What is the solution for y^'=-3y,y(0)=e^{-3t},y(0)=1 ?

The solution for y^'=-3y,y(0)=e^{-3t},y(0)=1 is y=e^{-3t+c_{1}}

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y^'=-3y,y(0)=e^{-3t},y(0)=1

y^{^{'}} = - 3 y, y (0) = e^{- 3 t}, y (0) = 1

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

Solution steps

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

Solve separable ODE:

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

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

\frac{d y}{d x} + ln (x) y = 3 ln (x)

y^{^{'}} = 2 y^{- 2}, y (0) = 3

(y + 2) d x + 25 - x^{2} d y = 0

2 x^{2} y + 3 x y^{^{'}} - 15 y = 0

- 5 \frac{d y}{d x} = 2 y - 150

What is the solution for y^'=-3y,y(0)=e^{-3t},y(0)=1 ?
The solution for y^'=-3y,y(0)=e^{-3t},y(0)=1 is y=e^{-3t+c_{1}}