We have an answer to "Solving differential equationdydx=13ey−1results in strange result?"

rijstmeel7d4t

Answered question

2022-04-03

Solving differential equation

\frac{d y}{d x} = \frac{1}{3 e^{y}} - 1

results in strange result?

armejantm925

Beginner2022-04-04Added 20 answers

Explanation:
In your work, you should have

3 e^{y} (1 + \frac{d y}{d x}) = 1

Hint: let

z = e^{y}, \frac{d z}{d x} = z \cdot \frac{d y}{d x}

and solve the linear ODE in z,

\frac{d z}{d x} = \frac{1}{3} - z

such that

z (0) = e^{0} = 1

, then go back to

y = \log (z)

Laylah Hebert

Beginner2022-04-05Added 15 answers

A direct approach:

\frac{d y}{d x} = {\frac{13}{e}}^{- y} - 1 = \frac{1 - 3 e^{y}}{3 e^{y}} \Rightarrow - \int \frac{- 3 e^{y}}{1 - 3 e^{y}} d y = \int d x \Rightarrow \ln | 1 - 3 e^{y} | = - x + C \Rightarrow

and then use the initial condition

0 = y (0)

to get

K = - 2

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