en

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

y^'=-ay^2+b

Solution

y^{^{'}} = - a y^{2} + b

Solution

y = - \frac{b e ^{2 a b t + 2 c_{1} a b}}{a ( 1 - e ^{2 a b (t + c_{1})} )} - \frac{b}{a ( 1 - e ^{2 a b (t + c_{1})} )}

Solution steps

y^{'} (t) = - a y (t)^{2} + b

Solve separable ODE:

y = - \frac{b e ^{2 a b t + 2 c_{1} a b}}{a ( 1 - e ^{2 a b (t + c_{1})} )} - \frac{b}{a ( 1 - e ^{2 a b (t + c_{1})} )}

y = - \frac{b e ^{2 a b t + 2 c_{1} a b}}{a ( 1 - e ^{2 a b (t + c_{1})} )} - \frac{b}{a ( 1 - e ^{2 a b (t + c_{1})} )}

Popular Examples

derivative of tan(1+e^{2x})

\frac{d}{d x} (tan (1 + e^{2 x}))

derivative of-2log_{e}(x)+x^2-4+e^x+x^e derivative of

- 2 lo g_{e} (x) + x^{2} - 4 + e^{x} + x^{e}

integral of 2csc^4(x)cot^6(x)

\int 2 csc^{4} (x) cot^{6} (x) d x

limit as x approaches-2 of 7-2x

x \to - 2 lim (7 - 2 x)

integral of-2xe^{x^2}

\int - 2 x e^{x^{2}} d x

Frequently Asked Questions (FAQ)

What is the solution for y^'=-ay^2+b ?
The solution for y^'=-ay^2+b is