Techniques of solving order 1 ODEs

Separable form

If a differential equation is of the form $\frac{dy}{dx} = \frac{P(x)}{Q(y)}$ (or equivalently $y’=P(x)Q(y)$), then we solve the equation by \(\int P(x)dx = \int Q(y)dy.\)

Integral factor (non separable form)

If a differential equation is of the form $y’+P(x)y=Q(x)$ (note that the leading coefficient is 1), then we will find its integral factor \(I=e^{\int P(x)dx}\) and the solution is \(y=\int IQ(x)dx.\)

Direction Field

Direction Field of the logistic model $P’ = k P(1-\frac{1}{P})$

P,t = var('P t') M = 3 k = 0.7 plot_slope_field(k * (1-P/M) * P, (t,0,5), (P,0,M*2))

Direction Field of the logistic model $y’ = \frac{y}{x}$

y,x = var('y x') plot_slope_field(y/x, (x,-5,5), (y,-5,5))

Direction Field of the logistic model $y’ = -\frac{x}{y}$

y,x = var('y x') plot_slope_field(-y/x, (x,-5,5), (y,-5,5))

Direction Field of the logistic model $y’ = \left(\frac{y}{x}\right)^2$

y,x = var('y x') plot_slope_field((y/x)^2, (x,-5,5), (y,-5,5))

Direction Field of the logistic model $y’ = -\left(\frac{x}{y}\right)^2$

y,x = var('y x') plot_slope_field(-(x/y)^2, (x,-5,5), (y,-5,5))

Orthogonal trajectories

Given a family of curves ${C_k}$ where $k$ is a parameter that can be expressed by $x$ and $y$, an orthogonal trajectory of the family is a curve $C$ such that every intersection of $C$ and $C_i$ are perpendicular. To find $C$, we follow the following steps:

Step 1 Find $\frac{dy}{dx}=f(x,y)$ of $C_k$

Step 2 Replace $k$ in $f(x,y)$ in terms of $x$ and $y$.

Step 3 Let $\frac{dy}{dx} = -\frac{1}{f(x,y)}$

Step 4 Solve the last ODE.