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Popular >

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

Solution

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

Solution

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

Solution steps

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

Solve separable ODE:

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

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

Popular Examples

(dy)/(dx)=\sqrt[3]{x(y-1)}

(dx)/(dt)=5000+0.04x,x=18000,t=0

\frac{d x}{d t} = 5000 + 0.04 x, x = 18000, t = 0

y+x(dy)/(dx)=2x^2y(dy)/(dx)+2xy^2

y + x \frac{d y}{d x} = 2 x^{2} y \frac{d y}{d x} + 2 x y^{2}

(dy)/(dt)=3-0.04y

\frac{d y}{d t} = 3 - 0.04 y

(xy+3x-y-3)/(xy-2x+4y-8)=(dy)/(dx)

\frac{x y + 3 x - y - 3}{x y - 2 x + 4 y - 8} = \frac{d y}{d x}

Frequently Asked Questions (FAQ)

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