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Popular >

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

Solution

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

Solution

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

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

Graph

Popular Examples

x^3ydx+x^2y^2dy=0

x^{3} y d x + x^{2} y^{2} d y = 0

(dy)/(dx)=2xy^2+(5y^2)/(sqrt(x))

\frac{d y}{d x} = 2 x y^{2} + \frac{5 y ^{2}}{x}

(dy)/(dx)y=ln(ln(x^5))

\frac{d y}{d x} y = ln (ln (x^{5}))

y^'=-3x^2(y+2),y(1)=-1

y^{^{'}} = - 3 x^{2} (y + 2), y (1) = - 1

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

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

Frequently Asked Questions (FAQ)

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