Section 10.3: Solving First-Order Linear Differential Equations: Integration Factors

A first order linear differential equation is a differential equation of the form

$$y' + a(t) y = b(t)$$

Example

Solve the differential equation

$$y' + t y = 0$$

Solution

In this case we can use the method of separation of variables.

If $y$ is constant, then $ty \equiv y' \equiv 0$ so that $y \equiv 0$.

Otherwise, we may express the equation as $\frac{y'}{y} = -t$. Let $C = y(0)$. Integrating with respect to $t$, we have

$$\begin{eqnarray*} -\frac{1}{2}T^2 & = & \int_{0}^T -t dt \\ & = & \int_0^T \fr... ...nt_C^{y(T)} \frac{dy}{y} \\ & = & \ln \vert\frac{y(T)}{C}\vert \end{eqnarray*}$$

Solution, continued

(As our solution must be continous and cannot take the value zero, the signs of $y(T)$ and $C = y(0)$ must agree. So, we may drop the absolute value bars.)

Exponentiating both sides of this equation and multiplying by $C$, we obtain $y(T) = C e^{\frac{-1}{2} T^2}$.

Another Example

Solve the differential equation

$$y' + y = 10 e^{-t}$$

Solution

In this case, we cannot apply the separation of variables technique.

However, as $e^t$ is never equal to zero, the solutions to the original equation and to the equation

$$e^t y' + e^t y = 10$$

are the same.

Observe that

$$\begin{eqnarray*} \frac{d}{dt} (e^t y) & = & e^t y' + e^t y \end{eqnarray*}$$

Solution, continued

Thus, if our differential equation holds, we have $\frac{d}{dt}(e^t y) = 10$.

We integrate with respect to $t$.

$$\begin{eqnarray*} e^T y(T) - y(0) & = & e^t y(t) \vert_{t = 0}^{t = T} \\ & = ... ...frac{d}{dt} (e^t y) dt \\ & = & \int_0^T 10 dt \\ & = & 10 T \end{eqnarray*}$$

So, if we write $C = y(0)$, then we have $y(T) = 10 e^{-T} T + C e^{-T}$.

A Third Example

Solve the differential equation

$$y' + \frac{1}{t}y = \cos(t)$$

Solution

In this case, multiplying by $t$ we may express the equation as $t y' + y = t \cos(t)$. Using the product rule we check that $\frac{d}{dt}(t y) = t y' + y$.

We integrate this expression.

Note: The original equation is singular at $t = 0$ in the sense that the function $\frac{1}{t}$ is not defined. We need to take for the lower limit of integration some other constant. The number $\pi$ is a convenient choice in this case.

Solution, continued

$$\begin{eqnarray*} T y(T) - \pi y(\pi) & = & \int_\pi^T \frac{d}{dt} (t y) dt \\ ... ... = t$\ and $dv = \cos(t) dt$} \\ & = & T \sin(T) + \cos(T) + 1 \end{eqnarray*}$$

Write $C := y(\pi)$. Then we conclude that

$y(T) = \sin(T) + \frac{1}{T} (\cos(T) + 1 + \pi C)$.

General Solution

In general, if $A'(t) = a(t)$, then

$$\begin{eqnarray*} \frac{d}{dt}(e^{A(t)} y) & = & e^{A(t)} y' + A'(t) e^{A(t)} y \\ & = & e^{A(t)} (y' + a(t) y) \end{eqnarray*}$$

Thus, a differential equation of the form $y' + a(t) y = b(t)$ may be expressed as $\frac{d}{dt} (e^{A(t)} y) = e^{A(t)} (y' + a(t)) = e^{A(t)} b(t)$.

General solution, continued

So, if $\alpha$ is in the domain of the functions $a(t)$ and $b(t)$, we have

$$\begin{eqnarray*} e^{A(T)} y(T) - e^{A(\alpha)} y(\alpha) & = & \int_\alpha^T \f... ...}{dt} (e^{A(t)} y) dt \\ & = & \int_\alpha^T e^{A(t)} b(t) dt \end{eqnarray*}$$

Set $C := e^{A(\alpha)} y(\alpha)$, then $Y(T) = e^{-A(T)} \int_\alpha^T e^{A(t)} b(t) dt + C e^{-A(T)}$.

An example reconsidered

In solving the equation $y' + \frac{1}{t} y = \cos(t)$, we multiplied by $t$ and then observed that $\frac{d}{dt} (t y) = t y' + y = t (y' + \frac{1}{t} y)$.

In terms of the general solution, $a(t) = \frac{1}{t}$ and if $A(t) = \ln\vert t\vert$, then we have $A'(t) = a(t)$.

Note that $e^{A(t)} = e^{\ln\vert t\vert} = \vert t\vert$. So, multiplying by $t$ is the same as multiplying by $e^{A(t)}$ for $t > 0$.

Our general method gives

$$\begin{eqnarray*} y(T) & = & e^{-A(T)} \int_\alpha^T e^{A(t)} b(t) dt + C e^{-A(... ...c{1}{T} \int_\alpha^T t \cos(t) dt + \frac{\alpha y(\alpha)}{T} \end{eqnarray*}$$

To finish, we must choose $\alpha$ and evaluate the above integral.