Exact 1st-order ODEs
\(\rightarrow\) Specific type of ODE
\(\rightarrow\) Method of solving exact \(1^\circ\) ODEs was developed by guessing and messing around.
\(\rightarrow\) Now, if we can classify as “exact”, we have a method to find the solution.
The \(1^\circ\) ODE \(\frac{dy}{dx} = f(x,y)\) is exact if:
It can be written as \(M(x,y)dx + N(x,y)dy = 0 \hspace{3cm} (*)\)
There is some function \(u(x,y)\) such that
(114)\[\begin{align}
\frac{\partial u}{\partial x} =M(x,y) && \text{and} && \frac{\partial u}{\partial y}=N(x,y)
\end{align}\]
\(\rightarrow\) plugging this into \((*)\), we see:
(115)\[\begin{align}
\frac{\partial u}{\partial x}dx + \frac{\partial u}{\partial y} dy = 0\\
\therefore du = 0
\end{align}\]
\(\rightarrow\) integrating this gives \(u(x,y) = c\)
\(\rightarrow\) Therefore if we have an exact equation \(y'=f(x,y)\), then the general solution can be written as \(u(x,y) = c \implies\) function of \(x,y\) gives a constant, not a derivative.
To prove exactness:
(116)\[\begin{align}
\frac{\partial u}{\partial x} = M(x,y) && \text{and} && \frac{\partial u}{\partial y} = N(x,y) \hspace{1cm} ?
\end{align}\]
This strategy can also be applied to higher order differential equations, but it’s not as helpful there so I’m not going to cover that in lectures.
(117)\[\begin{align}
\frac{\partial^2u}{\partial x\partial y} &= \frac{\partial^2u}{\partial y\partial x}\\
\frac{\partial}{\partial x}\left(\frac{\partial u}{\partial y} \right) &= \frac{\partial}{\partial y}\left(\frac{\partial u}{\partial x} \right)\\
\frac{\partial}{\partial x}N(x,y) &= \frac{\partial}{\partial y}M(x,y) && \leftarrow \text{if exact, this will be true}
\end{align}\]
Example
\[3xy^2y' + y^3 = 0\]
\(3xy^2y' + y^3 = 0 \rightarrow\) we can solve 2 ways: by separating and by exactness
Separate
(118)\[\begin{align}
3xy^2 \frac{dy}{dx} = -y^3\\
\frac{3}{y} dy = -\frac{1}{x}dx\\
3 \ln y = - \ln x + \beta\\
\ln (y^3) + \ln x = \beta\\
\ln(xy^3) = \beta\\
xy^3 = \exp \beta \\
xy^3 = \alpha &&\rightarrow \text{implicit solution(didn't solve for y) by separation}
\end{align}\]
Let’s check to see if we can solve by exactness:
a) Write as \(M(x,y)dx + N(x,y)dy=0\):
(119)\[\begin{align}
y^3dx + 3xy^2 dy = 0
\end{align}\]
b) Check if \(\frac{\partial}{\partial x}N(x,y) = \frac{\partial}{\partial y} M(x,y)\)
(120)\[\begin{align}
\frac{\partial}{\partial x} (3xy^2) &= \frac{\partial}{\partial y}(y^3)\\
3y^2 &= 3y^2 && \rightarrow \text{ODE is exact; we can continue}
\end{align}\]
c) Find \(u(x,y)\)
(121)\[\begin{align}
\text{we know that} && \frac{\partial u}{\partial x} = M(x,y) = y^3\\
\text{and} && \frac{\partial u}{\partial y} = N(x,y) = 3xy^2\\
\end{align}\]
Since these are partial derivatives, we can integrate them as follows:
(122)\[\begin{align}
\frac{\partial u}{\partial x} &= y^3\\
\int du &= \int y^3 dx\\
u(x,y) &= xy^3 + k(y)
\end{align}\]
we need to include the last term because its derivative wr.t. \(x=0\)
We can find \(k(y)\) by considering our second partial derivative:
(123)\[\begin{align}
\frac{\partial u}{\partial y} = 3xy^2\\
\frac{\partial}{\partial y}(xy^3 + k(y)) = 3xy^2\\
3xy^2 + \frac{dk}{dy} = 3xy^2\\
\frac{dk}{dy} = 0 \implies k(y) = k && \text{(constant)}\\
\therefore u(x,y) = xy^3 + k
\end{align}\]
d) Use \(u(x,y)\) to identify solution
(124)\[\begin{align}
u(x,y) = c = xy^3 + k \\
\implies xy^3 = \alpha
\end{align}\]
Implicit solution; same as when we separated.
