Differential Equations

The Geometric View of ODEs

01Direction Fields and the Geometric View of ODEs 02Integral Curves and the Isocline Method 03Perpendicular Isoclines: The Circles of y' = -x/y 04Isocline Corridors and a Trap for Solutions 05Existence and Uniqueness: One Curve Through a Point 06Where Existence and Uniqueness Fail Problem set0/10 Problem set 20/10 MIT problem set0/3 Practice∞

Numerical Methods

01Euler's Method and Its Three Equations 02Euler's Method by Hand: Building the Table 03Convexity and the Sign of Euler's Error 04First Order Error: Halve the Step, Halve the Error 05Improved Euler (RK2): Averaging Two Slopes 06Runge-Kutta 4 and the Price of Accuracy 07Singular Points: Where Every Method Fails Problem set0/10 Problem set 20/10 MIT problem set0/1 Practice∞

First-Order Linear Equations

01First Order Linear: Standard Form and the Sign of p 02The Conduction-Diffusion Model: Newton's Law of Cooling 03The Integrating Factor and the Four-Step Method 04Integrating Factors: Two Mistakes and a Trick Question 05Constant Coefficients: Steady State and Transient Problem set0/10 Problem set 20/10 MIT problem set0/1 Practice∞

Perpendicular Isoclines: The Circles of y' = -x/y

Why do the line elements along every isocline of $y' = -x/y$ stand perpendicular to the isocline itself?

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Your summary note

1. Isoclines as lines $y = -(1/C)x$
  Set $y = -(1/C)x$ , rewrite as $y = -(1/C)x$ , draw the $C = 1$ and $C = -1$ cases, and show the elements of slope $C$ are perpendicular to each isocline.
2. Recovering the axes: $Cy = -x$
  Rewrite the isocline as $Cy = -x$ to get the $y$ -axis at $C = 0$ with horizontal elements, and record that the $x$ -axis is excluded but included by convention as ' $C = \infty$ ' with vertical elements.
3. Integral curves $x^2 + y^2 = c_1^2$
  State that curves perpendicular to the rays are circles centred at the origin, and confirm by separating variables $y\,dy = -x\,dx$ and integrating to those circles.
4. Limited domain of $y = \sqrt{c_1^2 - x^2}$
  Write the explicit solution, note its domain runs only from $-c_1$ to $c_1$ , and record that a solution's domain cannot be read off the differential equation in advance.
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