What is the solution for (5y-2x)y^'-2y=0 ?

The solution for (5y-2x)y^'-2y=0 is y=(4x+sqrt(16x^2+c_{1)})/(10),y=(4x-sqrt(16x^2+c_{1)})/(10)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

(5y-2x)y^'-2y=0

(5 y - 2 x) y^{^{'}} - 2 y = 0

y = \frac{4 x + 16 x ^{2} + c _{1}}{10}, y = \frac{4 x - 16 x ^{2} + c _{1}}{10}

Solution steps

(5 y - 2 x) y^{'} (x) - 2 y = 0

Solve exact ODE:

y = \frac{4 x + 16 x ^{2} + c _{1}}{10}, y = \frac{4 x - 16 x ^{2} + c _{1}}{10}

y = \frac{4 x + 16 x ^{2} + c _{1}}{10}, y = \frac{4 x - 16 x ^{2} + c _{1}}{10}

f (x) = x x^{2} + 60, at x = 2

\frac{\partial}{\partial x} ((3 - x) (3 - y) (x + y - 3))

\frac{d ^{2}}{d x ^{2}} ((3 + 2 x) e^{- 3 x})

x (- 3 x^{- 4})

\int - 5 sin^{3} (x) cos^{2} (x) d x

What is the solution for (5y-2x)y^'-2y=0 ?
The solution for (5y-2x)y^'-2y=0 is y=(4x+sqrt(16x^2+c_{1)})/(10),y=(4x-sqrt(16x^2+c_{1)})/(10)