What is the solution for y'=y^2-4 ?

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

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y'=y^2-4

y^{^{'}} = y^{2} - 4

y = \frac{2 e ^{4 c_{1}}}{e ^{4 t} - e ^{4 c_{1}}} + \frac{2 e ^{4 t}}{e ^{4 t} - e ^{4 c_{1}}}

Solution steps

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

Solve for

t, y^{'} (t) = \frac{d y}{d t} = \frac{1}{\frac{d t}{d y}} = \frac{1}{t ^{^{'}}}

(\frac{1}{t ^{^{^{'}}}}) = y^{2} - 4

Isolate

t^{^{'}} : t^{^{'}} = \frac{1}{y ^{2} - 4}

f^{^{'}} (x) = g (x)

then

f (x) = \int g (x) d x

t = \int \frac{1}{y ^{2} - 4} d y

\int \frac{1}{y ^{2} - 4} d y = - \frac{1}{4} (ln (\frac{y}{2} + 1) - ln (\frac{y}{2} - 1)) + c_{1}

t = - \frac{1}{4} (ln (\frac{y}{2} + 1) - ln (\frac{y}{2} - 1)) + c_{1}

Isolate

y : y = \frac{2 e ^{4 c_{1}}}{e ^{4 t} - e ^{4 c_{1}}} + \frac{2 e ^{4 t}}{e ^{4 t} - e ^{4 c_{1}}}

y = \frac{2 e ^{4 c_{1}}}{e ^{4 t} - e ^{4 c_{1}}} + \frac{2 e ^{4 t}}{e ^{4 t} - e ^{4 c_{1}}}

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

\int x^{3} x^{2} + 40 d x

\frac{4}{x ^{2}}

\int \frac{( x - 95 )}{x ^{2} ( x + 5 )} d x

y^{^{'}} = (x - 3) (y + 1)^{(\frac{2}{3})}

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