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Popular >

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

Solution

\frac{1}{1 + y ^{2}} y^{^{'}} = x

Solution

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

Solution steps

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

Solve separable ODE:

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

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

Graph

Popular Examples

y^'+2yt=0,y(1)=((e+3))/2

y^{^{'}} + 2 y t = 0, y (1) = \frac{( e + 3 )}{2}

(dy)/(dx)=14e^{5x-4y}

\frac{d y}{d x} = 14 e^{5 x - 4 y}

(4x-8)dx=(6y-10)dy

(4 x - 8) d x = (6 y - 10) d y

e^x3yy^'=e^{-y}+e^{-4x-y}

e^{x} 3 y y^{^{'}} = e^{- y} + e^{- 4 x - y}

y^'=(9x)/(y+x^2y),y(0)=-7

y^{^{'}} = \frac{9 x}{y + x ^{2} y}, y (0) = - 7

Frequently Asked Questions (FAQ)

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