What is the solution for y^'+y^{1/2}=0 ?

The solution for y^'+y^{1/2}=0 is y=(t^2)/4-(c_{1}t)/2+(c_{1}^2)/4

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

y^'+y^{1/2}=0

y^{^{'}} + y^{\frac{1}{2}} = 0

y = \frac{t ^{2}}{4} - \frac{c _{1} t}{2} + \frac{c _{1}^{2}}{4}

Solution steps

y^{'} (t) + y^{\frac{1}{2}} = 0

Solve separable ODE:

y = \frac{t ^{2}}{4} - \frac{c _{1} t}{2} + \frac{c _{1}^{2}}{4}

y = \frac{t ^{2}}{4} - \frac{c _{1} t}{2} + \frac{c _{1}^{2}}{4}

(sin^{2} (t))^{^{'}}

\frac{d}{d x} (- 18 sin (3 x))

x \to 0 + lim (x^{4})

x \to - 3 lim (\frac{x + 1}{x + 3})

(19, 3), (20, 3)

What is the solution for y^'+y^{1/2}=0 ?
The solution for y^'+y^{1/2}=0 is y=(t^2)/4-(c_{1}t)/2+(c_{1}^2)/4