What is the solution for y^'=sqrt(-4y+17),y(0)=2 ?

The solution for y^'=sqrt(-4y+17),y(0)=2 is y=-t^2+3t+2

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

y^'=sqrt(-4y+17),y(0)=2

y^{^{'}} = - 4 y + 17, y (0) = 2

y = - t^{2} + 3 t + 2

Solution steps

y^{'} (t) = - 4 y + 17

Solve separable ODE:

y = - t^{2} + 3 t + 2

y = - t^{2} + 3 t + 2

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

\frac{d y}{d x} = (\frac{3 x ^{2}}{e ^{y} + 1})

\frac{d y}{d x} = \frac{( y + 1 )}{2 x}

3 y^{2} (x - 3)^{2} \frac{d y}{d x} 3 y^{2} (x - 3)^{2} = k

\frac{d y}{d x} = e^{4 y + 6 x}

What is the solution for y^'=sqrt(-4y+17),y(0)=2 ?
The solution for y^'=sqrt(-4y+17),y(0)=2 is y=-t^2+3t+2