Ordinary differential equations (ODE) are important in physical modeling and deduction.
Integrating factor method
For functions of two variables, the partial derivative is defined as,
\[ \frac{\partial f(x, y)}{\partial x} = \lim_{\Delta x \to 0} \frac{f(x + \Delta x, y) - f(x, y)}{\Delta x} \]
Similarly, the 2nd partial derivative is,
\[ \begin{aligned} \frac{\partial}{\partial y} \frac{\partial f(x, y)}{\partial x} &= \lim_{\Delta y \to 0} \frac{\frac{\partial}{\partial x}f(x, y+\Delta y) - \frac{\partial}{\partial x}f(x, y)}{\Delta y} \\ &= \lim_{\Delta y \to 0} \lim_{\Delta x \to 0} \frac{1}{\Delta x \Delta y} [ f(x+\Delta x, y+\Delta y) - f(x, y+\Delta y) - f(x+\Delta x, y) + f(x, y) ] \\ &= \frac{\partial}{\partial x} \frac{\partial f(x, y)}{\partial y} \end{aligned} \]
Thus, the 2nd partial derivative is symmetric,
\[ \frac{\partial^2 f}{\partial x \partial y} = \frac{\partial^2 f}{\partial y \partial x} \]
We use this criteria to judge whether the function is exact differential.
exact differential equations
If the ODE satisfies,
\[ A(x,y) dx + B(x,y) dy = 0 \text{, where } \frac{\partial A}{\partial y} = \frac{\partial B}{\partial x} \]
It's the exact differential equation which can be solved by,
\[ \begin{aligned} dU(x,y) &= A(x,y) dx + B(x,y) dy = 0 \\ U(x,y) &= \int A(x,y) \: dx + F(y) = c_{1} \\ \end{aligned} \]
The \(F(y)\) can be found by differentiating the above \(U(x,y)\) with respect to \(y\), which should yield \(B(x,y)\),
\[ F(y) = \int [ B(x,y) - \partial_y ( \int A(x,y) dx ) ] dy \]
inexact differential equations
If the ODE is,
\[ A(x,y) dx + B(x,y) dy = 0 \text{, where } \frac{\partial A}{\partial y} \neq \frac{\partial B}{\partial x} \]
We should introduce an integrating factor \(\mu\) where \(\mu\) satisfies,
\[ \frac{\partial (\mu A)}{\partial y} = \frac{\partial (\mu B)}{\partial x} \]
\(\lambda\) isn't unique so we can just arbitrarily defined \(\mu = \mu(x)\),
\[ \mu \frac{\partial A}{\partial y} = B \frac{\partial \mu}{\partial x} + \mu \frac{\partial B}{\partial x} \] \[ \mu = \exp[ \int \frac{1}{B} (\frac{\partial A}{\partial y} - \frac{\partial B}{\partial x}) \: dx ] \]
Then substitute the integrating factor into the ODE and solve it. It's also suitable for the linear equation,
\[ \frac{dy}{dx} + P(x)y = Q(x) \]
where the integrating factor \(\mu(x) = \exp[\int P(x) \: dx]\).
solutions to Langevin equation
The Langevin equation can be solved by integrating factor method as below. Let's say the Langevin equation is,
\[ \partial_{t} u = - \xi u + R(t) \]
where \(u\) is velocity, \(\xi\) is the friction constant and \(R(t)\) is random force independent of coordinates. We can easily get the integrating factor \(\mu = \exp[ \xi t ]\). Inserting the integrating factor into the differential equation,
\[ \exp[ \xi t ][ \xi u - R(t) ] \ dt + \exp[ \xi t ] \ du = 0 \]
Integrating the second term, we can get,
\[ U(t, u) = u \exp[ \xi t ] + F(t) = C \]
where \(F(t)\) is a sole function about \(t\), which can be determined by the derivative of \(U(t, u)\) with respect to \(t\),
\[ \begin{aligned} \xi u \exp[ \xi t ] + F'(t) &= \exp[ \xi t ] [ \xi u - R(t) ] \\ F'(t) &= - \exp[ \xi t ] R(t) \\ F(t) &= - \int_{0}^{t} \exp[ \xi \tau ] R(\tau) \ d{\tau} \end{aligned} \] \[ \]
Inserting \(F(t)\) into \(U(t, u)\) and considering the boundary condition \(u = u_{0}\) at the initial time \(t=0\), we get the solution \(u(t)\) to the Langevin equation,
\[ u = u_{0} \exp[- \xi t] + \exp[- \xi t] \int_{0}^{t} \exp[\xi \tau] R(\tau) \ d{\tau} \]
Substitution method to separate variables
homogeneous equations
For the \(n\)-degree homogeneous function \(f(x,y)\), we have \(f(\lambda x, \lambda y)=\lambda^{n} f(x,y)\). Thus for the ODE,
\[ \frac{dy}{dx} = \frac{A(x,y)}{B(x,y)} = F(\frac{y}{x}) \]
where \(A(x,y)\) and \(B(x,y)\) are both homogeneous functions of the same degree, it may then be solved by making the substitution \(y = \nu (x) x\). Because \(y\) equals a function of \(x\), \(\frac{y}{x}\) should also be a function of \(x\) as \(\nu (x)\). Differentiate \(y\) with respect to \(x\), we get,
\[\frac{dy}{dx} = \nu + x \frac{d\nu}{dx}\]
Thus the ODE can be solved as,
\[ \int \frac{d\nu}{F(\nu) - \nu} = \int \frac{dx}{x}\]
isobaric equations
More generally, if the ODE isn't homogeneous but it has the form,
\[ \frac{dy}{dx} = \frac{A(x,y)}{B(x,y)} \]
It may then be solved by making the substitution,
\[ y = \nu x^{m} \]
if the substitution makes the ODE dimensionally consistent,
\[ \pmb{[} A(x,y) \pmb{]} + 1 = \pmb{[} B(x,y) \pmb{]} + m \]
where the dimension of \(x\) and \(dx\) is \(1\) and of \(y\) and \(dy\) is \(m\). Then try to substitute it into the ODE, cancel the \(\nu\) factors containing \(x\) and make the ODE separable.
