What is the solution for 2y^'=e^{2x}y^4 ?

The solution for 2y^'=e^{2x}y^4 is y=-\sqrt[3]{4/(3(e^{2x)+c_{1})}}

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

2y^'=e^{2x}y^4

2 y^{^{'}} = e^{2 x} y^{4}

y = - 3 \frac{4}{3 ( e ^{2 x} + c _{1} )}

Solution steps

2 y^{'} (x) = e^{2 x} y^{4}

Solve separable ODE:

y = - 3 \frac{4}{3 ( e ^{2 x} + c _{1} )}

y = - 3 \frac{4}{3 ( e ^{2 x} + c _{1} )}

\frac{d x}{d t} = 1 - \frac{x}{25}

(1 + cos (x)) \frac{d y}{d x} = sin (x) (e - y + 1)

x \frac{d t}{d x} = \frac{a + a t ^{2}}{1 - a t}

sec (t) y^{^{'}} = - \frac{y ^{5}}{4}

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

What is the solution for 2y^'=e^{2x}y^4 ?
The solution for 2y^'=e^{2x}y^4 is y=-\sqrt[3]{4/(3(e^{2x)+c_{1})}}