7.3 Guess and Check
A final approach is called the guess and check method. Say for example we have the following equation that describes the rate of change above: The first approach if a function is a solution to a differential equation is the guess and check method, or by direct substitution.
\[ \frac{dS}{dt} = 0.7 S \]
We know that we can apply separation of variables, but instead let’s try to see if the function \(\tilde{S}(t) = 5 e^{0.7t}\) is a solution in order to do that, we need to differentiate \(\tilde{S}(t)\), which using our knowledge of calculus is \(0.7 \cdot 5 e^{0.7t}\). If we note that \(\displaystyle \frac{d\tilde{S}}{dt} = 0.7 \tilde{S} = 0.7 e^{0.7t}\) than the function \(\tilde{S}\) does solve the differential equation.
Super cool! Let’s try an example:
Example 7.2 Verify the following functions are solutions to the differential equation \(\displaystyle \frac{dS}{dt} = 0.7 S\):
- \(\tilde{R}(t) = 10e^{0.7t}\)
- \(\tilde{P}(t) = e^{0.7t}\)
- \(\tilde{Q}(t) = 5e^{0.7t}\)
- \(\tilde{F}(t)=3\)
- \(\tilde{G}(t)=0\)
Solution. Let’s apply direct differentiation to each of these functions:
- \(\tilde{R}(t) = 10e^{0.7t} \rightarrow \tilde{R}'(t) = 7e^{0.7t}\)
- \(\tilde{P}(t) = e^{0.7t} \rightarrow \tilde{P}'(t) = e^{0.7t}\)
- \(\tilde{Q}(t) = 5e^{0.7t} \rightarrow \tilde{Q}'(t) = 3.5e^{0.7t}\)
- \(\tilde{F}(t)=3 \rightarrow \tilde{F}'(t) = 0\)
- \(\tilde{G}(t)=0 \rightarrow \tilde{G}'(t) = 0\)
Now we will compare each of these solutions to the right hand side:
- \(0.7\tilde{R}(t) = 0.7 \cdot 10e^{0.7t} \rightarrow 7e^{0.7t}\)
- \(0.7\tilde{P}(t) = 0.7 e^{0.7t}\)
- \(0.7\tilde{Q}(t) = 0.7 \cdot 5e^{0.7t} \rightarrow = 3.5e^{0.7t}\)
- \(0.7 \tilde{F}(t)=0.7 \cdot 3 \rightarrow 2.1\)
- \(0.7 \tilde{G}(t)=0.7 \cdot 0 \rightarrow 0\)