What is the solution for y^'=13y-y^2-40 ?

The solution for y^'=13y-y^2-40 is y=-(8e^{3t+3c_{1}})/(1-e^{3(t+c_{1))}}+5/(1-e^{3(t+c_{1))}}

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y^'=13y-y^2-40

y^{^{'}} = 13 y - y^{2} - 40

y = - \frac{8 e ^{3 t + 3 c_{1}}}{1 - e ^{3 (t + c_{1})}} + \frac{5}{1 - e ^{3 (t + c_{1})}}

Solution steps

y^{'} (t) = 13 y - y^{2} - 40

Solve separable ODE:

y = - \frac{8 e ^{3 t + 3 c_{1}}}{1 - e ^{3 (t + c_{1})}} + \frac{5}{1 - e ^{3 (t + c_{1})}}

y = - \frac{8 e ^{3 t + 3 c_{1}}}{1 - e ^{3 (t + c_{1})}} + \frac{5}{1 - e ^{3 (t + c_{1})}}

\int \frac{1}{x 9 x ^{2} - 1} d x

x^{^{'}} = \frac{1}{1 + t ^{2}}

f (x) = 4 - 2 x^{2}, at x = 5

(\frac{- 2 x - 1}{- 4 x + 2})^{5}

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

What is the solution for y^'=13y-y^2-40 ?
The solution for y^'=13y-y^2-40 is y=-(8e^{3t+3c_{1}})/(1-e^{3(t+c_{1))}}+5/(1-e^{3(t+c_{1))}}