What is the solution for y(dy)/(dx)=(y^2-y)(2x+5),y(1)=-3 ?

The solution for y(dy)/(dx)=(y^2-y)(2x+5),y(1)=-3 is y=1-4/(e^{6-x^2-5x)}

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y(dy)/(dx)=(y^2-y)(2x+5),y(1)=-3

y \frac{d y}{d x} = (y^{2} - y) (2 x + 5), y (1) = - 3

y = 1 - \frac{4}{e ^{6 - x^{2} - 5 x}}

Solution steps

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

Solve linear ODE:

y = 1 - \frac{4}{e ^{6 - x^{2} - 5 x}}

y = 1 - \frac{4}{e ^{6 - x^{2} - 5 x}}

\int_{0}^{6} \frac{π}{4} (36 - x^{2}) d x

x \to 2 lim (\frac{x}{x - 1})

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

\int sin (θ) cos^{2} (θ) d θ

x \to 0 + lim (\frac{x + 4}{x})

What is the solution for y(dy)/(dx)=(y^2-y)(2x+5),y(1)=-3 ?
The solution for y(dy)/(dx)=(y^2-y)(2x+5),y(1)=-3 is y=1-4/(e^{6-x^2-5x)}