Question

What is the solution of differential equation Xdy/dx= y(logy-logx+1) ?

Answer 1

`y=xA^x, A>1`

Explanation:

Use the substitution `y =v x`, so that,

`(d y)/( d x)=v+x (dv)/(d x)`, and so, from the given differential equation,

`v+x(dv)/(dx)=v(ln v +1)`.

Simplifying, separating variables and integrating,

`int(dv)/(v ln v)=int(d ln v)/ln v=int (dx)/x`.Upon integration,

`ln(ln v)= ln x +ln C`, C in the integration constant`>0`,

Simplifying,

`ln(ln v/x) = ln C`. Further simplification and reversion to y=xv

gives

`ln(y/x)=Cx`. Inverting,

`y/x=e^(Cx)=(e^c)^x=A^x, A=e^C>1`, as `C>0`

Thus, `y = xA^x, A>1`

.