Find general solution and particular solution given a initial condition

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.

• Thank you for showing what you have done. I am working on an answer right now. Hold on for a bit. – 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 $x(t)$: $$x(t)=\ln\left( \frac{t^2}{2}+e\right)$$