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Popular >

y^'+x=0

Solution

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

Solution

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

Solution steps

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

Isolate

y^{'} (x) : 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

integral of sqrt(8-4x-4x^2)

\int 8 - 4 x - 4 x^{2} d x

limit as x approaches 0 of x^2+4x+2

x \to 0 lim (x^{2} + 4 x + 2)

sum from n=1 to infinity of 1/(ln(n+1))

n = 1 \sum \infty \frac{1}{ln ( n + 1 )}

y^'=k(y-9-10cos(2*pi*t))

y^{^{'}} = k (y - 9 - 10 cos (2 \cdot π \cdot t))

(d^2)/(dx^2)(2sin(-4x-3))

\frac{d ^{2}}{d x ^{2}} (2 sin (- 4 x - 3))

Frequently Asked Questions (FAQ)

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