2.4: Separable Differential Equations
- Page ID
- 383
\( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \) \( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)\(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\) \(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\)\(\newcommand{\AA}{\unicode[.8,0]{x212B}}\)
A differential equation is called separable if it can be written as
\[ f(y)\,dy = g(x)\,dx. \]
To solve a separable differential equation
- Get all the \(y\)'s on the left hand side of the equation and all of the \(x\)'s on the right hand side.
- Integrate both sides.
- Plug in the boundary conditions (e.g. given initial values) to find the constant of integration (\(C\)).
- Solve for \(y\).
Solve \( \dfrac{dy}{dx} = y(3 - x) \) with \( y(0 )= 5 \).
Solution
\[\begin{align*} \dfrac{dy}{y} &= (3 - x) dx \\[4pt] \int \dfrac{dy}{y} &= \int (3-x) dx \\[4pt] \ln\; y &= 3x - \dfrac{x^2}{2} + C \\[4pt] \ln 5 &= 0 + 0 + C \\[4pt] C &= \ln\; 5 \\[4pt] y &= e^{3x-\frac{x^2}{2} + \ln \; 5} \\[4pt] y &= e^{3x-\frac{x^2}{2}} \; e^{\ln \; 5} \\[4pt] &= 5e^{3x-\frac{x^2}{2}} \end{align*}\]
- \( \dfrac{dy}{dx} = \dfrac{x}{y} \) with \( y(0) = 1 \)
- \( \dfrac{dy}{dx} = x(x+1) \) with \( y(1) = 1 \)
- \( 2xy + \dfrac{dy}{dx} = x \) with \( y(0) = 2 \)
Contributors and Attributions
- Larry Green (Lake Tahoe Community College)
Integrated by Justin Marshall.