## mathlilly 2 years ago How do you solve x'+3x = t + e^-2t? Tried separation of variables, but didn't work

1. UnkleRhaukus

its 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 $\big(x\mu(t)\big)'=q(t)\mu(t)$

2. oldrin.bataku

Here we have a typical linear first-order ordinary differential equation. The idea is that we observe $$(\mu x)'=\mu x'+\mu'x$$, so if we can make our left-hand 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.

3. mathlilly

Thank you guys so much.

4. oldrin.bataku

Generally you can solve a linear first-order ODE of the form $$y'+Px=Q$$ where $$P,Q$$ are functions of $$x$$ as follows: $$y=\frac1\mu\int\mu Q\ dx$$ where $$\mu=e^{\int P\ dx}$$

5. oldrin.bataku

Oops, I meant $$y'+Py=Q$$