b|}=e^{a t+C} \notag\\
& \implies |a y+b|=e^C e^{a t} \notag \\
& \implies  a y+b= \pm\left(e^C e^{a t}\right) \notag\\
& \implies a y=C e^{a t}-b \notag\\
& \implies y=C e^{a t}-\frac{b}{a} \notag
\end{align}
\]

Thus \(  y=C e^{a t}-\frac{b}{a}\) is a general solution to the equation.

 

Note

To solve these types of problems, most of the time we are going to follow a general procedure:
1) Write the equation so that it is in the form \(y' + p(t) y = g(t)\)
2) Make the product rule appear using the integrating factor. 
3) Integrate and solve.

 

Example \(\PageIndex{6}\)

Solve \(y' + p(t) y = g(t)\).

Solution

First we have the integrating factor \(e^{\int p(t) dt}\) and then multiplying it on both sides of the equation we have
\[ \displaystyle \begin{align*}
e^{\int p(t) dt}(y' + p(t) y) &= e^{\int p(t) dt} g(t) \notag\\
e^{\int p(t) dt}y' + e^{\int p(t) dt}p(t) y &= e^{\int p(t) dt} g(t) \notag
\end{align*}
\]
Note that \(e^{\int p(t) dt}y' + e^{\int p(t) dt}p(t) y = (e^{\int p(t) dt}y)' \) by the product rule. Therefore, 
\[ \displaystyle \begin{align*}
(e^{\int p(t) dt}y)' &= e^{\int p(t) dt} g(t)dtd \notag\\
 \int (e^{\int p(t) dt}y)'dt &= \int e^{\int p(t) dt} g(t)dt \notag\\
e^{\int p(t) dt}y &= \int e^{\int p(t) dt} g(t)dt \notag\\
y &= e^{-\int p(t) dt} \cdot \int e^{\int p(t) dt} g(t)dt \notag
\end{align*}
\]

Caution

It is not recommended to use the final line above as a formula to solve linear first-order differential equations.

Example \(\PageIndex{7}\) (With Integrating Factor)

Determine the general solution of \(t^3y' +3t^2y = 4\). Why do this example again?

Solution

First we see that this is a First Order Linear Differential equation as it can be rewritten as \(y'+3t^{-1}y = 4t^{-3}.\) So we have \(p(t) = \frac{3}{t}\) and \(g(t) = \frac{4}{t^3}.\) Thus our integrating factor will be
\[e^{\int \frac{3}{t} dt} = e^{3 \ln|t| + C} = Ke^{\ln|t|^3} = K|t|^3, \notag\]
where \(K = e^C.\) Now we will multiply both sides of the equation by our integrating factor and get the following:
\[
\notag \begin{align*}
t^3(y'+3t^{-1}y) &= t^3(4t^{-3})\\
t^3y'+3t^{2}y &= 4\\
(t^3 y)' &= 4\\
\int (t^3 y)' dt &= \int 4 dt\\
t^3 y &= \int 4 dt\\
t^3 y &= 4t+C\\
y &= 4t^{-2}+Ct^{-3}\\
\end{align*}
\]
Note that our final answer using this method is the same as in our if we didn't use the integrating factor.