What is the solution for y^'=(y+2t)^2,y(0)=sqrt(3) ?

The solution for y^'=(y+2t)^2,y(0)=sqrt(3) is y=sqrt(2)tan(sqrt(2)t+0.88607…+pin)-2t

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

y^'=(y+2t)^2,y(0)=sqrt(3)

y^{^{'}} = (y + 2 t)^{2}, y (0) = 3

y = 2 tan (2 t + 0.88607 \dots + πn) - 2 t

Solution steps

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

Solve using substitution:

y = 2 tan (2 t + 0.88607 \dots + πn) - 2 t

y = 2 tan (2 t + 0.88607 \dots + πn) - 2 t

\frac{\partial}{\partial x} (x^{7} - 6 x y - 3 y^{2})

\int (sin (\frac{x}{2}) + cos (\frac{x}{2}))^{2} d x

\frac{2 ( s + 1 )}{s ( s ^{2} + s + 2 )}

n = 1 \sum \infty \frac{4}{( n + 1 ) !}

f (x) = 7 x^{2}

What is the solution for y^'=(y+2t)^2,y(0)=sqrt(3) ?
The solution for y^'=(y+2t)^2,y(0)=sqrt(3) is y=sqrt(2)tan(sqrt(2)t+0.88607…+pin)-2t