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Popular >

y^'=xy^2+x

Solution

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

Solution

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

Solution steps

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

Solve separable ODE:

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

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

Popular Examples

(dy)/(dx)=-2y+5

\frac{d y}{d x} = - 2 y + 5

x^3(dy)/(dx)=y^2(x-3)

x^{3} \frac{d y}{d x} = y^{2} (x - 3)

(dy)/(dx)-x/y =0

\frac{d y}{d x} - \frac{x}{y} = 0

(t^3+7t^3x^2)dt+xe^{t^4}dx=0,x(0)=0

(t^{3} + 7 t^{3} x^{2}) d t + x e^{t^{4}} d x = 0, x (0) = 0

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

y^{^{'}} = 3 (y - 1)^{\frac{2}{3}}

Frequently Asked Questions (FAQ)

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