9.3: Separable Differential Equations

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Throughout this module, if something does not exist, write DNE in the answer box.

Recap Video

Take a look at the following video which recaps the ideas from the section.

Example Video

Below is a video showing a worked example.

Problems

Solve the differential equation $\displaystyle \frac {dy}{dx} = 4xy^2$ .

To solve the differential equation, we will isolate the $y$ ’s and $dy$ on one side, and $x$ ’s and $dx$ on the other, with the goal of having $\text {something} dy = \text {something} dx.$ In this case, we can divide by $y^2$ and multiply by $dx$ to get $\answer {\frac {1}{y^2}}dy = \answer {4x}dx.$ Integrating both sides gives $\answer {-\frac {1}{y}}+C_1 = \answer {2x^2}+C_2.$ In this case, we can move all the constants to one side (e.g. by subtracting $C_1$ from both sides), and call the new constant $C = C_2-C_1$ , giving $\answer {-\frac {1}{y}} = 2x^2+C.$ We can now solve for $y$ , and we get which of the following?

$y = -\frac {1}{2x^2}+C$ $y = -\frac {1}{2x^2+C}$ $y = -\frac {1}{2x^2} - \frac {1}{C}$

If $\displaystyle \frac {dy}{dt} = -\frac {t}{y}$ , find a solution for $y$ which satisfies $y(0) = 1$ .

Which of the following represents a correct separation of the variables?

$-\frac {1}{y}dy = \frac {1}{t}dt$ $y\,dy = -t\,dt$ $\frac {1}{y}dy = -t\,dt$

Integrating both sides of the correct separation gives $\answer {\frac {y^2}{2}} = \answer {-\frac {t^2}{2}} + C,$ where, again, we combine the constants of integration on the right side to get just one “ $C$ .” To solve for $y$ , we can multiply both sides by $2$ to get $\answer {y^2} = \answer {-t^2}+D,$ where $D = 2C$ is just another constant. It might be easier to solve for the constant now. Plugging in $t = 0$ and $y=1$ into the above expression gives $D = \answer {1}$ . Finally, to solve for $y$ , we square root both sides to get $y = \pm \answer {\sqrt {1-t^2}}.$ Should we take the positive or negative root? PositiveNegative . Therefore, $y = \answer {\sqrt {1-t^2}}.$

Find a solution to the differential equation $\displaystyle \frac {dy}{dx} = y\cos (x)+\cos (x)$ which satisfies $y(0) = 2$ .

Separating the variables gives: $\answer {\frac {1}{y+1}}dy = \answer {\cos (x)}dx.$ The solution for $y$ is $y = \answer {3e^{\sin (x)}-1}.$

Find a solution to the differential equation $\displaystyle \frac {dy}{dx} = xy-2x-y+2$ which satisfies $y(2) = 4$ . (Hint: To separate the variables, you will need to factor the right side by grouping.)

Separating the variables gives: $\answer {\frac {1}{y-2}}dy = \answer {x-1}dx.$ The solution for $y$ is $y = \answer {2+2e^{\frac {x^2}{2}-x}}.$