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Differential Equations
The Geometric View of ODEs
01Direction Fields and the Geometric View of ODEs02Integral Curves and the Isocline Method03Perpendicular Isoclines: The Circles of y' = -x/y04Isocline Corridors and a Trap for Solutions05Existence and Uniqueness: One Curve Through a Point06Where Existence and Uniqueness FailProblem set0/10Problem set 20/10MIT problem set0/3Practice∞
Numerical Methods
01Euler's Method and Its Three Equations02Euler's Method by Hand: Building the Table03Convexity and the Sign of Euler's Error04First Order Error: Halve the Step, Halve the Error05Improved Euler (RK2): Averaging Two Slopes06Runge-Kutta 4 and the Price of Accuracy07Singular Points: Where Every Method FailsProblem set0/10Problem set 20/10MIT problem set0/1Practice∞
First-Order Linear Equations
01First Order Linear: Standard Form and the Sign of p02The Conduction-Diffusion Model: Newton's Law of Cooling03The Integrating Factor and the Four-Step Method04Integrating Factors: Two Mistakes and a Trick Question05Constant Coefficients: Steady State and TransientProblem set0/10Problem set 20/10MIT problem set0/1Practice∞

Where Existence and Uniqueness Fail

How can every line through $(0,1)$ solve one equation, while points on the $y$-axis have no solution at all?


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Your summary note

    1. 1

      Solving x dydx=y−1x\,\frac{dy}{dx} = y - 1xdxdy​=y−1 by separation

      Separate to dyy−1=dxx\frac{dy}{y-1} = \frac{dx}{x}y−1dy​=xdx​, integrate to ln⁡∣y−1∣=ln⁡∣x∣+c1\ln|y-1| = \ln|x| + c_1ln∣y−1∣=ln∣x∣+c1​, and absorb the absolute values with C=±ec1C = \pm e^{c_1}C=±ec1​ to reach y=1+Cxy = 1 + Cxy=1+Cx.

    2. 2

      The fan of lines through (0,1)(0,1)(0,1)

      Describe the solutions as all lines with yyy-intercept 111 and any slope, and note that off the yyy-axis exactly one line passes through each point.

    3. 3

      Where existence and uniqueness fail

      Record that on the yyy-axis no solution passes (existence fails) while at (0,1)(0,1)(0,1) every line passes (uniqueness fails).

    4. 4

      No violation of the Existence and Uniqueness Theorem

      Rewrite in standard form dydx=y−1x\frac{dy}{dx} = \frac{y-1}{x}dxdy​=xy−1​, note f(x,y)f(x,y)f(x,y) is undefined at x=0x=0x=0 so the hypotheses fail there, and state this is how failure happens in ordinary practice.

    Attempt 1 of 2