From a Rate Rule to a Particular Solution
A differential equation such as \(dy/dt=ky\) describes how a quantity changes: at each time, its rate of change is proportional to its current value. The equation alone does not say which particular curve to use. An initial condition, such as \(y(0)=50\), supplies the value at a specified starting time and selects one solution.
Solving this initial-value problem has two parts. First, use the differential equation to find the general exponential form. Then use the initial condition to determine the particular solution. Once that solution is known, substitute the requested time to find the model's predicted amount.
In an application, \(y\) might represent a population, a quantity of a substance, or another amount. If time is measured in days, for example, \(k\) has units of inverse days so that \(ky\) has the same units as \(dy/dt\). The time difference \(t-t_0\) must use the same time units as those associated with \(k\).
The Initial-Value Solution
For a constant \(k\), the solution that satisfies \(y(t_0)=y_0\) is
The shift \(t-t_0\) matters: the exponential factor equals \(1\) at the initial time, so the formula returns \(y_0\) there. When the initial time is \(0\), the solution simplifies to \(y(t)=y_0e^{kt}\).
Why the Formula Works
The formula can be derived from the differential equation, and its uniqueness can be checked without dividing by \(y\). That matters because dividing by \(y\) would not be valid for a solution whose initial value is zero.
For any differentiable solution \(y\), define \(z(t)=e^{-k(t-t_0)}y(t)\). This definition is valid whether \(y\) is zero or nonzero.
Since \(dy/dt=ky\), the derivative is \(dz/dt=-ke^{-k(t-t_0)}y+e^{-k(t-t_0)}ky=0\).
A function with derivative zero on the interval is constant there. At \(t=t_0\), \(z(t_0)=e^0y_0=y_0\), so \(z(t)=y_0\).
From \(e^{-k(t-t_0)}y(t)=y_0\), multiply by \(e^{k(t-t_0)}\) to obtain \(y(t)=y_0e^{k(t-t_0)}\).
This argument shows both that any solution must have the displayed form and that the form satisfies the initial condition. Direct differentiation gives \(dy/dt=ky_0e^{k(t-t_0)}=ky(t)\), confirming the differential equation as well.
Growth, Decay, and Constant Solutions
The exponential factor \(e^{k(t-t_0)}\) is positive. For an amount that starts positive, the sign of \(k\) determines how the amount changes as time moves forward from \(t_0\):
| Condition | Behavior for \(y_0>0\) and \(t>t_0\) |
|---|---|
| \(k>0\) | The amount increases: exponential growth. |
| \(k<0\) | The amount decreases toward zero: exponential decay. |
| \(k=0\) | The amount stays constant at \(y_0\). |
| \(y_0=0\) | The solution remains zero for every \(k\). |
The qualification \(y_0>0\) is useful when \(y\) represents an amount. For a negative mathematical initial value, the sign of the derivative \(ky\) also depends on the sign of \(y\); the solution formula remains valid, but the usual language of a positive amount growing or decaying may not apply.
Worked Examples
Worked Example: Growth from a Known Initial Amount
A quantity satisfies \(dy/dt=0.4y\), with \(y(0)=50\). Find the amount at \(t=3\), and state whether the model grows or decays.
Here \(k=0.4\), \(t_0=0\), and \(y_0=50\). Substitution into the initial-value formula gives
Since \(k=0.4>0\) and the initial amount is positive, this is growth for \(t>0\). At \(t=3\),
Rounded to the nearest tenth, the predicted amount is \(166.0\). To check the solution, differentiate \(y(t)=50e^{0.4t}\):
Also, \(y(0)=50e^0=50\), as required.
Worked Example: Decay with an Initial Time Other Than Zero
A quantity satisfies \(dy/dt=-0.2y\), with \(y(2)=80\). Find its value at \(t=7\), rounded to the nearest tenth, and classify its behavior after \(t=2\).
The initial time is \(t_0=2\), not \(0\). Thus
Because \(k=-0.2<0\) and \(y(2)=80>0\), the quantity decays as time increases. At \(t=7\), the elapsed time is \(7-2=5\), so
Rounded to the nearest tenth, \(y(7)\approx29.4\). Check the differential equation by differentiating:
The initial condition also holds: \(y(2)=80e^{-0.2(2-2)}=80e^0=80\).
Worked Example: A Zero Initial Amount
Solve \(dy/dt=3y\) with \(y(4)=0\), and find \(y(6)\).
Use \(k=3\), \(t_0=4\), and \(y_0=0\) in the formula:
Therefore, \(y(6)=0\). This is the constant zero solution. Check it directly: its derivative is \(dy/dt=0\), and \(3y=3(0)=0\), so the differential equation holds. At the initial time, \(y(4)=0\), as required.
This case is also why it is not safe to divide by \(y\) before considering the initial value: division by \(y\) would be invalid when \(y=0\).
Worked Example: Applying a Condition at a Nonzero Time
A model satisfies \(dy/dt=\frac{\ln 3}{2}y\) and \(y(1)=10\). Find \(y(3)\) exactly.
Here \(k=\frac{\ln 3}{2}\), \(t_0=1\), and \(y_0=10\). The solution is
Substitute \(t=3\):
To check, \(y(1)=10e^0=10\). Differentiating the solution gives
so it satisfies the differential equation and the initial condition.
A Common Initial-Time Mistake
A frequent error is to use \(y_0e^{kt}\) even when the initial condition is specified at a time other than zero. For the decay problem \(dy/dt=-0.2y\), \(y(2)=80\), that incorrect setup would be \(y=80e^{-0.2t}\). Substituting the stated initial time shows the problem:
The expression fails the initial condition, so it cannot be the required solution. The correction is to measure elapsed time from the initial time:
Always substitute the initial time into your proposed formula before evaluating a later time. This quick check catches a missing time shift, a sign error, or an incorrectly applied initial value.
A Reliable Solution Routine
For a constant-rate exponential initial-value problem, organize the calculation in this order:
- Identify the constant \(k\), the initial time \(t_0\), and the initial value \(y_0\).
- Write \(y(t)=y_0e^{k(t-t_0)}\), keeping the parentheses around the time difference.
- Use the sign of \(k\) and the initial amount to describe growth, decay, or a constant solution.
- Substitute the requested time, retain the exact exponential expression, and round only the final numerical value if needed.
- Check the initial condition and, when appropriate, differentiate the proposed solution to verify the equation.
These checks connect the model's algebra to its defining conditions: a valid answer must obey both the rate rule and the specified starting value.
Check Your Understanding
For each question, use the initial-value formula and check that your result is consistent with the stated condition.
- Solve \(dy/dt=0.3y\) with \(y(0)=20\). Find \(y(4)\) and state whether the amount grows or decays.
- Solve \(dy/dt=-0.5y\) with \(y(1)=12\). Find \(y(3)\) exactly.
- For \(dy/dt=ky\), explain why \(y_0=0\) gives a solution even though separating variables by dividing by \(y\) would not apply.
- A proposed solution is \(y(t)=7e^{0.2t}\) for \(dy/dt=0.2y\) with \(y(3)=7\). Does it satisfy the initial condition? If not, write the correct solution.
- For a positive initial amount, what behavior does \(k=0\) produce?
