What is the solution for (dy)/(dt)=e^{-3t},y(0)=10 ?

The solution for (dy)/(dt)=e^{-3t},y(0)=10 is y=-1/3 e^{-3t}+31/3

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(dy)/(dt)=e^{-3t},y(0)=10

\frac{d y}{d t} = e^{- 3 t}, y (0) = 10

y = - \frac{1}{3} e^{- 3 t} + \frac{31}{3}

Solution steps

\frac{d y}{d t} = e^{- 3 t}

Substitute

\frac{d y}{d t}

with

y^{'} (t)

y^{^{'}} (t) = e^{- 3 t}

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

then

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

y = \int e^{- 3 t} d t

\int e^{- 3 t} d t = - \frac{1}{3} e^{- 3 t} + c_{1}

y = - \frac{1}{3} e^{- 3 t} + c_{1}

Apply initial conditions:

y = - \frac{1}{3} e^{- 3 t} + \frac{31}{3}

y = - \frac{1}{3} e^{- 3 t} + \frac{31}{3}

(x - 1)^{2} (2 x + 1)^{4}

\frac{\partial}{\partial x} (e^{x^{2} - y^{2}})

y^{^{'}} + 2 x^{2} y = x^{2}

\int \frac{2 - 2 x}{1 - x} d x

\frac{d y}{d x} + \frac{y}{x} = 2 x^{8} y^{2}

What is the solution for (dy)/(dt)=e^{-3t},y(0)=10 ?
The solution for (dy)/(dt)=e^{-3t},y(0)=10 is y=-1/3 e^{-3t}+31/3