Solving the Continuous Growth Equation
Our goal is to solve the differential equation that describes continuous growth, like a bank account with interest \(a\):
\[\dot{x} = ax\]We need to find the original function \(x(t)\) that makes this rule true.
1. The Method: Separation of Variables
First, we rewrite \(\dot{x}\) as \(\frac{dx}{dt}\) and get all the \(x\) terms on one side and all the \(t\) terms on the other.
\[\frac{dx}{dt} = ax\] \[\frac{1}{x} dx = a \ dt\]2. The Integration Step
Now, we integrate both sides:
\[\int \frac{1}{x} dx = \int a \ dt\]Right Side (The Easy Part): The integral of the constant \(a\) with respect to \(t\) is just \(at\).
Left Side (The “Undefined”): We know the Power Rule \(\int x^n dx\) fails for \(x^{-1}\) (it gives a \(\frac{1}{0}\) error).
Finding the Integral of \(\frac{1}{x}\)
To solve this, we must first find the function whose derivative is \(\frac{1}{x}\). Let’s find the derivative of \(y = \ln(x)\).
We know \(y = \ln(x)\).
To “undo” the \(\ln\), we apply its inverse, \(e\):
\[e^y = e^{\ln(x)}\]Because \(e\) and \(\ln\) are inverses, they “self-solve” (or “reduce”), leaving:
\[e^y = x\]Now, we take the derivative of both sides with respect to \(x\):
\[\frac{d}{dx}(e^y) = \frac{d}{dx}(x)\]The right side is \(1\). The left side requires the Chain Rule:
\[e^y \cdot \frac{dy}{dx} = 1\]Now, we solve for \(\frac{dy}{dx}\):
\[\frac{dy}{dx} = \frac{1}{e^y}\]And from our substitution in Step 3, we know \(e^y = x\):
\[\frac{dy}{dx} = \frac{1}{x}\]Conclusion: We have proven that the derivative of \(\ln(x)\) is \(\frac{1}{x}\). Therefore, the integral of \(\frac{1}{x} dx\) must be \(\ln|x| + C\).
3. Solving for \(x\)
Now that we’ve solved the “Undefined” part, we can resume solving our main equation:
\[\int \frac{1}{x} dx = \int a \ dt\]This becomes:
\[\ln|x| + C_1 = at + C_2\] \[\ln|x| = at + \underbrace{C_2 - C_1}_{C}\](Where \(C\) is the constant of integration.)
Now, we must solve for \(x\). To “undo” the \(\ln\), we use its inverse: the exponential function \(e\). We make both sides the exponent of \(e\):
\[e^{\ln|x|} = e^{at + C}\]Left Side: The \(e\) and \(\ln\) “undo” each other, leaving just \(x\).
\[x = e^{at + C}\]Right Side: Using the exponent rule \(b^{m+n} = b^m \cdot b^n\), we can split the right side:
\[x = e^{at} \cdot e^C\]4. The Final Solution
Since \(C\) is just a constant, \(e^C\) is also just a constant. Let’s rename this new constant \(K\) to keep things clean.
\[x(t) = K e^{at}\]This is the general solution. But what is \(K\)?
\(K\) is the initial state of the function at \(t=0\). We can prove this by plugging in \(t=0\):
\[x(0) = K e^{a \cdot 0}\] \[x(0) = K e^0\] \[x(0) = K \cdot 1 = K\]So, the constant \(K\) is just the starting value, \(x(0)\).
This gives us the final, descriptive solution:
\[x(t) = x(0) e^{at}\]