What is the solution for x^'+2e^{-4t}x=0,x(0)=5 ?

The solution for x^'+2e^{-4t}x=0,x(0)=5 is x=5e^{1/2 e^{-4t}-1/2}

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x^'+2e^{-4t}x=0,x(0)=5

x^{^{'}} + 2 e^{- 4 t} x = 0, x (0) = 5

x = 5 e^{\frac{1}{2} e^{- 4 t} - \frac{1}{2}}

Solution steps

x^{'} (t) + 2 e^{- 4 t} x = 0

Solve separable ODE:

x = 5 e^{\frac{1}{2} e^{- 4 t} - \frac{1}{2}}

x = 5 e^{\frac{1}{2} e^{- 4 t} - \frac{1}{2}}

\frac{d y}{d x} = \frac{3 x ^{4}}{y ^{4}}, y (0) = 4

\frac{d y}{d x} + 3 e x^{2} y = 6 x^{2}

\frac{d y}{d x} = 2 y^{2}, x = 2

\frac{d y}{d x} = 4 x^{2} y + 6 y

P^{^{'}} = (P) (t) (1 - \frac{P}{10})

What is the solution for x^'+2e^{-4t}x=0,x(0)=5 ?
The solution for x^'+2e^{-4t}x=0,x(0)=5 is x=5e^{1/2 e^{-4t}-1/2}