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Popular >

xy^'=yln(xy)

Solution

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

Solution

y = \frac{e ^{c_{1} x - 1}}{x}

Solution steps

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

Solve using substitution:

y = \frac{e ^{c_{1} x - 1}}{x}

y = \frac{e ^{c_{1} x - 1}}{x}

Popular Examples

limit as x approaches 0+of 2xln(4x^2)

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

(dy)/(dx)= x/(3y^2)

\frac{d y}{d x} = \frac{x}{3 y ^{2}}

sum from n=1 to infinity of (25n!)/(n^n)

n = 1 \sum \infty \frac{25 n !}{n ^{n}}

derivative of x/({f(x)})

\frac{d}{d x} (\frac{x}{f ( x )})

2yy^'-xy^'=y-2x

2 y y^{^{'}} - x y^{^{'}} = y - 2 x

Frequently Asked Questions (FAQ)

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