From Proportional Rate to an Exponential Function
A differential equation can express how a quantity changes without giving its value at every time. In the equation \( \frac{dy}{dt}=ky \), the rate of change of \(y\) is proportional to its current value. The number \(k\) is a constant: it sets the proportional rate, including its sign.
This equation is a basic model of growth and decay. If \(k>0\), a positive solution increases; if \(k<0\), a positive solution decreases; and if \(k=0\), the rate is zero. The equation itself does not specify one curve. Its solutions form a family, and deriving that family requires keeping track of the constant introduced by integration—and of what happens when \(y=0\).
Deriving the Exponential Form When \(y\) Is Nonzero
First consider a solution that is nonzero throughout an interval. Dividing the equation by \(y\) is then allowed. Separating the \(y\)- and \(t\)-expressions gives \( \frac{1}{y}\frac{dy}{dt}=k \). Integrating both sides with respect to \(t\) gives a logarithm on the left and a linear expression on the right.
Here \(A\) is an arbitrary constant of integration. The absolute value is essential: \( \int \frac{1}{y}\,dy=\ln|y|+A \) applies for either positive or negative \(y\), as long as \(y\ne 0\). Exponentiating yields \( |y|=e^{kt+A}=e^A e^{kt} \). Since a continuous, nonzero function on an interval cannot change sign, the sign can be absorbed into a single nonzero constant \(C\). Thus \(y=Ce^{kt}\), with \(C\ne0\) for this part of the derivation.
The integration constant does not disappear: exponentiating turns the additive constant \(A\) into the positive factor \(e^A\), and the possible sign of \(y\) is included in \(C\). Together, these give an arbitrary nonzero real multiplier.
Accounting for the Zero Solution
The separation step divided by \(y\), so it cannot describe a solution that equals zero. But \(y(t)=0\) is itself a solution: its derivative is zero, and \(k\cdot 0=0\). It is included in the same formula by allowing \(C=0\).
To make sure the separated calculation has not missed a solution that reaches zero partway through its interval, use a product-rule argument. This also gives a concise proof that the stated family is complete.
For any solution \(y\), differentiate \(e^{-kt}y(t)\) using the product rule.
Substitute \(y'=ky\). The two terms cancel, so the product has derivative zero throughout the interval.
A differentiable function with derivative zero on an interval is constant. Therefore \(e^{-kt}y(t)=C\), which gives \(y(t)=Ce^{kt}\).
In equations, the argument is:
It follows that \(e^{-kt}y(t)=C\) for a constant \(C\), and multiplying by \(e^{kt}\) gives \(y(t)=Ce^{kt}\). This reasoning applies whether \(y\) is positive, negative, zero, or changes none of those statuses across the interval. In particular, if \(y(t_0)=0\) at even one point \(t_0\), then \(C=e^{-kt_0}y(t_0)=0\), so \(y\) is identically zero on the interval.
For the converse, take any real constant \(C\) and differentiate \(y(t)=Ce^{kt}\). The derivative is \(y'(t)=kCe^{kt}=ky(t)\), so the function satisfies the equation. This verification includes \(C=0\) and \(k=0\).
Worked Examples
Worked Example: A Positive Proportional Rate
Derive the solution family for \( \frac{dy}{dt}=\frac{1}{2}y \), first for nonzero solutions and then including the zero solution.
For a nonzero solution, divide by \(y\) and separate variables:
Integrate both sides:
Exponentiating and absorbing the sign into a constant gives \(y=Ce^{t/2}\) for \(C\ne0\). The zero function also satisfies the original equation, so the full family allows \(C=0\):
Check by differentiation: \(y'=\frac{1}{2}Ce^{t/2}\). The right side of the equation is \(\frac{1}{2}y=\frac{1}{2}Ce^{t/2}\), which equals \(y'\). The family satisfies the equation for every real \(C\).
Worked Example: A Negative Proportional Rate
Derive and verify the solution family for \( \frac{dy}{dt}=-3y \).
For \(y\ne0\), divide by \(y\), separate, and integrate:
Exponentiating gives \( |y|=e^A e^{-3t} \). Incorporating either sign in the constant, and then including the zero solution, gives:
Differentiate: \(y'=-3Ce^{-3t}\). Substitute into the right side: \(-3y=-3Ce^{-3t}\). These expressions agree, including when \(C=0\). For \(C>0\), the function is positive and decreases as \(t\) increases; for \(C<0\), it is negative, while still satisfying the same differential equation.
Worked Example: Zero Rate and Constant Solutions
Solve \( \frac{dy}{dt}=0y \) without losing the zero solution.
Here \(k=0\), so the equation is \(y'=0\). For a nonzero solution, separating and integrating gives:
Exponentiating shows that \(y\) is a nonzero constant; including the zero function gives every constant solution. The general exponential formula says the same thing, since \(e^{0t}=1\):
Check: differentiating \(y=C\) gives \(y'=0\), and the right side is \(0y=0\). Thus every real constant, including zero, satisfies the equation.
Common Pitfall: Dividing by \(y\) and Losing a Solution
A common incomplete calculation starts by dividing \(y'=ky\) by \(y\) and then announces that the result describes every solution. That step assumes \(y\ne0\). For example, if \(k=2\), division gives \(y'/y=2\), but substituting \(y=0\) into this divided equation is impossible: its left side would involve division by zero. Yet \(y=0\) satisfies the original equation \(y'=2y\).
The correction is to state the nonzero condition before dividing, derive \(y=Ce^{kt}\) for \(C\ne0\), and then test \(y=0\) in the original equation. Allowing \(C=0\) brings that solution into the final family. Another error is writing \(\ln y\) without first restricting to positive solutions. Using \(\ln|y|\) handles both signs; the zero case is still treated separately because the logarithm is undefined there.
Why the Formula Matters
The formula \(y=Ce^{kt}\) connects a simple statement about a rate to a specific family of curves. The equation says that at every time, the rate is the current value multiplied by the same constant \(k\). The exponential function has exactly this property: its derivative is a constant multiple of itself. The arbitrary factor \(C\) lets the family include different sizes and signs of solutions, while \(C=0\) accounts for equilibrium.
In a model where \(t\) measures time, \(k\) has units of inverse time: \(ky\) must have the same units as the rate \(dy/dt\). This is consistent with the exponent \(kt\), which must be dimensionless. The formula is useful because it displays the full solution family directly, before any particular value of \(y\) is specified.
Check Your Understanding
Use separation, the zero-solution check, and substitution to answer these questions.
- When separating \( \frac{dy}{dt}=5y \), what expression appears after integrating the separated equation?
- Why is the absolute value needed in the logarithm when deriving the family for nonzero solutions?
- What solution could be missed if you divide the differential equation by \(y\) without checking \(y=0\)?
- Differentiate \(y=Ce^{-t}\) and verify that it satisfies \( \frac{dy}{dt}=-y \).
- What does the general formula become when \(k=0\), and why does it still include the zero solution?