bernoulli's equation
Bernoulli's equation is the form,
\[ \frac{dy}{dx} + P(x)y = Q(x)y^{n} \text{, where } n \neq 0 \text{ or } 1 \]
which can be transformed by dividing \(y^{n}\) on both side,
\[ \frac{1}{1-n} \frac{d(y^{1-n})}{dx} + P(x) y^{1-n} = Q(x) \]
Thus the ODE can be solved as a linear equation described above by making the substitution,
\[ \nu = y^{1-n} \]
Miscellaneous equations
If the ODE has the form,
\[ \frac{dy}{dx} = F(ax+by+c) \]
we can solve the equation by just making the substitution,
\[ \nu = ax + by +c \]
Thus the ODE is transformed to be,
\[ \frac{d\nu}{dx} = a + bF(\nu) \]
Otherwise, if the ODE has the form,
\[ \frac{dy}{dx} = \frac{ax+by+c}{ex+fy+g} \]
we can cancel the constants in both the numerator and denominator by linear transformation,
\[ \begin{aligned} X &= x + \alpha \\ Y &= y + \beta \end{aligned} \]
where \(\alpha\) and \(\beta\) satisfy,
\[ \begin{bmatrix} a & b \\ e & f \\ \end{bmatrix} \begin{bmatrix} \alpha \\ \beta \\ \end{bmatrix} = - \begin{bmatrix} c \\ g \\ \end{bmatrix} \]
Then we get the homogeneous ODE, which can be solved by substitution,
\[ \frac{dY}{dX} = \frac{aX+bY}{eX+fY} \]
Higher-degree first-order ODEs
In higher-degree ODEs it's usually hard to separate variables because three variables (\(x,y,y'\)) are in nonlinear relationship in general. So an intuitive method is to eliminate one variable and degenerate the ODE to the form we are familiar with by the differential relation \(\frac{dy}{dx} = y'\).
equations soluble for y'
If the ODE can be factorised as,
\[ \prod_{i=1}^{n} (y' - F_{i}(x, y)) = 0 \]
The solution for \(y' - F_{i}(x, y) = 0\) is \(y_{i}\),
\[ y_{i} = H_{i}(x) \] \[ y'_{i} = F_{i}(x, y_{i})\] \[ G_{i}(x, y_{i}) = y_{i} - H_{i} = 0 \]
The general solution for the ODE is,
\[ \prod_{i=1}^{n} G_{i}(x, y) = 0 \]
Here is why,
\[ \frac{d}{dx} \prod_{i=1}^{n} G_{i}(x, y) = \sum_{i=1}^{n} \prod_{\substack{j=1 \\ j \neq i}}^{n} G_{j}(x, y) (y' - F_{i}) = 0 \]
\[ \begin{aligned} \sum_{i=1}^{n} \prod_{\substack{j=1 \\ j \neq i}}^{n} G_{j}(x, y) y' &= \sum_{i=1}^{n} \prod_{\substack{j=1 \\ j \neq i}}^{n} G_{j}(x, y) F_{i} \\ y' &= \frac{\sum_{i=1}^{n} \prod_{\substack{j=1 \\ j \neq i}}^{n} G_{j}(x, y) F_{i}}{\sum_{i=1}^{n} \prod_{\substack{j=1 \\ j \neq i}}^{n} G_{j}(x, y)} \\ y' - F_{i} &= G_{i} A_{i} \end{aligned} \]
\(y'\) subtracts \(F_{i}\) cancelled the factor without \(G_{i}\) in the numerator. Thus we can extract the factor \(G_{i}\) from \(p - F_{i}\),
\[ \prod_{i=1}^{n} [ p - F_{i}(x, y) ] = \prod_{i=1}^{n} G_{i} \prod_{i=1}^{n} A_{i} = 0 \]
equations soluble for x
If the ODE can be written as,
\[ x = F(y, y') \]
We can eliminate \(x\) by differentiating both sides with respect to \(y\),
\[ \frac{dx}{dy} = \frac{1}{y'} = \frac{\partial F}{\partial y} + \frac{\partial F}{\partial y'} \frac{dy'}{dy} \]
This results in an equation in the form of \(G(y, y') = 0\), which is likely to be separable.
equations soluble for y
If the ODE can be written as,
\[ y = F(x, y') \]
We can eliminate \(y\) by differentiating both sides with respect to \(x\),
\[ \frac{dy}{dx} = y' = \frac{\partial F}{\partial x} + \frac{\partial F}{\partial y'} \frac{dy'}{dx} \]
This results in an equation in the form of \(G(x, y') = 0\), which is likely to be separable.