Lesson 3: Slope Fields and Solution Curves (1.3)

Original Notes

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Lesson 3: Slope Fields and Solution Curves (1.3)

Review eg: Solve the IVP \(x'' = \cos t, \quad x'(0) = 1, \quad x(0) = -1\)

\[ \frac{d^2x}{dt^2} = \cos t \]

integrate \(\leadsto \frac{dx}{dt} = \sin t + c\)

integrate \(\leadsto x(t) = -\cos t + ct + d\)

General solution: \(x(t) = -\cos t + ct + d\)

Plug in initial conditions to find \(c\) & \(d\) values

\(x'(0) = 1 \leadsto 1 = x'(0) = \sin(0) + c \Rightarrow c = 1\)

\(x(0) = -1 \leadsto -1 = x(0) = -1 + 0 + d \Rightarrow d = 0\)

\(x(t) = -\cos t + t\) is solution to IVP

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e.g. \(\frac{dy}{dx} = 10 - 2y \leadsto \frac{dy}{dx} = 10 - 2y(x)\)

Cannot really integrate \(10 - 2y\) w.r.t \(x\)

\(y(x) = 5 \leadsto \frac{dy}{dx} = 0 = 10 - 2(5) \leadsto y(x) = 5\) is a solution

\(y > 5 \leadsto \frac{dy}{dx} = 10 - 2y < 0 \leadsto\) Solution curve is decreasing

\(y < 5 \leadsto \frac{dy}{dx} = 10 - 2y > 0 \leadsto\) Solution curve is increasing

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\[ \frac{dy}{dx} = 10 - 2y \]

Slope field and solution curves for the autonomous differential equation dy/dx = 10 - 2y on the Cartesian plane with x from -4 to 4 and y from -4 to 10. Horizontal slope segments of slope 0 appear along the line y = 5, with negative slopes for y > 5 and positive slopes for y < 5. Three solution trajectories are shown: a red horizontal line representing the equilibrium solution y = 5; a green curve passing through (0, 8) that decreases toward the asymptote y = 5 as x approaches infinity and tends to infinity as x approaches negative infinity; and a purple curve passing through (0, 0) that increases toward the asymptote y = 5 as x approaches infinity and tends to negative infinity as x approaches negative infinity.
Visual Description: Slope field and solution curves for the autonomous differential equation dy/dx = 10 - 2y on the Cartesian plane with x from -4 to 4 and y from -4 to 10. Horizontal slope segments of slope 0 appear along the line y = 5, with negative slopes for y > 5 and positive slopes for y < 5. Three solution trajectories are shown: a red horizontal line representing the equilibrium solution y = 5; a green curve passing through (0, 8) that decreases toward the asymptote y = 5 as x approaches infinity and tends to infinity as x approaches negative infinity; and a purple curve passing through (0, 0) that increases toward the asymptote y = 5 as x approaches infinity and tends to negative infinity as x approaches negative infinity.

For \( y(0) = y_0 > 5 \):

\[ \frac{dy}{dx} = 10 - 2y < 0 \]

\[ x \to \infty \Rightarrow y(x) \to 5 \]

\[ x \to -\infty \Rightarrow y(x) \to \infty \]

\( y = 5 \) is a solution curve

For \( y(0) = y_0 < 5 \):

\[ \frac{dy}{dx} = 10 - 2y > 0 \]

\[ x \to \infty \Rightarrow y(x) \to 5 \]

\[ x \to -\infty \Rightarrow y(x) \to -\infty \]

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ODE: \( \frac{dy}{dx} = f(x, y) \)

IVP: \( y(x_0) = y_0 \)

Sketch illustrating a solution curve y(x) drawn in blue passing through a designated point labeled (x_0, y_0), intersected by a straight black tangent line at (x_0, y_0).
Visual Description: Sketch illustrating a solution curve y(x) drawn in blue passing through a designated point labeled (x_0, y_0), intersected by a straight black tangent line at (x_0, y_0).

