What is the solution for x^'=1-3x+2x^2,x(0)=2 ?

The solution for x^'=1-3x+2x^2,x(0)=2 is x=-(e^t)/(3-2e^t)+3/(3-2e^t)

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x^'=1-3x+2x^2,x(0)=2

x^{^{'}} = 1 - 3 x + 2 x^{2}, x (0) = 2

x = - \frac{e ^{t}}{3 - 2 e ^{t}} + \frac{3}{3 - 2 e ^{t}}

Solution steps

x^{'} (t) = 1 - 3 x + 2 x^{2}

Solve separable ODE:

x = - \frac{e ^{t}}{3 - 2 e ^{t}} + \frac{3}{3 - 2 e ^{t}}

x = - \frac{e ^{t}}{3 - 2 e ^{t}} + \frac{3}{3 - 2 e ^{t}}

3 y + 5 + (3 x - 7) \frac{d y}{d x} = 0

y^{^{'}} = \frac{7 x ^{2} - 1}{7 + 4 y}

\frac{d y}{d x} y = ln (arccos (x^{2} + 4))

y^{^{'}} + y sin (x) = 3 sin (x)

(\frac{5 y}{3}) y^{^{'}} = x

What is the solution for x^'=1-3x+2x^2,x(0)=2 ?
The solution for x^'=1-3x+2x^2,x(0)=2 is x=-(e^t)/(3-2e^t)+3/(3-2e^t)