# `ylnx - xy' = 0` Find the general solution of the differential equation For the given problem: `yln(x)-xy'=0` , we can evaluate this by applying variable separable differential equation in which we express it in a form of `f(y) dy = f(x)dx` .

to able to apply direct integration:  `int f(y) dy = int f(x)dx` .

Rearranging the problem:

`yln(x)-xy'=0`

`yln(x)=xy'`  or `xy' =...

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For the given problem: `yln(x)-xy'=0` , we can evaluate this by applying variable separable differential equation in which we express it in a form of `f(y) dy = f(x)dx` .

to able to apply direct integration:  `int f(y) dy = int f(x)dx` .

Rearranging the problem:

`yln(x)-xy'=0`

`yln(x)=xy'`  or `xy' = y ln(x)`

`(xy')/(yx) = (y ln(x))/(yx)`

`(y') /y = ln(x)/x`

Applying direct integration, we denote `y' = (dy)/(dx)` :

`int (y') /y = int ln(x)/x`

`int 1 /y (dy)/(dx) = int ln(x)/x`

`int 1 /y (dy)= int ln(x)/x dx`

For the left side, we apply the basic integration formula for logarithm: `int (du)/u = ln|u|+C`

`int 1 /y (dy) = ln|y|`

For the right side, we apply u-substitution by letting `u= ln(x)` then `du = 1/x dx` .

`int ln(x)/x dx=int udu`

Applying the Power Rule for integration : `int x^n= x^(n+1)/(n+1)+C` .

`int udu=u^(1+1)/(1+1)+C`

`=u^2/2+C`

Plug-in `u = ln(x)` in `u^2/2+C` , we get:

`int ln(x)/x dx =(ln(x))^2/2+C`

Combining the results, we get the general solution for differential equation `(yln(x)-xy'=0)`  as:

`ln|y|=(ln|x|)^2/2+C`

The general solution:` ln|y|=(ln|x|)^2/2+C` can be expressed as:

`y = C_1e^((ln|x|)^2/2)+C` .

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