Example
(125)\[\begin{align}
y' = \frac{-2xy^3 - 2}{3x^2y^2 + e^y} && \text{(not separable)}
\end{align}\]
Put into the form \(M(x,y)dx + N(x,y)dy = 0\)
(126)\[\begin{align}
(2xy^3 + 2) dx + (3x^2y^2 + e^y)dy = 0
\end{align}\]
Check for exactness
(127)\[\begin{align}
\frac{\partial}{\partial x}N(x,y) &= \frac{\partial}{\partial y}M(x,y) && \text{(must be true)}\\
\frac{\partial}{\partial x} (3x^2y^2 + e^y) &= \frac{\partial}{\partial y} (2xy^3 + 2)\\
3y^2\frac{\partial}{\partial x} (x^2) &= 2x \frac{\partial}{\partial y}(y^3)\\
6y^2x &=6xy^2 &&\text{(exact)}
\end{align}\]
Now find \(u(x,y)\) such that
(128)\[\begin{align}
\frac{\partial u}{\partial y} = N(x,y)=3x^2y^2 + e^y && \text{and} && \frac{\partial u}{\partial x} = M(x,y)=2xy^3 + 2 && \text{(can start with either)}
\end{align}\]
(129)\[\begin{align}
\int \partial u = \int (3x^2y^2 + e^y) dy\\
u(x,y) = x^2y^3 + e^y + k(x)
\end{align}\]
\(\rightarrow\) determine \(k(y)\) with second partial derivative:
(130)\[\begin{align}
\frac{\partial}{\partial x}(x^2y^3 + e^y + k(x)) &= 2xy^3 + 2\\
2xy^3 + \frac{dk}{dx} &= 2xy^3 + 2\\
\frac{dk}{dx} &= 2\\
k(x) &= 2x + k
\end{align}\]
\(\therefore u(x,y) = x^2y^3 + e^y + 2x + k\)
4. Find implicit solution using \(u(x,y) = c\)
(131)\[\begin{align}
c = x^2y^3 + e^y + 2x + k\\
x^2 y^3 + e^y + 2x = \alpha && \text {general solution}
\end{align}\]
Integrating Factors
(132)\[\begin{align}
P(x,y)dx + Q(x,y) dy = 0
\end{align}\]
Where \(\frac{\partial P}{\partial y} \neq \frac{\partial Q}{\partial x} \implies\) not exact
We can try to find an I.F., \(F(x,y)\), s.t. \(F(x,y)P(x,y)dx + F(x,y)Q(x,y)dy = 0\) is an exact equation. This would require:
(133)\[\begin{align}
\frac{\partial}{\partial y}[F(x,y)P(x,y)] &= \frac{\partial}{\partial x} [F(x,y)Q(x,y)]\\
F(x,y)\frac{\partial P}{\partial y} + P(x,y)\frac{\partial F}{\partial y} &= F(x,y)\frac{\partial Q}{\partial x} + Q(x,y)\frac{\partial F}{\partial x}
\end{align}\]
Knowing only P(x,y) and Q(x,y) would make this nearly impossible to solve. Can use the following strategy:
First assume that \(F(x,y) = F(x)\) only. Now,
(134)\[\begin{align}
\frac{\partial F}{\partial y} = 0 && \text{and} && \frac{\partial F}{\partial x} = \frac{dF}{dx} && \text{(partial becomes full)}
\end{align}\]
(135)\[\begin{align}
\therefore F(x) \frac{\partial P}{\partial y} + P(x,y) \frac{\partial F}{\partial y} &= F(x) \frac{\partial Q}{\partial x} + Q(x,y) \frac{dF}{dx}\\
F(x)\left[\frac{\partial P}{\partial y} - \frac{\partial Q}{\partial x} \right] &= Q(x,y)\frac{dF}{dx}\\
\frac{1}{F(x)}\frac{dF}{dx} &= \frac{1}{Q(x,y)}\left[\frac{\partial P}{\partial y} - \frac{\partial Q}{\partial x} \right]
\end{align}\]
we know the RHS term, call it \(R(x,y)\)
\(\rightarrow\) If \(R(x,y) = R(x)\) only, then you can find \(F(x)\) from integration
(136)\[\begin{align}
\frac{\partial F}{F} = R(x)dx \implies F(x) = \exp\left[\int R(x)dx \right]
\end{align}\]
\(\rightarrow\) If \(R(x,y)\neq R(x)\), then out assumption that \(F(x,y)=F(x)\) is not true. Then try step 2.
