What is the solution for y^'=-y^2+15y-50 ?

The solution for y^'=-y^2+15y-50 is y=-(10e^{5t-5c_{1}})/(1-e^{5(t-c_{1))}}+5/(1-e^{5(t-c_{1))}}

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y^'=-y^2+15y-50

y^{^{'}} = - y^{2} + 15 y - 50

y = - \frac{10 e ^{5 t - 5 c_{1}}}{1 - e ^{5 (t - c_{1})}} + \frac{5}{1 - e ^{5 (t - c_{1})}}

Solution steps

y^{'} (t) = - y^{2} + 15 y - 50

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} + 15 y - 50

Isolate

t^{^{'}} : t^{^{'}} = \frac{1}{- y ^{2} + 15 y - 50}

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

then

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

t = \int \frac{1}{- y ^{2} + 15 y - 50} d y

\int \frac{1}{- y ^{2} + 15 y - 50} d y = \frac{1}{5} (ln (\frac{2}{5} y - 2) - ln (\frac{2}{5} y - 4)) + c_{1}

t = \frac{1}{5} (ln (\frac{2}{5} y - 2) - ln (\frac{2}{5} y - 4)) + c_{1}

Simplify

t = \frac{1}{5} ln (\frac{y - 5}{y - 10}) + c_{1}

Isolate

y : y = - \frac{10 e ^{5 t - 5 c_{1}}}{1 - e ^{5 (t - c_{1})}} + \frac{5}{1 - e ^{5 (t - c_{1})}}

y = - \frac{10 e ^{5 t - 5 c_{1}}}{1 - e ^{5 (t - c_{1})}} + \frac{5}{1 - e ^{5 (t - c_{1})}}

\frac{d y}{d x} = \frac{( y x + 3 x - y - 3 )}{x y - 2 x + 4 y - 8}

y^{^{'}} = e^{- y} + 1

x^{3} y d x + x^{2} y^{2} d y = 0

\frac{d y}{d x} = 2 x y^{2} + \frac{5 y ^{2}}{x}

\frac{d y}{d x} y = ln (ln (x^{5}))

What is the solution for y^'=-y^2+15y-50 ?
The solution for y^'=-y^2+15y-50 is y=-(10e^{5t-5c_{1}})/(1-e^{5(t-c_{1))}}+5/(1-e^{5(t-c_{1))}}