en

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

Solution

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

Solution

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

Solution steps

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

Solve separable ODE:

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

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

Popular Examples

tangent of y=t^3-36t+4,(6,4)tangent of

y = t^{3} - 36 t + 4, (6, 4)

limit as x approaches 0-of e^{4/(x^5)}

x \to 0 - lim (e^{\frac{4}{x ^{5}}})

integral of (x^2-6)/((x+2)(2x-1))

\int \frac{x ^{2} - 6}{( x + 2 ) ( 2 x - 1 )} d x

area y=x^2,y=sqrt(6+x)area

y = x^{2}, y = 6 + x

derivative of log_{e}(4x)

\frac{d}{d x} (lo g_{e} (4 x))

Frequently Asked Questions (FAQ)

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