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Popular >

2xy^'ln(x)=-2y

Solution

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

Solution

y = \frac{c _{1}}{ln ( x )}

Solution steps

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

Solve separable ODE:

y = \frac{c _{1}}{ln ( x )}

y = \frac{c _{1}}{ln ( x )}

Popular Examples

y^'sec(t)=(-y^{23})/(22)

y^{^{'}} sec (t) = \frac{- y ^{23}}{22}

(dy)/(dx)6cos(4x)sin(2y)=1

\frac{d y}{d x} 6 cos (4 x) sin (2 y) = 1

x*(dy)/(dx)+x^3y=0

x \cdot \frac{d y}{d x} + x^{3} y = 0

(dy)/(dx)=y^2-13y+42

\frac{d y}{d x} = y^{2} - 13 y + 42

y^'= y/(36t^2+1)

y^{^{'}} = \frac{y}{36 t ^{2} + 1}

Frequently Asked Questions (FAQ)

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