What is the solution for (e^tx+1)dt+(e^t-1)dx=0 ?

The solution for (e^tx+1)dt+(e^t-1)dx=0 is x=(c_{1})/(-1+e^t)-t/(-1+e^t)

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(e^tx+1)dt+(e^t-1)dx=0

(e^{t} x + 1) d t + (e^{t} - 1) d x = 0

x = \frac{c _{1}}{- 1 + e ^{t}} - \frac{t}{- 1 + e ^{t}}

Solution steps

(e^{t} x + 1) d t + (e^{t} - 1) d x = 0

Solve exact ODE:

x = \frac{c _{1}}{- 1 + e ^{t}} - \frac{t}{- 1 + e ^{t}}

x = \frac{c _{1}}{- 1 + e ^{t}} - \frac{t}{- 1 + e ^{t}}

\int x (x + 4)^{2} d x

\int \frac{x ^{2}}{7 + x} d x

\int e^{s x} d x

\int x^{2} sec^{2} (x^{3}) d x

5 x e^{x} csc (x)

What is the solution for (e^tx+1)dt+(e^t-1)dx=0 ?
The solution for (e^tx+1)dt+(e^t-1)dx=0 is x=(c_{1})/(-1+e^t)-t/(-1+e^t)