Lecture3 - integrating factor, separable equations, and exact solution,#
Integrating factor#
Consider the first-order ODE:
Note that now a is a function of t. We solve it by an integrating factor M (M(0)=1):
$$ \frac{dM}{dt} = -a(t)M. $$
If a is a constant, then $$ M = e^{-at}. $$
If a is a function of t, then $$ M = e^{-\int_{0}^{t}a(s)ds}. $$
The idea is to consider My and make it in the form of a total derivative: $$ \frac{dMy}{dt} = \frac{dy}{dt}M - a(t)My = Mq $$
Integrate both side from 0 to t, we can get: $$ M(t)y(t) - M(0)y(0) = \int^{t}_{0} M(s)q(s)ds $$
$$ \rightarrow M(t)y(t) = y(0) + \int^{t}_{0} M(s)q(s)ds $$
Divided by M on both sides: $$ y(t) = y(0)e^{\int^{t}_{0} a(s)ds} + \frac{\int^{t}_0{M(s)q(s)ds}}{e^{-\int^{t}_0{a(s)ds}}} $$
$$ \rightarrow y(t) = y(0)e^{\int^{t}_{0} a(s)ds} + \int^{t}_0{e^{-\int^{s}_0{a(T)dT}}q(s)ds}\cdot e^{\int^{t}_0{a(k)dk}} $$
$$ \rightarrow y(t) = y(0)e^{\int^{t}_{0} a(s)ds} + \int^{t}_0{e^{\int^{t}_0{a(k)dk - \int^{s}_0{a(T)dT}}}q(s)ds} $$
Note
If a is a constant, above equation gives “the formula” for very particular solution: $$ y_{vp}(t) = \int^{t}_0{e^{a(t-s)}q(s)ds} $$
And the complete solution is: $$ y(t) = y(0)e^{at} + \int^{t}_0{e^{a(t-s)}q(s)ds} $$
Note
In engineering school or discipline, people use different notations.
“Geometrically, the general solution of an ODE is a family of infinitely many solution curves, one for each value of the constant c. If we choose a specific c, we obtain what is called a particular solution of the ODE. A particular solution does not contain any arbitrary constants.” from Advanced Engineering Mathematics by Erwin Kreyszig.
We call them [general solution] and [particular solution], respectively.
Example 1
Using integrating factor to solve $$ t\frac{dy}{dt}-y=t^{3} $$
Rearrange equation: $$\frac{dy}{dt}-\frac{1}{t}y = t^{2} \rightarrow \frac{dy}{dt} = \frac{1}{t}y + t^{2}$$
Integrating factor: M(t) = $$e^{-\int \frac{1}{t}dt} = e^{-ln(t)} = \frac{1}{t}$$
Rearrange equation: $$\frac{1}{t}\frac{dy}{dt} - \frac{1}{t^2}y = t $$
Take integration: $$\frac{d\frac{y}{t}}{dt} = t \rightarrow \frac{y}{t} = \frac{1}{2}t^{2}+C \rightarrow y=\frac{t^3}{2}+Ct$$
Use initial condition to get C.
Separable equations#
Consider ODE in the form:
$$ \frac{dy}{dt} = \frac{g(t)}{f(y)}. $$
Then we can get and solve for y(t): $$ \int_{y(0)}^{y(t)} f(y)dy = \int_{0}^{t}g(s)ds. $$
Example 2
$$ \frac{dy}{dt} = \frac{t}{y} \rightarrow \int_{y(0)}^{y(t)} ydy = \int_{0}^{t}tdt = \frac{1}{2}t^2 $$
$$ \rightarrow \frac{1}{2}y(t)^{2} - \frac{1}{2}y(0)^{2} = \frac{1}{2}t^2 $$
$$ \rightarrow y(t)^{2} = y(0)^{2} + t^{2} \rightarrow y(t) = \pm\sqrt{y(0)^{2} + t^{2}} $$
Note that y(0) allows y=t and y=-t.
Exact solution#
Consider ODE in the form: $$ \frac{dy}{dt} = \frac{g(y,t)}{f(y,t)} \rightarrow f(y,t)dy = g(y,t)dt \rightarrow f(y,t)dy - g(y,t)dt = 0 $$
The idea is that if we can write and treat y and t independent variables: $$ \frac{\partial \tilde{F}}{\partial y} = f(y,t) $$ $$ \frac{\partial \tilde{F}}{\partial t} = -g(y,t) $$ The we can have a total deriviative form of this problem and can solve it: $$ D\tilde{F} = \frac{\partial \tilde{F}}{\partial y}dy + \frac{\partial \tilde{F}}{\partial t}dt = 0 $$
Step 1: $$ \int f(y,t)dy = \tilde{F}(y,t) + \tilde{C}(t) \rightarrow \tilde{F}(y,t) = \int f(y,t)dy + C(t) = F(y,t) + C(t) $$ We let $$ \int f(y,t)dy = F(y,t) $$
Step 2, choose C(t) such that: $$ \frac{\partial}{\partial t} (F(y,t)+C(t)) = -g(y,t) $$
Step 3: $$ \frac{dy}{dt} = \frac{g(y,t)}{f(y,t)} $$ is solved by $$ F(y,t) + C(t) = constant $$
Example 3
$$ \frac{dy}{dt} = \frac{2yt-1}{y^{2}-t^{2}}, $$ $$ g(y,t) = 2yt-1, f(y,t) = y^{2}-t^{2} $$
Step 1: $$ \int fdy = \int y^{2}-t^{2}dy = \frac{1}{3}y^{3}-t^{2}y + C(t) $$
Step 2: $$ \frac{\partial}{\partial t} (\frac{1}{3}y^{3}-t^{2}y+C(t)) = -2ty + C’(t) = -2yt+1 = -g(y,t) $$ $$ \rightarrow C’(t) = 1 \rightarrow C(t) = t $$
Step 3: $$ \frac{1}{3}y^{3}-t^{2}y+t = constant $$ is a solution in implicit form.