What is the solution for y^'=(3y-4x)/(2y-3x) ?

The solution for y^'=(3y-4x)/(2y-3x) is y=(3x+sqrt(x^2+c_{1)})/2 ,y=(3x-sqrt(x^2+c_{1)})/2

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y^'=(3y-4x)/(2y-3x)

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

y = \frac{3 x + x ^{2} + c _{1}}{2}, y = \frac{3 x - x ^{2} + c _{1}}{2}

Solution steps

y^{'} (x) = \frac{3 y - 4 x}{2 y - 3 x}

Solve exact ODE:

y = \frac{3 x + x ^{2} + c _{1}}{2}, y = \frac{3 x - x ^{2} + c _{1}}{2}

y = \frac{3 x + x ^{2} + c _{1}}{2}, y = \frac{3 x - x ^{2} + c _{1}}{2}

\int x y d y

y = 3 x, x + y = 4, y = 1, y = 0

\int \frac{3}{x ^{2} + 3} d x

\frac{d y}{d x} = x^{4} - 1

g (x) = 1 + 6 x + 3 x^{2}

What is the solution for y^'=(3y-4x)/(2y-3x) ?
The solution for y^'=(3y-4x)/(2y-3x) is y=(3x+sqrt(x^2+c_{1)})/2 ,y=(3x-sqrt(x^2+c_{1)})/2