What is the solution for ydx-(4x^2y+x)dy=0 ?

The solution for ydx-(4x^2y+x)dy=0 is y=-(1+sqrt(1-8c_{1)x^2})/(4x),y=-(1-sqrt(1-8c_{1)x^2})/(4x)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

ydx-(4x^2y+x)dy=0

y d x - (4 x^{2} y + x) d y = 0

y = - \frac{1 + 1 - 8 c _{1} x ^{2}}{4 x}, y = - \frac{1 - 1 - 8 c _{1} x ^{2}}{4 x}

Solution steps

y d x - (4 x^{2} y + x) d y = 0

Solve exact ODE:

y = - \frac{1 + 1 - 8 c _{1} x ^{2}}{4 x}, y = - \frac{1 - 1 - 8 c _{1} x ^{2}}{4 x}

y = - \frac{1 + 1 - 8 c _{1} x ^{2}}{4 x}, y = - \frac{1 - 1 - 8 c _{1} x ^{2}}{4 x}

\frac{d}{d x} (ln (x e^{- 7 x}))

\int_{1.5}^{2.5} 9800 (7 - y) (1) d y

\frac{1}{s ^{2} + 16}

\frac{1}{s ( s ^{2} + s + \frac{5}{4} )}

e^{x - 1}

What is the solution for ydx-(4x^2y+x)dy=0 ?
The solution for ydx-(4x^2y+x)dy=0 is y=-(1+sqrt(1-8c_{1)x^2})/(4x),y=-(1-sqrt(1-8c_{1)x^2})/(4x)