What is the solution for y^'=y^2-9 ?

The solution for y^'=y^2-9 is y=(3e^{6c_{1}})/(e^{6t)-e^{6c_{1}}}+(3e^{6t})/(e^{6t)-e^{6c_{1}}}

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y^'=y^2-9

y^{^{'}} = y^{2} - 9

y = \frac{3 e ^{6 c_{1}}}{e ^{6 t} - e ^{6 c_{1}}} + \frac{3 e ^{6 t}}{e ^{6 t} - e ^{6 c_{1}}}

Solution steps

y^{'} (t) = y^{2} - 9

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} - 9

Isolate

t^{^{'}} : t^{^{'}} = \frac{1}{y ^{2} - 9}

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

then

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

t = \int \frac{1}{y ^{2} - 9} d y

\int \frac{1}{y ^{2} - 9} d y = - \frac{1}{6} (ln (\frac{y}{3} + 1) - ln (\frac{y}{3} - 1)) + c_{1}

t = - \frac{1}{6} (ln (\frac{y}{3} + 1) - ln (\frac{y}{3} - 1)) + c_{1}

Isolate

y : y = \frac{3 e ^{6 c_{1}}}{e ^{6 t} - e ^{6 c_{1}}} + \frac{3 e ^{6 t}}{e ^{6 t} - e ^{6 c_{1}}}

y = \frac{3 e ^{6 c_{1}}}{e ^{6 t} - e ^{6 c_{1}}} + \frac{3 e ^{6 t}}{e ^{6 t} - e ^{6 c_{1}}}

(x y + 2 y) d x + x d y = 0

8 y^{2} (x d y + y d x) = y d x - x d y

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

d r = b (cos (θ) d r + r sin (θ) d θ)

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

What is the solution for y^'=y^2-9 ?
The solution for y^'=y^2-9 is y=(3e^{6c_{1}})/(e^{6t)-e^{6c_{1}}}+(3e^{6t})/(e^{6t)-e^{6c_{1}}}