What is the solution for y^'=-(y^2)/((0.05+y)),y(0)=9 ?

The solution for y^'=-(y^2)/((0.05+y)),y(0)=9 is -(0.05)/y+ln(y)=-t-(0.05)/9+2ln(3)

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y^'=-(y^2)/((0.05+y)),y(0)=9

y^{^{'}} = - \frac{y ^{2}}{( 0.05 + y )}, y (0) = 9

- \frac{0.05}{y} + ln (y) = - t - \frac{0.05}{9} + 2 ln (3)

Solution steps

y^{'} (t) = - \frac{y ^{2}}{( 0.05 + y )}

Solve separable ODE:

- \frac{0.05}{y} + ln (y) = - t - \frac{0.05}{9} + 2 ln (3)

- \frac{0.05}{y} + ln (y) = - t - \frac{0.05}{9} + 2 ln (3)

\int \frac{1}{2} x^{\frac{3}{4}} d x

\int (9) d x

y = x^{2} + 19 x^{9}

x \to \infty lim (8 + \frac{1}{x})

\frac{d}{d x} (ln (9 - x^{2}))

What is the solution for y^'=-(y^2)/((0.05+y)),y(0)=9 ?
The solution for y^'=-(y^2)/((0.05+y)),y(0)=9 is -(0.05)/y+ln(y)=-t-(0.05)/9+2ln(3)