What is the solution for 2xy+6x+(x^2-4)y^'=0,y(3)=2 ?

The solution for 2xy+6x+(x^2-4)y^'=0,y(3)=2 is y=(25)/(x^2-4)-3

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2xy+6x+(x^2-4)y^'=0,y(3)=2

2 x y + 6 x + (x^{2} - 4) y^{^{'}} = 0, y (3) = 2

y = \frac{25}{x ^{2} - 4} - 3

Solution steps

2 x y + 6 x + (x^{2} - 4) y^{'} (x) = 0

Solve separable ODE:

y = \frac{25}{x ^{2} - 4} - 3

y = \frac{25}{x ^{2} - 4} - 3

\frac{d y}{d x} = - \frac{0.6}{cos ( y )}, y (0) = \frac{π}{4}

\frac{d y}{d x} = x^{5} e^{- y}

2 (x + 1) y y^{^{'}} - y^{2} = 2, y (5) = 2

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

\frac{d v}{d t} + 2 v = 9, v (0) = 0

What is the solution for 2xy+6x+(x^2-4)y^'=0,y(3)=2 ?
The solution for 2xy+6x+(x^2-4)y^'=0,y(3)=2 is y=(25)/(x^2-4)-3