What is the solution for y^'(sqrt(x^2+2x+5))=(1+y^2)(x+1) ?

The solution for y^'(sqrt(x^2+2x+5))=(1+y^2)(x+1) is y=tan((x^2)/2+x+c_{1})

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

y^'(sqrt(x^2+2x+5))=(1+y^2)(x+1)

y^{^{'}} (x^{2} + 2 x + 5) = (1 + y^{2}) (x + 1)

y = tan (\frac{x ^{2}}{2} + x + c_{1})

Solution steps

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

Solve separable ODE:

y = tan (\frac{x ^{2}}{2} + x + c_{1})

y = tan (\frac{x ^{2}}{2} + x + c_{1})

\int e^{5 x^{2}} d x

f (x) = ln (1 - x^{2} + 2 x^{4}), at x = 1

\int \frac{1}{( 3 x - 2 ) ^{2}} d x

n = 1 \sum \infty \frac{4}{n ( n + 3 )}

\frac{\partial}{\partial y} (ln (x - 2 y))

What is the solution for y^'(sqrt(x^2+2x+5))=(1+y^2)(x+1) ?
The solution for y^'(sqrt(x^2+2x+5))=(1+y^2)(x+1) is y=tan((x^2)/2+x+c_{1})