What is the solution for y^{1/2}y^'=1-y^{3/2} ?

The solution for y^{1/2}y^'=1-y^{3/2} is y=(-e^{-(3t+c_{1})/2}+1)^{2/3}

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

y^{1/2}y^'=1-y^{3/2}

y^{\frac{1}{2}} y^{^{'}} = 1 - y^{\frac{3}{2}}

y = (- e^{- \frac{3 t + c _{1}}{2}} + 1)^{\frac{2}{3}}

Solution steps

y^{\frac{1}{2}} y^{'} (t) = 1 - y^{\frac{3}{2}}

Solve separable ODE:

y = (- e^{- \frac{3 t + c _{1}}{2}} + 1)^{\frac{2}{3}}

y = (- e^{- \frac{3 t + c _{1}}{2}} + 1)^{\frac{2}{3}}

\frac{d y}{d x} = (x^{2} y)

\frac{d y}{d x} = \frac{( 5 x ^{6} )}{y ^{2}}, y (0) = 1

e^{x^{2}} \frac{d y}{d x} + e^{x^{2}} 2 x y = e^{x^{2}} 2 x

y^{^{'}} = - 2 \cdot t \cdot y^{2}, y (0) = 1

x^{\frac{1}{3}} y^{^{'}} + y = 0

What is the solution for y^{1/2}y^'=1-y^{3/2} ?
The solution for y^{1/2}y^'=1-y^{3/2} is y=(-e^{-(3t+c_{1})/2}+1)^{2/3}