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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

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}

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