What is the solution for gx^'=sqrt(2x-9) ?

The solution for gx^'=sqrt(2x-9) is x=(t^2+2c_{1}t+c_{1}^2)/(2g^2)+9/2

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gx^'=sqrt(2x-9)

g x^{^{'}} = 2 x - 9

x = \frac{t ^{2} + 2 c _{1} t + c _{1}^{2}}{2 g ^{2}} + \frac{9}{2}

Solution steps

g x^{'} (t) = 2 x - 9

Solve separable ODE:

x = \frac{t ^{2} + 2 c _{1} t + c _{1}^{2}}{2 g ^{2}} + \frac{9}{2}

x = \frac{t ^{2} + 2 c _{1} t + c _{1}^{2}}{2 g ^{2}} + \frac{9}{2}

y^{^{'}} = 5 x y^{6}

7 y d y = \frac{3}{x} d x

x^{2} + 6 d x - \frac{1}{x csc ( - 8 y )} d y = 0

y^{^{'}} = \frac{y ( y - 2 )}{x ( y - 1 )}

x \cdot \frac{d x}{d t} = e^{3 x} t

What is the solution for gx^'=sqrt(2x-9) ?
The solution for gx^'=sqrt(2x-9) is x=(t^2+2c_{1}t+c_{1}^2)/(2g^2)+9/2