Int Factors, step unknown

Question

how does the professor go from e^-5t = -1/6 e ^-6t + c to

x(t) = e^5t - 1/6 e^-t

im confused about this step

Answer 1

Here is the general case: we have the differential equation $$x'+f(t)x=g(t)$$ $$x'e^{\int f(t)dt}+e^{\int f(t)dt}f(t)x=e^{\int f(t)dt}g(t)$$ The value $e^{\int f(t)dt}$ is known as the integrating factor and is quite useful as now we can apply product rule to get. $$(x*e^{\int f(t)dt})'=e^{\int f(t)dt}g(t)$$ $$x=\frac{\int e^{\int f(t)dt}g(t)dt}{e^{\int f(t)dt}}$$ Applying this here we have $f(t)=-5$ and $g(t)=e^{-t}$. Thus we have $$x=\frac{\int{e^{-5t}*e^{-t}dt}}{e^{-5t}}=ce^{5t}-\frac{e^{-t}}{6}$$ And since we know that $x(0)=\frac{3}{4}$, so we can solve for c. $$\frac{3}{4}=c-\frac{1}{6} \rightarrow c=\frac{11}{12}$$