Separable differential equations

Solving separable differential equations dy/dx = f(x)g(y) by separating the variables and integrating both sides, finding constant solutions first, handling absolute values, and using initial conditions.

The method

Example 1●●

Solve $\dfrac{dy}{dx} = \dfrac{x}{y}$, given $y = 3$ when $x = 0$.

$\displaystyle\int y\,dy = \int x\,dx$, so $\tfrac12 y^2 = \tfrac12 x^2 + C$, that is, $y^2 = x^2 + D$.

The initial condition gives $D = 9$. Since $y \gt 0$ at the start, $y = \sqrt{x^2 + 9}$.

Example 2: an absolute value●●●

Solve $\dfrac{dy}{dx} = 2y$, given $y = 3$ when $x = 0$.

$y = 0$ is a constant solution, but it doesn't meet the initial condition. Otherwise, $\displaystyle\int \frac{1}{y}\,dy = \int 2\,dx$.

$\ln|y| = 2x + C$, so $|y| = e^{C}e^{2x}$, that is, $y = Ae^{2x}$ where $A = \pm e^C$.

At $x = 0$, $A = 3$, so $y = 3e^{2x}$.

Example 3●●●

Solve $\dfrac{dy}{dx} = y^2$, given $y = 1$ when $x = 0$.

$\displaystyle\int y^{-2}\,dy = \int dx$, so $-\dfrac1y = x + C$.

At $x = 0$: $-1 = C$. So $-\dfrac1y = x - 1$, giving $y = \dfrac{1}{1 - x}$, for $x \lt 1$.

Flashcards

Separable DE

$\dfrac{dy}{dx} = f(x)\,g(y)$

Separating

$\displaystyle\int \frac{1}{g(y)}\,dy = \int f(x)\,dx$

Before dividing by $g(y)$

Look for constant solutions where $g(y) = 0$.

$ln|y| = kx + C$

$y = Ae^{kx}$

Practice

Solve $\dfrac{dy}{dx} = \dfrac{x}{y}$ with $y = 3$ when $x = 0$. Find $y$ when $x = 4$.
Solve $\dfrac{dy}{dx} = 2y$ with $y = 3$ when $x = 0$. Find $y$ when $x = \ln 2$.
Solve $\dfrac{dy}{dx} = e^{-y}$ with $y = 0$ when $x = 0$. For what $x$ is $y = \ln 5$?
Solve $\dfrac{dy}{dx} = y^2$ with $y = 1$ when $x = 0$. Find $y$ when $x = \tfrac12$.
Solve $\dfrac{dy}{dx} = 3x^2 y$ with $y = 2$ when $x = 0$. Find the exact value of $y$ when $x = 1$.
Find the constant solution of $\dfrac{dy}{dx} = x(y - 1)$.
Which DE is not separable?
$\ln|y| = 4x + C$ can be written as: