Solving DEs: The Method of Separation of Variables
Core Concepts
Focuses on the core technique for solving first-order separable differential equations. Learn to identify, separate, and integrate to find the general solution, which often appears in implicit form.
Solving DEs: The Method of Separation of Variables
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What defines a 'separable' first-order differential equation?
A differential equation that can be algebraically manipulated into the form \(f(y)dy = g(x)dx\). This means all \(y\) terms (including \(dy\)) can be moved to one side of the equation, and all \(x\) terms (including \(dx\)) to the other.
Is the differential equation \(\frac{dy}{dx} = x + y\) separable?
No. It is impossible to factor the expression \(x+y\) into a product of a function of \(x\) and a function of \(y\), so you cannot separate the variables \(x\) and \(y\).
Is the differential equation \(\frac{dy}{dx} = xy^2\) separable? If so, show the separated form.
Yes, it is separable. To separate the variables, divide by \(y^2\) and multiply by \(dx\) to get \(\frac{1}{y^2}dy = x \, dx\).
What are the two main steps to solving a separable differential equation after identifying it as separable?
1. Separate the variables to get the form \(f(y)dy = g(x)dx\).
2. Integrate both sides of the equation: \(\int f(y)dy = \int g(x)dx\).
When integrating both sides of a separated equation, why do we typically only add one constant of integration, \(C\)?
Integrating both sides yields \(\int f(y)dy = F(y) + C_1\) and \(\int g(x)dx = G(x) + C_2\). We can combine the two constants into a single constant \(C = C_2 - C_1\) and write \(F(y) = G(x) + C\) for simplicity.
Find the general solution to \(\frac{dy}{dx} = \frac{x^2}{y}\).
1. Separate variables: \(y \, dy = x^2 \, dx\).
2. Integrate both sides: \(\int y \, dy = \int x^2 \, dx\).
3. The result is the implicit solution: \(\frac{1}{2}y^2 = \frac{1}{3}x^3 + C\), or \(3y^2 - 2x^3 = C_1\).
Find the general solution to \(\frac{dy}{dx} = e^{x-y}\).
First, rewrite as \(\frac{dy}{dx} = e^x e^{-y}\). Then separate variables: \(e^y dy = e^x dx\). Integrate both sides: \(\int e^y dy = \int e^x dx\), which gives the solution \(e^y = e^x + C\).
What is the difference between an 'implicit' and an 'explicit' solution?
An explicit solution is one where the dependent variable (usually \(y\)) is isolated on one side, in the form \(y = f(x)\). An implicit solution defines a relationship between the variables, like \(x^2 + y^2 = C\), without isolating \(y\).
After solving a DE, you get \(\ln|y| = x^2 + C\). Express this general solution in explicit form.
Exponentiate both sides: \(e^{\ln|y|} = e^{x^2 + C}\). This simplifies to \(|y| = e^{x^2}e^C\). Since \(e^C\) is an arbitrary positive constant, we can let \(A = \pm e^C\) to absorb the absolute value, giving the explicit solution \(y = Ae^{x^2}\).
Solve the differential equation \(xy' = 2y\).
Rewrite as \(x \frac{dy}{dx} = 2y\). Separate variables: \(\frac{1}{y} dy = \frac{2}{x} dx\). Integrate: \(\int \frac{1}{y} dy = \int \frac{2}{x} dx\). This gives \(\ln|y| = 2\ln|x| + C = \ln(x^2) + C\). In explicit form, \(y = Ax^2\).