\( y(x) = \) solution curve through \( (x_0, y_0) \)

has slope of tangent line at \( (x_0, y_0) = \frac{dy}{dx} = f(x_0, y_0) \)

Isoclines: Curves \( f(x, y) = \text{Constant} \)

Equilibrium Solution: \( f(x, y) = 0 \) and satisfy \( \frac{dy}{dx} = 0 \)

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eg:

\[ \frac{dy}{dx} = x^2 + y^2 \]

isoclines:

\[ x^2 + y^2 = c \implies \text{circles of radius } \sqrt{c} \]

Hand-drawn sketch showing three concentric circles centered around a central point at the origin, illustrating circular isoclines of the form x^2 + y^2 = c for varying values of c.
Visual Description: Hand-drawn sketch showing three concentric circles centered around a central point at the origin, illustrating circular isoclines of the form x^2 + y^2 = c for varying values of c.

equilibrium solution

\[ x^2 + y^2 = 0 \implies \underline{(0,0)} \] \[ y(x) = 0, \, y'(0) = 0 \]

is a solution to \( \frac{dy}{dx} = x^2 + y^2 \)

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eg:

\[ \frac{dy}{dx} = x - y \]

isoclines:

\[ x - y = c \]

Coordinate plane showing Cartesian axes with several parallel diagonal lines of slope 1. A line passing through the origin is drawn in red and labeled y = x. Another parallel line shifted downward to the right is drawn in blue and labeled y = x - 1. Additional parallel blue lines are sketched above and below representing isoclines.
Visual Description: Coordinate plane showing Cartesian axes with several parallel diagonal lines of slope 1. A line passing through the origin is drawn in red and labeled y = x. Another parallel line shifted downward to the right is drawn in blue and labeled y = x - 1. Additional parallel blue lines are sketched above and below representing isoclines.

\(c = 1 \leadsto y = x - 1 \implies \frac{dy}{dx} = 1 \Bigg\}\) Solution to ODE but not equilibrium solution

\(c = 0 \leadsto y = x \leadsto \frac{dy}{dx} = 1\), but \(x - y = 0 \implies y = x\) is Not solution

No equilibrium solution

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Example: \(\frac{dy}{dx} = y(y-2)\)

  1. find equilibrium solutions
  2. draw slope fields
  3. What does solution curve through \((0,1)\) look like

equilibrium solution: \(y(y-2) = 0 \Rightarrow y=0 \text{ and } y=2\)

Slope field and solution curves plot for the differential equation dy/dx = y(y - 2). The Cartesian grid features two horizontal red lines representing equilibrium solutions at y = 2 and y = 0. Blue dashed slope lines show the direction field across different regions: positive slopes for y > 2 and y < 0, and negative downward slopes for 0 < y < 2. Red solution curves illustrate the behavior: one curve above y = 2 bends upwards away from the equilibrium line; a curve in the band 0 < y < 2 passes through the initial point (0, 1) and slopes downwards asymptotically approaching y = 0; and a curve below y = 0 curves sharply downwards into the negative y direction.
Visual Description: Slope field and solution curves plot for the differential equation dy/dx = y(y - 2). The Cartesian grid features two horizontal red lines representing equilibrium solutions at y = 2 and y = 0. Blue dashed slope lines show the direction field across different regions: positive slopes for y > 2 and y < 0, and negative downward slopes for 0 < y < 2. Red solution curves illustrate the behavior: one curve above y = 2 bends upwards away from the equilibrium line; a curve in the band 0 < y < 2 passes through the initial point (0, 1) and slopes downwards asymptotically approaching y = 0; and a curve below y = 0 curves sharply downwards into the negative y direction.

\(y > 2 \Rightarrow y(y-2) > 0\)

\(0 < y_0 < 2 \Rightarrow y(y-2) < 0\)

\(y < 0 \Rightarrow y(y-2) > 0\)

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\[ \frac{dy}{dx} = y(y-2) \]

Direction field (slope field) on Cartesian coordinates for the autonomous first-order differential equation dy/dx = y(y-2). The horizontal x-axis ranges from -4 to 4 and the vertical y-axis ranges from -2 to 4. Constant equilibrium solutions are marked at the horizontal lines y = 2 and y = 0. In the region y > 2, slopes are positive and increase with y; a sketched red solution curve rises sharply upward to the right, approaching the asymptote y = 2 as x goes to negative infinity. In the region 0 < y < 2, slopes are negative; a solution trajectory passing through the point (0, 1) decreases from near y = 2 towards y = 0 as x increases. In the region y < 0, slopes are positive; a red solution curve approaches y = 0 as x increases to the right and plunges downward steeply toward negative infinity as x decreases to the left.
Visual Description: Direction field (slope field) on Cartesian coordinates for the autonomous first-order differential equation dy/dx = y(y-2). The horizontal x-axis ranges from -4 to 4 and the vertical y-axis ranges from -2 to 4. Constant equilibrium solutions are marked at the horizontal lines y = 2 and y = 0. In the region y > 2, slopes are positive and increase with y; a sketched red solution curve rises sharply upward to the right, approaching the asymptote y = 2 as x goes to negative infinity. In the region 0 < y < 2, slopes are negative; a solution trajectory passing through the point (0, 1) decreases from near y = 2 towards y = 0 as x increases. In the region y < 0, slopes are positive; a red solution curve approaches y = 0 as x increases to the right and plunges downward steeply toward negative infinity as x decreases to the left.

