What is the solution for y^'=(x-3)/(y+5) ?

The solution for y^'=(x-3)/(y+5) is y=-5+sqrt(x^2-6x+c_{1)+25},y=-sqrt(x^2-6x+c_{1)+25}-5

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y^'=(x-3)/(y+5)

y^{^{'}} = \frac{x - 3}{y + 5}

y = - 5 + x^{2} - 6 x + c_{1} + 25, y = - x^{2} - 6 x + c_{1} + 25 - 5

Solution steps

y^{'} (x) = \frac{x - 3}{y + 5}

Solve separable ODE:

y = - 5 + x^{2} - 6 x + c_{1} + 25, y = - x^{2} - 6 x + c_{1} + 25 - 5

y = - 5 + x^{2} - 6 x + c_{1} + 25, y = - x^{2} - 6 x + c_{1} + 25 - 5

\frac{d}{d x} (- 2 e^{4 x})

\int \frac{1}{5 ( 2 x - 1 )} d x

f (x) = x x

x \cdot cos (x), 1

x \to 3 lim (\frac{x ^{2} - 9}{3})

What is the solution for y^'=(x-3)/(y+5) ?
The solution for y^'=(x-3)/(y+5) is y=-5+sqrt(x^2-6x+c_{1)+25},y=-sqrt(x^2-6x+c_{1)+25}-5