A separable differential equation is one that can be written in the form $\frac{dy}{dx} = f(x)g(y)$, where the right-hand side factors into a product of a function of $x$ and a function of $y$. The method of separation of variables allows us to rewrite the equation so that all $y$ terms are on one side and all $x$ terms on the other, then integrate both sides.
🎯 In this section you will learn
- How to identify separable differential equations
- The separation of variables technique
- How to find general solutions (with constant $C$)
- How to find particular solutions using initial conditions
📌 The Method
$$ \frac{dy}{dx} = f(x)g(y) \quad \Longrightarrow \quad \frac{dy}{g(y)} = f(x) \, dx $$
Separate, then integrate both sides
💡 Step-by-Step Process
- Rewrite the equation so that all $y$ terms are on the left and all $x$ terms on the right.
- Integrate both sides: $\int \frac{dy}{g(y)} = \int f(x) dx$.
- Add the constant of integration $C$ (usually on one side).
- Solve for $y$ explicitly if possible.
- If given an initial condition, substitute to find $C$.
Solve $\displaystyle \frac{dy}{dx} = 2xy$.
①
Separate variables
$\frac{dy}{y} = 2x \, dx$
②
Integrate both sides
$\int \frac{1}{y} dy = \int 2x dx$
③
Evaluate integrals
$\ln|y| = x^2 + C$
④
Solve for $y$
$y = e^{x^2 + C} = e^C e^{x^2} = A e^{x^2}$, where $A = e^C$
Solve $\displaystyle \frac{dy}{dx} = 3x^2 y$ with the initial condition $y(0) = 5$.
①
Separate variables
$\frac{dy}{y} = 3x^2 dx$
②
Integrate
$\int \frac{1}{y} dy = \int 3x^2 dx$ → $\ln|y| = x^3 + C$
③
Apply initial condition $y(0)=5$
$\ln|5| = 0^3 + C$ → $C = \ln 5$
④
Solve for $y$
$\ln|y| = x^3 + \ln 5$ → $\ln|y| - \ln 5 = x^3$ → $\ln\left|\frac{y}{5}\right| = x^3$ → $\frac{y}{5} = e^{x^3}$ → $y = 5e^{x^3}$
Solve $\displaystyle \frac{dy}{dx} = \frac{2x}{y}$.
①
Separate variables
$y \, dy = 2x \, dx$
②
Integrate
$\int y dy = \int 2x dx$ → $\frac{y^2}{2} = x^2 + C$
③
Solve for $y$
$y^2 = 2x^2 + 2C$ → $y = \pm \sqrt{2x^2 + C_1}$ (where $C_1 = 2C$)
Solve $\displaystyle \frac{dy}{dx} = 4y + 8$ with $y(0) = 1$.
①
Separate variables
$\frac{dy}{4y+8} = dx$ → $\frac{dy}{4(y+2)} = dx$
②
Integrate
$\frac{1}{4} \int \frac{1}{y+2} dy = \int dx$ → $\frac{1}{4} \ln|y+2| = x + C$
③
Apply $y(0)=1$
$\frac{1}{4} \ln|3| = 0 + C$ → $C = \frac{1}{4}\ln 3$
④
Solve for $y$
$\frac{1}{4} \ln|y+2| = x + \frac{1}{4}\ln 3$ → $\ln|y+2| = 4x + \ln 3$ → $\ln\left|\frac{y+2}{3}\right| = 4x$ → $\frac{y+2}{3} = e^{4x}$ → $y = 3e^{4x} - 2$
⚠️ Common Mistakes
- Not separating correctly: Make sure all $y$ terms are on one side and all $x$ terms on the other before integrating.
- Forgetting the absolute value: When integrating $1/y$, you get $\ln|y|$, not $\ln y$.
- Losing solutions: When dividing by $g(y)$, check if $g(y)=0$ gives any equilibrium solutions.
- Missing the constant $C$: Always include the constant of integration.
🔍 Key Takeaways
- A separable DE can be written as $\frac{dy}{dx} = f(x)g(y)$.
- Separate: $\frac{dy}{g(y)} = f(x) dx$, then integrate both sides.
- General solutions include an arbitrary constant $C$.
- Particular solutions are found using an initial condition.
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