We consider linear first order differential equations of the form:
A first order linear differential equation can be defined in the form $$\frac{dy}{dt} + P(t)\,y(t) = g(t)$$ where $P(t)$ and $g(t)$ are functions of $t$.
Example. The falling body problem $\;\dfrac{dv}{dt} = 9.8 - \dfrac{v}{5}\;$ can be written as
$$\frac{dv}{dt} + \frac{v}{5} = 9.8$$which is in the above form with $P(t) = \dfrac{1}{5}$ and $g(t) = 9.8$ (both constants).
We illustrate the Integration Factor (IF) method by the following examples.
Consider $\;\dfrac{dv}{dt} + \dfrac{v}{5} = 9.8.$
Multiply both sides by a function $\mu(t)$:
$$\mu(t)\frac{dv}{dt} + \mu(t)\frac{v}{5} = 9.8\,\mu(t). \tag{1}$$We demand that the left side equals $\dfrac{d}{dt}\!\big[\mu(t)\,v(t)\big]$:
$$\big(\mu(t)\,y(t)\big)' = 9.8\,\mu(t). \tag{2}$$Expanding the left side via the product rule (recall $(uv)' = u'v + uv'$):
$$\mu(t)\frac{dv}{dt} + \frac{d\mu(t)}{dt}\,v(t) = 9.8\,\mu(t). \tag{3}$$Comparing equations (1) and (3):
This implies
$$\mu'(t) = \frac{1}{5}\,\mu(t)$$must hold for the relevant $t$. Solving (by observation or separation of variables):
$$\mu(t) = Ae^{t/5}$$for some $A \neq 0$. We may choose $A = 1$ since we can divide both sides of (3) by $A$. Substituting back into (2):
$$\big(e^{t/5}\,y(t)\big)' = 9.8\,e^{t/5}.$$Integrating both sides:
$$e^{t/5}\,y(t) = \int 9.8\,e^{t/5}\,dt = 9.8 \times 5\,e^{t/5} + c.$$Rearranging:
which is the same expression obtained by separation of variables earlier (apart from the $|\cdot|$ sign issue, but the final solution with initial condition $v(0) = 0$ is the same for both methods).
Solve $\;\dfrac{dy}{dt} + \dfrac{1}{2}\,y = \dfrac{1}{2}\,e^{t/3}.$
Multiply both sides by $\mu(t)$:
$$\mu(t)\frac{dy}{dt} + \mu(t)\frac{1}{2}\,y = \mu(t)\frac{1}{2}\,e^{t/3}$$and demand the left side equals $\dfrac{d}{dt}\!\big[\mu(t)\,y(t)\big]$. Comparing via the product rule:
So we need $\;\dfrac{d\mu(t)}{dt} = \dfrac{1}{2}\,\mu(t).$ Solving:
$$\frac{t}{2} + c = \int \frac{1}{2}\,dt = \int \frac{\mu'}{\mu}\,dt = \ln|\mu(t)|.$$Thus $\mu(t) = Ce^{t/2}$. Choosing $C=1$:
$$\frac{d}{dt}\big(e^{t/2}\,y\big) = \frac{1}{2}\,e^{t/2}\cdot e^{t/3} = \frac{1}{2}\,e^{5t/6}.$$Integrating:
$$e^{t/2}\,y = \int \frac{1}{2}\,e^{5t/6}\,dt = \frac{3}{5}\,e^{5t/6} + c.$$Hence
Solve $\;\dfrac{dy}{dt} - 2y = 4 - t.$
Multiplying by $\mu$ and comparing via the product rule gives $\mu' = -2\mu$, i.e.
$$\frac{\mu'}{\mu} = -2 \implies \ln|\mu| = -2t + c \implies \mu = Ce^{-2t}.$$Choosing $C = 1$:
$$\frac{d}{dt}\big(e^{-2t}\,y\big) = e^{-2t}(4-t).$$Integrating (using integration by parts for $\int te^{-2t}\,dt$):
$$e^{-2t}\,y(t) = \int e^{-2t}(4-t)\,dt = \int 4e^{-2t}\,dt - \int te^{-2t}\,dt.$$Computing each integral:
$$\int 4e^{-2t}\,dt = -2e^{-2t}$$For $\int te^{-2t}\,dt$, let $u = t$, $dv = e^{-2t}\,dt$, so $du = dt$, $v = -\tfrac{1}{2}e^{-2t}$:
$$\int te^{-2t}\,dt = -\frac{1}{2}te^{-2t} - \int -\frac{1}{2}e^{-2t}\,dt = -\frac{1}{2}te^{-2t} - \frac{1}{4}e^{-2t}.$$Combining:
$$e^{-2t}\,y = -2e^{-2t} - \Big(-\frac{1}{2}te^{-2t} - \frac{1}{4}e^{-2t}\Big) + k = -\frac{7}{4}e^{-2t} + \frac{1}{2}te^{-2t} + k.$$Thus
For the general equation $\;\dfrac{dy}{dt} + P(t)\,y = g(t)$, multiply by $\mu(t)$:
$$\mu(t)\frac{dy}{dt} + \mu(t)P(t)\,y(t) = \mu(t)\,g(t).$$We demand that $\mu(t)\,P(t) = \mu'(t)$, i.e.
$$\frac{\mu'(t)}{\mu(t)} = P(t).$$Integrating:
$$\ln|\mu(t)| = \int P\,dt + c, \qquad |\mu(t)| = e^c \cdot e^{\int P\,dt}.$$Hence $\mu(t) = Ce^{\int P\,dt}$. Choosing $C = 1$:
$$\frac{d}{dt}\!\Big(e^{\int P\,dt}\,y(t)\Big) = e^{\int P\,dt}\,g(t).$$A solution of $\;\dfrac{dy}{dt} + P(t)\,y = g(t)\;$ can be written as $$\boxed{\;y(t) = e^{-\int P\,dt}\!\left(\int e^{\int P\,dt}\,g(t)\,dt + c\right).\;}$$ This is called the Integration Factor formula.
