What is the solution for y^'=(3(e^x+1))/(2y) ?

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

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y^'=(3(e^x+1))/(2y)

y^{^{'}} = \frac{3 ( e ^{x} + 1 )}{2 y}

y = 3 (e^{x} + x) + c_{1}, y = - 3 (e^{x} + x) + c_{1}

Solution steps

y^{'} (x) = \frac{3 ( e ^{x} + 1 )}{2 y}

Solve separable ODE:

y = 3 (e^{x} + x) + c_{1}, y = - 3 (e^{x} + x) + c_{1}

y = 3 (e^{x} + x) + c_{1}, y = - 3 (e^{x} + x) + c_{1}

\frac{d}{dd} (\frac{a x + b}{c x + d})

\frac{d}{d x} (- \frac{1}{2} x^{4} cos (x))

\int \frac{2}{X} d X

f (x) = 5 e^{5 x}

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