What is the solution for y^'=t^3(y^2+3y+16) ?

The solution for y^'=t^3(y^2+3y+16) is y=(tan((sqrt(55)(t^4+c_{1}))/8)sqrt(55))/(2)-3/2

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

y^'=t^3(y^2+3y+16)

y^{^{'}} = t^{3} (y^{2} + 3 y + 16)

y = \frac{tan ( \frac{55 ( t ^{4} + c _{1} )}{8} ) 55}{2} - \frac{3}{2}

Solution steps

y^{'} (t) = t^{3} (y^{2} + 3 y + 16)

Solve separable ODE:

y = \frac{tan ( \frac{55 ( t ^{4} + c _{1} )}{8} ) 55}{2} - \frac{3}{2}

y = \frac{tan ( \frac{55 ( t ^{4} + c _{1} )}{8} ) 55}{2} - \frac{3}{2}

What is the solution for y^'=t^3(y^2+3y+16) ?
The solution for y^'=t^3(y^2+3y+16) is y=(tan((sqrt(55)(t^4+c_{1}))/8)sqrt(55))/(2)-3/2