Given this differential equation:

$$ \frac{dx}{dt} = te^{-x} $$

I want to;

(1) Find the general solution.

(2) Find the particular solution given the initial condition $x(0)=1$

So this is how i proceed:

$$ \int\frac{dx}{e^{-x}} = \int tdt \implies e^x = \frac{t^2}{2} $$

So if i take that last equation as a general solution for the initial condition $x(0)=1$ i have that: $$ 0 = ln \frac{1}{2}$$

So i think i am not doing it right and i would thank any kind of help.

  • $\begingroup$ Thank you for showing what you have done. I am working on an answer right now. Hold on for a bit. $\endgroup$ – FundThmCalculus May 31 '15 at 1:05

Given (with $x$ being a function of $t$): $$\frac{dx}{dt}=te^{-x}$$ Separate, as you have done: $$\frac{dx}{e^{-x}}=t*dt$$ Integrate both sides, as you have done: $$\int \frac{dx}{e^{-x}}= \int t*dt$$ Notice that I added a constant of integration. This is where you forgot something. $$\int e^x dx = e^x = \frac{1}{2} t^2 +C$$ Solve for the general solution: $$x(t)=\ln\left( \frac{t^2}{2}+C\right)$$ Now plug in the initial conditions to determine the value of the constant of integration: $$e^1=\frac{1}{2} 0^2+C \rightarrow C=e$$ Now plug in for $x(t)$: $$x(t)=\ln\left( \frac{t^2}{2}+e\right)$$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.