Separable differential equations initial value problems
Separable differential equations initial value problems
We already know how to separate variables in a separable differential equation in order to find a general solution to the differential equation.
When we’re given a differential equation and an initial condition to go along with it, we’ll solve the differential equation the same way we would normally, by separating the variables and then integrating.
The constant of integration ???C??? that’s left over from the integration is the value we’ll be able to solve for using the initial condition.
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Solving the separable differential equation initial value problem
Example
Solve the separable differential equation initial value problem, if ???u(0)=2???.
???u'=\frac{12t-3e^t}{6u}???
The first thing we want to recognize is that ???u'??? is the same thing as ???du/dt???, so that we can replace ???u'??? with ???du/dt???.
???u'=\frac{12t-3e^t}{6u}???
???\frac{du}{dt}=\frac{12t-3e^t}{6u}???
Now we can begin to separate variables by multiplying both sides of the equation by ???dt???.
???du=\frac{12t-3e^t}{6u}\ dt???
Now we’ll multiply both sides by ???6u???.
???6u\ du=12t-3e^t\ dt???
When we’re given a differential equation and an initial condition to go along with it, we’ll solve the differential equation the same way we would normally, by separating the variables and then integrating.
At this point, we’ve got all of the ???u???’s on the left side and all of the ???x???’s on the right side, so we’ve successfully separated our variables. Which means we can integrate both sides of the equation.
???\int6u\ du=\int12t-3e^t\ dt???
???3u^2+C_1=6t^2-3e^t+C_2???
???3u^2=6t^2-3e^t+C_2-C_1???
We can say ???C=C_2-C_1??? to get
???3u^2=6t^2-3e^t+C???
???u^2=2t^2-e^t+\frac13C???
???u=\pm\sqrt{2t^2-e^t+\frac13C}???
The ???1/3??? can be absorbed into the ???C???, so we can simplify the equation to
???u=\pm\sqrt{2t^2-e^t+C}???
Now we’ll plug the initial condition ???u(0)=2??? into the equation to solve for ???C???.
???2=\pm\sqrt{2(0)^2-e^{0}+C}???
???2=\pm\sqrt{-1+C}???
???4=-1+C???
???5=C???
Plugging this back into equation, we get
???u=\pm\sqrt{2t^2-e^t+5}???
If we plug in our initial condition again, we get
???2=\pm\sqrt{2(0)^2-e^{0}+5}???
???2=\pm\sqrt{-1+5}???
???2=\pm\sqrt{4}???
???2=\pm2???
We can see that ???2=+2??? and not ???2=-2???, which means only the positive square root can be a solution. So the specific solution to the separable differential equation is
???u=\sqrt{2t^2-e^t+5}???
