What is the solution for y^'y^{''}=1 ?

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

y^'y^{''}=1

y^{^{'}} y^{^{''}} = 1

y = \frac{1}{3} (2 t + c_{1})^{\frac{3}{2}} + c_{2}, y = - \frac{1}{3} (2 t + c_{1})^{\frac{3}{2}} + c_{2}

Solution steps

y^{'} (t) y^{''} (t) = 1

Solve using substitution:

y = \frac{1}{3} (2 t + c_{1})^{\frac{3}{2}} + c_{2}, y = - \frac{1}{3} (2 t + c_{1})^{\frac{3}{2}} + c_{2}

y = \frac{1}{3} (2 t + c_{1})^{\frac{3}{2}} + c_{2}, y = - \frac{1}{3} (2 t + c_{1})^{\frac{3}{2}} + c_{2}

x \to 2 lim (2)

\int x^{77} ln (x) d x

\int \frac{1}{2 t} d t

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

\int (4 x + 3) e^{3 x + 1} d x

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