Unit 3 · Lesson 1 of 6

A differential equation specifies how a state changes.

A population rate law links growth to the current amount. Instead of prescribing the curve, the equation prescribes its slope.

Change the input. Follow the mathematics.

What you'll be able to do

Set up the model and calculate its solution value.
Connect the equation to the changing visual.
Explain the assumptions and identify a misleading interpretation.

Predict before exploring

What will change if you increase time? Predict the direction, then explain the response.

A model you can changeMove one control at a time.

3.644 · illustrative relative amount

Solution

Time2 hInitial amount2 units

solution value

y(2)=2e^{0.3(2)}=3.64424

The equation gives a family of solutions. The initial condition selects one trajectory through the slope field.

Assumptions, units, and limits

A first-order initial-value problem with constant proportional growth. The derivative specifies a rate, not the value itself.

Each segment gives the local rate at that time and state. The red trajectory follows the chosen initial condition; Euler uses finite tangent steps.

The controls specify a new scenario. Curves are calculated from the equations; physical illustrations are schematic.

From the situation to an equation

A first-order differential equation relates a function to its first derivative. Verify a proposed solution by differentiating and substituting. An initial value selects a trajectory from the family; equilibrium solutions have zero rate.

y'=0.3y,\quad y(0)=y_0,\quad y(t)=y_0e^{0.3t}

Keep the reasoning visible

y'=f(t,y),\quad y(t_0)=y_0

The equation gives a family of solutions. The initial condition selects one trajectory through the slope field.

What makes this model useful?

A first-order initial-value problem with constant proportional growth. The derivative specifies a rate, not the value itself.

A differential equation is not just an algebraic formula for y. Solutions must satisfy both the rate law and the initial condition.

A worked example

Calculate the starting scenario

Use time = 2 h; initial amount = 2 units.
$y(2)=2e^{0.3(2)}=3.64424$
The equation gives a family of solutions. The initial condition selects one trajectory through the slope field. The checked quantity is solution value.

Make the idea your own

Calculate. Compare. Explain.

Numeric checks allow 1% rounding error. Values near zero use a small absolute tolerance. Work through the setup before opening a solution.

Calculate and explain

Rebuild the starting calculation

Calculate the solution value using time 2 h, initial amount 2 units. State the meaning of its sign and units.

Checked quantity: solution value · units. Follow the definition for finite-cutoff and numerical-model questions.

Hint

Explore the visual solution

3.644 · illustrative relative amount

Solution

Time2 hInitial amount2 units

solution value

y(2)=2e^{0.3(2)}=3.64424

The equation gives a family of solutions. The initial condition selects one trajectory through the slope field.

Assumptions, units, and limits

A first-order initial-value problem with constant proportional growth. The derivative specifies a rate, not the value itself.

Each segment gives the local rate at that time and state. The red trajectory follows the chosen initial condition; Euler uses finite tangent steps.

The controls specify a new scenario. Curves are calculated from the equations; physical illustrations are schematic.

The controls explore variations. These written steps retain the original question's inputs.

y'=0.3y,\quad y(0)=y_0,\quad y(t)=y_0e^{0.3t}

Calculate and explain

Apply the idea to a new scenario

With initial amount 3 units, find y at $t=4$ h for y′=0.3y.

Checked quantity: solution value · units. Follow the definition for finite-cutoff and numerical-model questions.

Hint

Explore the visual solution

9.96 · illustrative relative amount

Solution

Time4 hInitial amount3 units

solution value

y(4)=3e^{0.3(4)}=9.96035

The equation gives a family of solutions. The initial condition selects one trajectory through the slope field.

Assumptions, units, and limits

A first-order initial-value problem with constant proportional growth. The derivative specifies a rate, not the value itself.

Each segment gives the local rate at that time and state. The red trajectory follows the chosen initial condition; Euler uses finite tangent steps.

The controls specify a new scenario. Curves are calculated from the equations; physical illustrations are schematic.

The controls explore variations. These written steps retain the original question's inputs.

y'=0.3y,\quad y(0)=y_0,\quad y(t)=y_0e^{0.3t}

Challenge the interpretation

Which explanation survives the model?

Choose the statement you can justify, then use the visual to explain why the others fail.

Test it with the visual

3.644 · illustrative relative amount

Solution

Time2 hInitial amount2 units

solution value

y(2)=2e^{0.3(2)}=3.64424

The equation gives a family of solutions. The initial condition selects one trajectory through the slope field.

Assumptions, units, and limits

A first-order initial-value problem with constant proportional growth. The derivative specifies a rate, not the value itself.

Each segment gives the local rate at that time and state. The red trajectory follows the chosen initial condition; Euler uses finite tangent steps.

The controls specify a new scenario. Curves are calculated from the equations; physical illustrations are schematic.

The reasoning. A first-order differential equation relates a function to its first derivative. Verify a proposed solution by differentiating and substituting. An initial value selects a trajectory from the family; equilibrium solutions have zero rate.

Your learning path

Can you explain the connection?

Mark complete when you can set up the calculation and explain what the visual is showing. Opening a page never completes it automatically.

Completion is saved on this browser. This is separate from account practice progress.