Solving the following nonlinear differential equation: $x=\frac{3}{2}tx'+e^{x'}$ Can someone help me solving the following differential equation? $$x=\frac{3}{2}tx'+e^{x'}$$
I have tried it to solve using d'Alembert's method by first taking derivative of given equation and the taking $u=x'$.
Then, I solved it by integrating factor so in the solution I got:
$$tu^3=-2e^u (u^2-2u+2)+C$$
To get this solution back to $x$ and $t$ variables I used my previous assumption $u=x'$ which gives me: $$t(x')^3=-2e^{x'}[(x')^2-2x'+2]+C$$
Is it correct? 
 A: So far, your method and your working is correct. I have verified it multiple times.
What you have to do now is to use your original ODE:
$$x=\frac{3}{2}tx'+e^{x'} \tag{1}$$
And rearrange this to get a closed-form solution for $x'$ in terms of the Lambert W function. Substituting this into the solution you obtained will give you the general solution in implicit form.

To make $x'$ the subject on equation $(1)$, a valid substitution would be:
$$v=\frac{2x}{3t}-x' \iff x'=-v+\frac{2x}{3t}$$
This gives:
$$-\frac{3}{2}tv+e^{-v+\frac{2x}{3t}}=0$$
$$\frac{e^{\frac{2x}{3t}}}{e^v}=\frac{3}{2}tv$$
$$ve^v=\frac{2e^{\frac{2x}{3t}}}{3t}$$
Using the definition of the Lambert W gives:
$$v=W\left(\frac{2e^{\frac{2x}{3t}}}{3t}\right)$$
Substituting back gives:
$$x'=\frac{2x}{3t}-W\left(\frac{2e^{\frac{2x}{3t}}}{3t}\right) \tag{2}$$

Therefore, it just remains to substitute $(2)$ into the solution you obtained, given below:
$$t(x')^3=-2e^{x'}\cdot [(x')^2-2x'+2]+C \tag{3}$$
Doing this gives the general solution in implicit form:
$$\scriptsize \bbox[5px,border:2px solid #C0A000]{t\left(\frac{2x}{3t}-W\left(\frac{2e^{\frac{2x}{3t}}}{3t}\right)\right)^3=-2e^{\frac{2x}{3t}-W\left(\frac{2e^{\frac{2x}{3t}}}{3t}\right)}\cdot \left[\left(\frac{2x}{3t}-W\left(\frac{2e^{\frac{2x}{3t}}}{3t}\right)\right)^2-2\left(\frac{2x}{3t}-W\left(\frac{2e^{\frac{2x}{3t}}}{3t}\right)\right)+2\right]+C} \tag{4}$$
This can probably be simplified slightly, though it will still be very long and ugly.
