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Popular >

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

Solution

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

Solution

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

Solution steps

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

Solve separable ODE:

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

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

Popular Examples

inverse oflaplace 1/(s-10)inverse laplace

\frac{1}{s - 10}

(\partial)/(\partial x)(cos(y)sin(x))

\frac{\partial}{\partial x} (cos (y) sin (x))

derivative of (x+1/(x+2))

\frac{d}{d x} (\frac{x + 1}{x + 2})

integral of 1/((2x+5))

\int \frac{1}{( 2 x + 5 )} d x

derivative of 1/(sqrt(x^2))

\frac{d}{d x} (\frac{1}{x ^{2}})

Frequently Asked Questions (FAQ)

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