What is the solution for (dy)/(dx)=e^{-ka*x},y(0)=0 ?

The solution for (dy)/(dx)=e^{-ka*x},y(0)=0 is y=-(e^{-kax})/(ka)+1/(ka)

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(dy)/(dx)=e^{-ka*x},y(0)=0

\frac{d y}{d x} = e^{- ka \cdot x}, y (0) = 0

y = - \frac{e ^{- ka x}}{ka} + \frac{1}{ka}

Solution steps

\frac{d y}{d x} = e^{- ka \cdot x}

Substitute

\frac{d y}{d x}

with

y^{'} (x)

y^{^{'}} (x) = e^{- ka x}

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

then

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

y = \int e^{- ka x} d x

\int e^{- ka x} d x = - \frac{1}{ka} e^{- ka x} + c_{1}

y = - \frac{1}{ka} e^{- ka x} + c_{1}

Simplify

y = - \frac{e ^{- ka x}}{ka} + c_{1}

Apply initial conditions:

y = - \frac{e ^{- ka x}}{ka} + \frac{1}{ka}

y = - \frac{e ^{- ka x}}{ka} + \frac{1}{ka}

\int \frac{x ^{2} - x + 12}{x ^{3} + 6 x} d x

\frac{\partial}{\partial x} (e^{3 x} sin (cy))

n = 0 \sum \infty \frac{3 ^{n}}{2 ^{n}}

f (x) = \frac{1}{2 - 3 x}

\frac{d}{d x} (ln (\frac{1 + x}{x ^{4}}))

What is the solution for (dy)/(dx)=e^{-ka*x},y(0)=0 ?
The solution for (dy)/(dx)=e^{-ka*x},y(0)=0 is y=-(e^{-kax})/(ka)+1/(ka)