A community for students.
Here's the question you clicked on:
 0 viewing
anonymous
 4 years ago
How do you solve x'+3x = t + e^2t? Tried separation of variables, but didn't work
anonymous
 4 years ago
How do you solve x'+3x = t + e^2t? Tried separation of variables, but didn't work

This Question is Closed

UnkleRhaukus
 3 years ago
Best ResponseYou've already chosen the best response.1its a linear equation ie, of the form \[x'+p(t)x=q(t)\] so find an integrating factor \[\mu(t)=e^{∫p(t)dt}\] then \[\big(x\mu(t)\big)'=q(t)\mu(t)\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Here we have a typical linear firstorder ordinary differential equation. The idea is that we observe \((\mu x)'=\mu x'+\mu'x\), so if we can make our lefthand side (\(x'+3x\)) look like \(\mu x'+\mu'x\), then we can integrate both sides to yield \(\mu x\) and therefore \(x\). We can do this using multiplication, so essentially we need to find some \(\mu\) such that \(\mu x'+3\mu x=\mu x'+\mu' x\), i.e. \(3\mu=\mu'\); it should be clear then that \(\mu=e^{3t}\). Multiply throughout to yield \(\mu x'+3\mu x=\mu t+\mu e^{2t}\), which is equivalent to \((\mu x)'=\mu t+\mu e^{2t}\). Integrate both sides and solve for \(x\) to yield \(x=\frac1\mu\int(\mu t+\mu e^{2t})dt\). Substitute in \(\mu=e^{3t}\) and simplify.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Thank you guys so much.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Generally you can solve a linear firstorder ODE of the form \(y'+Px=Q\) where \(P,Q\) are functions of \(x\) as follows: \(y=\frac1\mu\int\mu Q\ dx\) where \(\mu=e^{\int P\ dx}\)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Oops, I meant \(y'+Py=Q\)
Ask your own question
Sign UpFind more explanations on OpenStudy
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.