What is the solution for y^'=1+sqrt(y-2x+3) ?

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

y^'=1+sqrt(y-2x+3)

y^{^{'}} = 1 + y - 2 x + 3

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

Solution steps

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

Solve using substitution:

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

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

\int (x - 1)^{2} (x + 1)^{2} d x

\int 5 - x d x

\frac{d}{d x} (- 3 cot (2 x) sec (2 x))

\int x^{2} + 8 e^{x} + C d x

y^{^{''}} + 3 y^{^{'}} + 2 y = e^{x} + e^{2 x}

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