Approximation of Differential Equation Solution: x = c1*t + c2*t^2

[Nusc]

Homework Statement

$$\frac{dx}{dt} =\frac{ -x}{(t-1+e^{-x})}$$

Show that an approximate solution leads to,

$$\frac{dx}{dt} = -\frac{ 1}{1-c1} [c1+(c2 + \frac{c2-c1/2}{1-c1})*t + O(t^3)]$$

Homework Equations

The Attempt at a Solution

The first equation is not separable.

To approximate, assume
$$x = c1*t+c2*t^2 + O(t^3)$$
Hence
$$dx/dt = c1 + 2*c2*t + O(t^2).$$

If I equate

$$dx/dt = c1 + 2*c2*t + O(t^2).$$

and

$$\frac{dx}{dt} =\frac{ -x}{(t-1+e^{-x})}$$

Should I immediately taylor expand the exponential?

 

[lippyka]

e^-x = 1 - x + O(x^2)So \frac{dx}{dt} =\frac{ -x}{(t-1+1 - x + O(x^2))}\frac{dx}{dt} = -\frac{ 1}{1-c1} [c1+(c2 + \frac{c2-c1/2}{1-c1})*t + O(t^3)]Would that be the right approach?