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Popular Calculus >

y^'+2x=0

Solution

y^{^{'}} + 2 x = 0

Solution

y = - x^{2} + c_{1}

Solution steps

y^{'} (x) + 2 x = 0

Isolate

y^{'} (x) : y^{'} (x) = - 2 x

If

f^{^{'}} (x) = g (x)

then

f (x) = \int g (x) d x

y = \int - 2 x d x

\int - 2 x d x = - x^{2} + c_{1}

y = - x^{2} + c_{1}

Graph

Popular Examples

integral of (1+2x)^2

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

derivative of (-3x-2)/(2x^2+x+3)

d er i v a t i v e \frac{- 3 x - 2}{2 x ^{2} + x + 3}

derivative of (e^x/(x^3))

\frac{d}{d x} (\frac{e ^{x}}{x ^{3}})

(\partial)/(\partial y)(-2xe^{xy})

\frac{\partial}{\partial y} (- 2 x e^{x y})

y^{^{'}} = (x - 14)^{2}

Frequently Asked Questions (FAQ)

What is the solution for y^'+2x=0 ?
The solution for y^'+2x=0 is y=-x^2+c_{1}