What is the solution for 2y+3+(2x-3)(dy)/(dx)=0 ?

The solution for 2y+3+(2x-3)(dy)/(dx)=0 is y=(c_{1})/(2(2x-3))-3/2

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2y+3+(2x-3)(dy)/(dx)=0

2 y + 3 + (2 x - 3) \frac{d y}{d x} = 0

y = \frac{c _{1}}{2 ( 2 x - 3 )} - \frac{3}{2}

Solution steps

2 y + 3 + (2 x - 3) \frac{d y}{d x} = 0

Solve separable ODE:

y = \frac{c _{1}}{2 ( 2 x - 3 )} - \frac{3}{2}

y = \frac{c _{1}}{2 ( 2 x - 3 )} - \frac{3}{2}

(4 + x^{2}) \frac{d y}{d x} + 2 x y = 4 x

y^{^{'}} - \frac{3}{4} y = x^{4} \frac{y ^{1}}{3}

y^{^{'}} = \frac{( ln ^{4} ( x ) )}{( x \cdot y ) ^{2}}, y (1) = 2

y^{^{'}} + 16 x y = 0

\frac{d y}{d x} + y cos (x) = 8 cos (x), y (0) = 10

What is the solution for 2y+3+(2x-3)(dy)/(dx)=0 ?
The solution for 2y+3+(2x-3)(dy)/(dx)=0 is y=(c_{1})/(2(2x-3))-3/2