What is the solution for f^'(x)-2f(x)(x^2e^x)=0 ?

The solution for f^'(x)-2f(x)(x^2e^x)=0 is f(x)=e^{2(x^2e^x-2(e^xx-e^x))+c_{1}}

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f^'(x)-2f(x)(x^2e^x)=0

f^{^{'}} (x) - 2 f (x) (x^{2} e^{x}) = 0

f (x) = e^{2 (x^{2} e^{x} - 2 (e^{x} x - e^{x})) + c_{1}}

Solution steps

f^{'} (x) - 2 f (x) (x^{2} e^{x}) = 0

Solve separable ODE:

f (x) = e^{2 (x^{2} e^{x} - 2 (e^{x} x - e^{x})) + c_{1}}

f (x) = e^{2 (x^{2} e^{x} - 2 (e^{x} x - e^{x})) + c_{1}}

(\frac{d y}{d x}) = (- \frac{1}{( x ^{2} y ^{2} )})

\frac{d y}{d x} = y x, y (1) = 7

\frac{d x}{d t} = 0.08 - \frac{0.15 x}{50}, t = 18

\frac{d y}{d x} = \frac{( 4 x e ^{x^{2}} )}{y}, y (0) = 3

\frac{d y}{d x} = - y (x^{2} - 7)

What is the solution for f^'(x)-2f(x)(x^2e^x)=0 ?
The solution for f^'(x)-2f(x)(x^2e^x)=0 is f(x)=e^{2(x^2e^x-2(e^xx-e^x))+c_{1}}