Using the formula directly: $P(t) = \tfrac{1}{2}$, so
$$\mu(t) = e^{\int \frac{1}{2}\,dt} = e^{t/2}.$$Hence
$$y(t) = e^{-t/2}\!\left(\int e^{t/2}\cdot\frac{1}{2}\,e^{t/3}\,dt + c\right) = \frac{3}{5}\,e^{t/3} + c\,e^{-t/2},$$as expected.
We immediately have $\;\mu(t) = e^{\int P\,dt} = e^{\int -2\,dt} = e^{-2t}.$ Hence
$$y(t) = e^{2t}\!\left(\int e^{-2t}(4-t)\,dt + c\right) = -\frac{7}{4} + \frac{t}{2} + k\,e^{2t}.$$Here $P(t) = 2/t$, so
$$\mu(t) = e^{\int 2/t\,dt} = e^{2\ln t} = t^2.$$The IF formula gives
$$y(t) = t^{-2}\!\left(\int t^2 \cdot 4t\,dt + c\right) = t^{-2}\!\left(t^4 + c\right) = t^2 + \frac{c}{t^2}.$$Applying $y(1) = 2$:
$$2 = y(1) = 1^2 + \frac{c}{1^2} = 1 + c \implies c = 1.$$The direction field plot for this DE shows that at $x = 0$ the field becomes "vertical" — i.e. $y'$ is infinite or undefined. This is because the differential equation is undefined when $x = 0$.
Rewrite as $\;y' + \dfrac{t}{2}\,y = 1.$ Here $P(t) = t/2$, so
$$\mu(t) = e^{\int t/2\,dt} = e^{t^2/4}.$$The IF formula gives
$$y(x) = e^{-t^2/4}\!\left(\int_0^t e^{s^2/4}\cdot 1\,ds + c\right) = e^{-t^2/4}\int_0^t e^{s^2/4}\,ds + c\,e^{-t^2/4}$$since we do not know how to find the primitive of $\int e^{t^2/4}\,dt$ in closed form (it is a function of $t$). Applying the initial condition $y(0) = 1$:
$$1 = y(0) = 0 + c \implies c = 1.$$Solve and draw the direction field of
$$ty' + 2y = \sin t \quad (t > 0).$$Investigate what happens when $t \to \infty$.
Solve the following first order DEs:
Solve the differential equation:
$$\frac{dy}{dt} + 3y = 6t$$What method should we use to solve this DE?
What's the first step?
What's the next step?
After finding $\mu(t)$, what's next?
Now we have $\frac{d}{dt}(\mu y) = \mu g$. What's next?
What is the integrating factor $\mu(t) = e^{\int P\,dt} = e^{\int 3\,dt}$?
After multiplying by $\mu = e^{3t}$, the left side becomes $\dfrac{d}{dt}(e^{3t}y)$. What integral do we evaluate on the right?
Evaluate $\int 6te^{3t}\,dt$ by parts. Let $u = 6t$, $dv = e^{3t}\,dt$. Then $du = 6\,dt$, $v = \frac{1}{3}e^{3t}$. The integral is:
We have $e^{3t}y = 2te^{3t} - \frac{2}{3}e^{3t} + c$. Solving for $y$:
Solve the initial value problem:
$$y' + \frac{2}{t}\,y = 4t, \quad y(1) = 2$$What method should we use?
First step?
After finding $\mu(t) = t^2$, what's next?
After integrating $\frac{d}{dt}(t^2 y) = 4t^3$ to get the general solution, what's the final step?
$\mu(t) = e^{\int 2/t\,dt} = e^{2\ln t} = \;?$
Multiply: $t^2 y' + 2ty = 4t^3$. The left side is $\frac{d}{dt}(t^2 y)$. Integrating: $t^2 y = \int 4t^3\,dt = \;?$
So $t^2 y = t^4 + c$, hence $y = t^2 + \frac{c}{t^2}$. Apply $y(1) = 2$:
The particular solution is:
For each DE, identify the correct integrating factor $\mu(t)$.
$\;y' + \dfrac{1}{t}\,y = t^2$. What is $\mu(t)$?
$\;y' - \dfrac{2}{t}\,y = t^3$. What is $\mu(t)$?
$\;y' + (\cos t)\,y = \sin t$. What is $\mu(t)$?
$\;2y' + ty = 2$ (rewrite as $y' + \tfrac{t}{2}y = 1$). What is $\mu(t)$?
Solve the differential equation (from lecture):
$$\frac{dy}{dt} - 2y = 4 - t$$What method should we use?
First, identify $P(t)$ and compute $\mu(t)$:
After multiplying by $\mu = e^{-2t}$, we'll need to integrate $\int e^{-2t}(4-t)\,dt$. What technique is needed?
For $\int te^{-2t}\,dt$ by parts, what should we choose for $u$ and $dv$?
$\displaystyle\int 4e^{-2t}\,dt = \;?$
For $\int te^{-2t}\,dt$: let $u = t$, $dv = e^{-2t}\,dt$, so $du = dt$, $v = -\frac{1}{2}e^{-2t}$. By IBP:
$e^{-2t}y = \int e^{-2t}(4-t)\,dt = -2e^{-2t} - (-\frac{1}{2}te^{-2t} - \frac{1}{4}e^{-2t}) + k$. Simplify:
Dividing by $e^{-2t}$ (i.e., multiplying by $e^{2t}$):
— End of Integration Factors Notes —