What is the solution for (y-x)y^'=y-x+8 ?

The solution for (y-x)y^'=y-x+8 is y=(2x+sqrt(64x+c_{1)})/2 ,y=(2x-sqrt(64x+c_{1)})/2

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

(y-x)y^'=y-x+8

(y - x) y^{^{'}} = y - x + 8

y = \frac{2 x + 64 x + c _{1}}{2}, y = \frac{2 x - 64 x + c _{1}}{2}

Solution steps

(y - x) y^{'} (x) = y - x + 8

Solve exact ODE:

y = \frac{2 x + 64 x + c _{1}}{2}, y = \frac{2 x - 64 x + c _{1}}{2}

y = \frac{2 x + 64 x + c _{1}}{2}, y = \frac{2 x - 64 x + c _{1}}{2}

x \to 0 lim (2 x + 1 - 1)

\frac{d}{d x} ((ln (3 x))^{5})

2 x^{2} - 7 x + k

\int (6 x) d x

y = - 3 x^{3} - 3 x^{2} - x + 3, (- 1, 4)

What is the solution for (y-x)y^'=y-x+8 ?
The solution for (y-x)y^'=y-x+8 is y=(2x+sqrt(64x+c_{1)})/2 ,y=(2x-sqrt(64x+c_{1)})/2