What is the solution for f^'(x)+2xf(x)=0,f(0)=3 ?

The solution for f^'(x)+2xf(x)=0,f(0)=3 is f(x)=3e^{-x^2}

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f^'(x)+2xf(x)=0,f(0)=3

f^{^{'}} (x) + 2 x f (x) = 0, f (0) = 3

f (x) = 3 e^{- x^{2}}

Solution steps

f^{'} (x) + 2 x f (x) = 0

Solve separable ODE:

f (x) = 3 e^{- x^{2}}

f (x) = 3 e^{- x^{2}}

- 2 y \frac{d y}{d x} + x^{2} + 2 x^{4} = 0

\frac{d y}{d x} = \frac{- 0.8}{cos ( y )}, y (0) = \frac{π}{2}

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

y^{^{'}} sin (2 y) - cos (x) = 0

\frac{df}{d t} = f (t)

What is the solution for f^'(x)+2xf(x)=0,f(0)=3 ?
The solution for f^'(x)+2xf(x)=0,f(0)=3 is f(x)=3e^{-x^2}