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

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

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

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

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}}

x \to 1 lim (lo g_{\frac{1}{e}} (x))

t \cdot e^{5 t} sin (3 t)

\frac{\partial}{\partial y} (1 - x y)

\frac{1}{s ( s ^{2} + 4 )} + 1

x \to 0 lim (\frac{\frac{4}{x - 1} + 4}{x})

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