What is the solution for (x-13)(x-3)y^'=y ?

The solution for (x-13)(x-3)y^'=y is y=(c_{1}\sqrt[10]{x-13})/(\sqrt[10]{x-3)}

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(x-13)(x-3)y^'=y

(x - 13) (x - 3) y^{^{'}} = y

y = \frac{c _{1} 10 x - 13}{10 x - 3}

Solution steps

(x - 13) (x - 3) y^{'} (x) = y

Solve separable ODE:

y = \frac{c _{1} 10 x - 13}{10 x - 3}

y = \frac{c _{1} 10 x - 13}{10 x - 3}

\frac{d x}{d y} = 3 e^{2 x - 3 y}

x d y - (y + 4) d x = 0

\frac{d y}{d x} = 14 e^{5 x - y}, y (0) = 0

\frac{d y}{d t} = \frac{( 3 t ^{2} + 4 t + 2 )}{2 ( y - 1 )}

(x^{2} + y^{2}) (x d y + y d x) - x y (x d x + y d y) = 0

What is the solution for (x-13)(x-3)y^'=y ?
The solution for (x-13)(x-3)y^'=y is y=(c_{1}\sqrt[10]{x-13})/(\sqrt[10]{x-3)}