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Popular Calculus >

y^'=e^{-y}-1

Solution

y^{^{'}} = e^{- y} - 1

Solution

y = ln (- e^{c_{1} - t} + 1)

Solution steps

y^{'} (t) = e^{- y} - 1

Solve for

t, y^{'} (t) = \frac{d y}{d t} = \frac{1}{\frac{d t}{d y}} = \frac{1}{t ^{^{'}}}

(\frac{1}{t ^{^{^{'}}}}) = e^{- y} - 1

Isolate

t^{^{'}} : t^{^{'}} = \frac{e ^{y}}{1 - e ^{y}}

If

f^{^{'}} (x) = g (x)

then

f (x) = \int g (x) d x

t = \int \frac{e ^{y}}{1 - e ^{y}} d y

\int \frac{e ^{y}}{1 - e ^{y}} d y = - ln (1 - e^{y}) + c_{1}

t = - ln (1 - e^{y}) + c_{1}

Isolate

y : y = ln (- e^{c_{1} - t} + 1)

y = ln (- e^{c_{1} - t} + 1)

Graph

Popular Examples

y^'+y=1,-\H(t-2)y(0)=2

(dx)/(dy)=(5y)/(6x)

(dy)/(dx)=(1+y^2)/(2y)

y^'sec(x)=6y^2,y((9pi)/2)= 1/7

Frequently Asked Questions (FAQ)

What is the solution for y^'=e^{-y}-1 ?
The solution for y^'=e^{-y}-1 is y=ln(-e^{c_{1}-t}+1)