What is the solution for x^'=(x-4)^2 ?

The solution for x^'=(x-4)^2 is x=-1/(t+c_{1)}+(4t)/(t+c_{1)}+(4c_{1})/(t+c_{1)}

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x^'=(x-4)^2

x^{^{'}} = (x - 4)^{2}

x = - \frac{1}{t + c _{1}} + \frac{4 t}{t + c _{1}} + \frac{4 c _{1}}{t + c _{1}}

Solution steps

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

Solve separable ODE:

x = - \frac{1}{t + c _{1}} + \frac{4 t}{t + c _{1}} + \frac{4 c _{1}}{t + c _{1}}

x = - \frac{1}{t + c _{1}} + \frac{4 t}{t + c _{1}} + \frac{4 c _{1}}{t + c _{1}}

\frac{( e ^{2 p} + 1 )}{3 e ^{2 p}}

x \to 7 + lim (- 2 x^{2} + 2)

sin (9 x) + sin^{2} (9 x)

\int (4 x + 5)^{2} d x

x \to \frac{1}{3} lim ((9 x - 3)^{3})

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