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Popular >

y^'=x

Solution

y^{^{'}} = x

Solution

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

Solution steps

y^{'} (x) = x

If

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

then

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

y = \int x d x

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

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

Popular Examples

derivative of f(x)=(-2x^2+4)^4 derivative of

f (x) = (- 2 x^{2} + 4)^{4}

tangent of y=(((x+1))/((x+3)^2))tangent of

y = (\frac{( x + 1 )}{( x + 3 ) ^{2}})

(dy)/(dt)=e^{t-y}

\frac{d y}{d t} = e^{t - y}

integral from 0 to 5 of e^xsin(x)

\int_{0}^{5} e^{x} sin (x) d x

tangent of f(x)=2x^2-x-3 tangent of

f (x) = 2 x^{2} - x - 3

Frequently Asked Questions (FAQ)

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