What is the solution for y^'=(y+2x)^2,y(0)=sqrt(3) ?

The solution for y^'=(y+2x)^2,y(0)=sqrt(3) is y=sqrt(2)tan(sqrt(2)x+0.88607…+pin)-2x

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

y^'=(y+2x)^2,y(0)=sqrt(3)

y^{^{'}} = (y + 2 x)^{2}, y (0) = 3

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

Solution steps

y^{'} (x) = (y + 2 x)^{2}

Solve using substitution:

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

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

4 x^{4} + 3 x^{3} + 2 x^{2} + x + 11

\frac{d}{d x} (\frac{- 1}{x})

\frac{d ^{2}}{d x ^{2}} (\frac{x ^{2}}{2 + x})

\frac{d y}{d x} - 3 \cdot y = e^{2 \cdot x}

\frac{d}{d x} (460 (1 + \frac{x}{x ^{2} + 80}))

What is the solution for y^'=(y+2x)^2,y(0)=sqrt(3) ?
The solution for y^'=(y+2x)^2,y(0)=sqrt(3) is y=sqrt(2)tan(sqrt(2)x+0.88607…+pin)-2x