Separable and Linear ODEs

$\newenvironment {prompt}{}{} \newcommand {\ungraded }[0]{} \newcommand {\bigmath }[1]{$\displaystyle #1$} \newcommand {\choicebreak }[0]{} \newcommand {\type }[0]{} \newcommand {\notes }[0]{} \newcommand {\keywords }[0]{} \newcommand {\offline }[0]{} \newcommand {\comments }[0]{\begin {feedback}} \newcommand {\multiplechoice }[0]{\begin {multipleChoice}} \newcommand {\HyperFirstAtBeginDocument }[0]{\AtBeginDocument }$

We learn techniques to solve first-order linear and separable ODEs.

Online Texts

Examples

The ODE $y' = x^2 e^y$ is separable.

To separate, we move all factors involving $y$ to the left-hand side and all factors involving $x$ to the right-hand side. We also formally write $y'$ as $dy/dx$ and verify that we can end up with $dy$ on the left-hand side and $dx$ on the right hand side: $\answer {e^{-y}} dy = \answer {x^2} dx.$
Apply an integral on both sides: $\int \answer {e^{-y}} dy = \int \answer {x^2} dx \Rightarrow \answer {-e^{-y}} = \answer {\frac {x^3}{3}} + C.$ That gives our general solution.
If we had initial conditions $y(0) = 1$ , we would plug in $x=\answer {0}$ , $y=\answer {1}$ to our general solution to determine that $C = \answer {-e^{-1}}$ . We can plug in this value for $C$ and solve for $y$ to determine that $y = \answer {- \ln \left ( e^{-1} - \frac {x^3}{3} \right )}.$

The ODE $x y' = -x^2 y + e^{-x^2/2}$ is linear.

To begin, we must write the ODE in standard form: $y' + \answer {x} y = \answer {e^{-x^2/2}/x}.$
We give the name $P(x)$ to the coefficient of $y$ (note that it doesn’t have to depend on $x$ . Writing $P(x)$ is only meant to indicate that it cannot depend on $y$ ). We give the name $Q(x)$ to the right-hand side.
The integrating factor is the exponential of the integral of $P$ . In this case, we can take $I(x) = e^{x^2/2}$ . Multiply both sides of the standard form equation by the integrating factor: $\answer {e^{x^2/2}} y' + \answer {x e^{x^2/2}} y = \answer {\frac {1}{x}}.$
The integrating factor gets its name because it is now possible to integrate the left-hand side. By the product rule, we know that $\left ( e^{x^2/2} y \right )' = \answer {e^{x^2/2}} y' + \answer {x e^{x^2/2}} y,$ which is exactly the right-hand side. Thus we have $\left ( e^{x^2/2} y \right )' = \answer {\frac {1}{x}}.$
Integrating the left-hand side gives $e^{x^2/2} y$ ; integrating both sides gives $e^{x^2/2} y = \answer {\ln |x|} + C$ (don’t forget absolute values in the logarithm). Solving for $y$ gives $y = \answer {e^{-x^2/2} \ln |x|} + C e^{-x^2/2}$ as the general solution.