What is the solution for y^'=sqrt(2y+39) ?

The solution for y^'=sqrt(2y+39) is y=(t^2)/2+c_{1}t+(c_{1}^2)/2-39/2

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y^'=sqrt(2y+39)

y^{^{'}} = 2 y + 39

y = \frac{t ^{2}}{2} + c_{1} t + \frac{c _{1}^{2}}{2} - \frac{39}{2}

Solution steps

y^{'} (t) = 2 y + 39

Solve separable ODE:

y = \frac{t ^{2}}{2} + c_{1} t + \frac{c _{1}^{2}}{2} - \frac{39}{2}

y = \frac{t ^{2}}{2} + c_{1} t + \frac{c _{1}^{2}}{2} - \frac{39}{2}

\frac{d}{d x} (\frac{x ^{\frac{3}{5}} x ^{3} + 2 x - 6}{( 4 x - 3 ) ^{8}})

f (x) = \frac{4}{x - 2}

x \to 4 - lim (e^{\frac{3}{4 - x}})

\int x^{5} (x - 9) d x

\frac{40 s + 20}{s ^{2} + 9 s + 14}

What is the solution for y^'=sqrt(2y+39) ?
The solution for y^'=sqrt(2y+39) is y=(t^2)/2+c_{1}t+(c_{1}^2)/2-39/2