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e.g. Solve IVP

\[ \frac{dy}{dx} = \frac{1}{x}, \quad y(0) = 0. \]

integrate

\[ y(x) = \ln|x| + C \]

CANNOT PLUGIN \(y(0) = 0\)

IVP has no solution

Solution curves do not pass through \(x = 0\)

Cartesian coordinate graph displaying solution curves for y = ln|x| + C. A horizontal x-axis is intersected by two symmetric blue logarithmic curves on either side of the vertical axis x = 0. As x approaches 0 from both the left and right, the curves plunge downward toward negative infinity along vertical dashed lines that indicate the vertical asymptote at x = 0. The curves never cross or meet at x = 0, illustrating why the initial value problem with y(0) = 0 has no solution.
Visual Description: Cartesian coordinate graph displaying solution curves for y = ln|x| + C. A horizontal x-axis is intersected by two symmetric blue logarithmic curves on either side of the vertical axis x = 0. As x approaches 0 from both the left and right, the curves plunge downward toward negative infinity along vertical dashed lines that indicate the vertical asymptote at x = 0. The curves never cross or meet at x = 0, illustrating why the initial value problem with y(0) = 0 has no solution.

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Eg: Solve IVP \( \frac{dy}{dx} = 3 y^{2/3} \), \( y(0) = 0 \)

Slope field plot with multiple solution curves. The Cartesian plane displays a grid of slope field segments with positive slope for y > 0 and y < 0, and horizontal segments along the line y = 0. Overlaid on the slope field are three solution curves corresponding to the initial condition y(0) = 0: a blue curve showing the cubic function y(x) = x^3 passing through the origin; a green horizontal line showing the constant solution y(x) = 0 along the x-axis; and a red curve tracing along y = 0 for x < 2 before branching upwards as y(x) = (x - 2)^3 for x >= 2.
Visual Description: Slope field plot with multiple solution curves. The Cartesian plane displays a grid of slope field segments with positive slope for y > 0 and y < 0, and horizontal segments along the line y = 0. Overlaid on the slope field are three solution curves corresponding to the initial condition y(0) = 0: a blue curve showing the cubic function y(x) = x^3 passing through the origin; a green horizontal line showing the constant solution y(x) = 0 along the x-axis; and a red curve tracing along y = 0 for x < 2 before branching upwards as y(x) = (x - 2)^3 for x >= 2.

  1. Solution 1: \( y(x) = x^3 \)

    \[ \frac{dy}{dx} = 3x^2 = 3 \left( y^{1/3} \right)^2 = 3 y^{2/3} \]

    \( y(0) = 0 \)

  2. Solution 2: \( y(x) = 0 \)

    \[ \frac{dy}{dx} = 0 = 3 y^{2/3} \]

    \( y(0) = 0 \)

  3. Solution 3:

    \[ y(x) = \begin{cases} 0 & x < 2 \\ (x-2)^3 & x \ge 2 \end{cases} \]

    \[ \frac{dy}{dx} = 3 y^{2/3} \]

\( \Rightarrow \) More than one solution to IVP

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In general IVP need not have a solution or have more than one solution.

Q: When do we have exactly one solution?

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Existence & Uniqueness Theorem:

IVP

\[ \frac{dy}{dx} = f(x, y), \quad y(x_0) = y_0 \]

\( R \) is some rectangle containing \( (x_0, y_0) \)

If

  1. \( 1) f(x, y) \) is continuous on \( R \)
  2. \( 2) \frac{\partial f}{\partial y} \) is continuous on \( R \),

then solution to IVP exists and is unique.

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Next week:

  • Seperable equations
  • Linear \( 1^{\text{st}} \) Order equations
  • Some Applications

Online HW Due: Monday

Handwritten HW Due: Wednesday