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Popular >

2xy^'+y=0

Solution

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

Solution

y = \frac{c _{1}}{x}

Solution steps

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

Solve separable ODE:

y = \frac{c _{1}}{x}

y = \frac{c _{1}}{x}

Graph

Popular Examples

(-1+y)dx+(x-1)dy=0

(- 1 + y) d x + (x - 1) d y = 0

y y^{^{'}} = 4 x + 9

(df)/(dt)f(t)=(sin(t^2)+1)^4

\frac{df}{d t} f (t) = (sin (t^{2}) + 1)^{4}

(dy}{dx}=\frac{15x^2)/y

\frac{d y}{d x} = \frac{15 x ^{2}}{y}

(dy)/(dx)=((x+2))/(2y),y(0)=1

\frac{d y}{d x} = \frac{( x + 2 )}{2 y}, y (0) = 1

Frequently Asked Questions (FAQ)

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