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

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

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y^'=sqrt(y^2-9)

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

y = \frac{3}{2 e ^{t + c_{1}}} + \frac{3 e ^{t + c_{1}}}{2}

Solution steps

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

Solve separable ODE:

y = \frac{3}{2 e ^{t + c_{1}}} + \frac{3 e ^{t + c_{1}}}{2}

y = \frac{3}{2 e ^{t + c_{1}}} + \frac{3 e ^{t + c_{1}}}{2}

\frac{1}{x ^{2} + 1}

\int cos (x) (8 + 4 sin^{2} (x)) d x

y = cos (x^{3})

x \to 5 lim (\frac{x ^{2} - 3 x - 10}{x - 5})

\frac{d}{d x} ((7 x - 8)^{5})

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