Assume \(F(x,y) = F(y)\) only. Then,
(137)\[\begin{align}
\frac{\partial F}{\partial x} = 0 && \text{and} && \frac{\partial F}{\partial y} = \frac{dF}{dy}
\end{align}\]
Partial derivative equation simplifies to:
(138)\[\begin{align}
\frac{1}{F} \frac{dF}{dy} = \frac{1}{P(x,y)}\left[\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} \right]
\end{align}\]
Again, let RHS = \(R(x,y)\)
\(\rightarrow\) If \(R(x,y) = R(y)\) only then \(F(y) = \exp \left[\int R(y)dy \right]\)
\(\rightarrow\)If \(R(x,y) \neq R(y)\), then \(F(x,y)\) = function of both \(x\) and \(y\) and you will need some other clue to solve.
Example
(139)\[\begin{align}
\frac{dy}{dx} = \frac{-2xy}{4y + 3x^2}
\end{align}\]
rewrite as: \((2xy)dx + (4y + 3x^2) dy = 0\)
check for exactness:
(140)\[\begin{align}
\frac{\partial P}{\partial y} &= \frac{\partial Q}{\partial x}\\
2x &\neq 6x \implies \text{not exact}
\end{align}\]
Try to find an I.F. \(F(x,y)=F(x)\)
(141)\[\begin{align}
\frac{\partial (FP)}{\partial y} &= \frac{\partial (FQ)}{\partial x}\\
F\frac{\partial P}{\partial y} + P\frac{\partial F}{\partial y} &= F \frac{\partial Q}{\partial x} + Q\frac{\partial F}{\partial x}\\
\therefore \frac{1}{F} \frac{dF}{dx} &= \frac{1}{Q}\left(\frac{\partial P}{\partial y} - \ \frac{\partial Q}{\partial x}\right)\\
&=\frac{1}{4y + 3x^2}(2x-6x)\\
&=-\frac{4x}{4y+3x^2} \implies \text{($R(x,y)$ depends on $x$ and $y\rightarrow$ no good)}
\end{align}\]
Try to find I.F. \(F(x,y) = F(y)\)
(142)\[\begin{align}
\frac{1}{F} \frac{dF}{dy} &= \frac{1}{P}\left(\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} \right)\\
&=\frac{1}{2xy}(6x-2x) = \frac{2}{y} = R(y) && \text{(good!)}
\end{align}\]
Continuing,
(143)\[\begin{align}
\int \frac{dF}{F} &= \int \frac{2}{y}dy\\
\ln F(y) &= 2 \ln y\\
F(y) &= y^2
\end{align}\]
We can now make our equation exact using \(F(y)\)
(144)\[\begin{align}
F(y)[P(x,y)dx + Q(x,y)dy] = 0\\
y^2 [(2xy)dx + (4y + 3x^2)dy] = 0\\
2xy^3 dx + (4y^3 + 3x^2y^2)dy = 0
\end{align}\]
Check for exactness:
(145)\[\begin{align}
\frac{\partial}{\partial y}(2xy^3) &= \frac{\partial}{\partial x}(4y^3 + 3x^2y^2)\\
6xy^2 &=6xy^2 && \text{exact}
\end{align}\]
Find \(u(x,y)\)
(146)\[\begin{align}
\frac{\partial u}{\partial x} = 2xy^3 \implies u(x,y) = x^2y^3 + k(y)\\
\frac{\partial u}{\partial y} = \frac{\partial}{\partial y} (x^2y^3 + k(y)) = 4y^3 + 3x^2y^2\\
3x^2y^2 + \frac{dk}{dy} = 4y^3 + 3x^2y^2\\
k(y) = y^4 + k
\end{align}\]
Find implicit solution using \(u(x,y) = c\)
(147)\[\begin{align}
x^2y^3 + y^4 + k = c\\
x^2 y^3 + y^4 = \alpha && \text{general solution}
\end{align}\]
