Question

Solve the system for the exact special solution y = y(x): (keep the fraction and the square root without decimals.) 1. ydx + x[ In(x) - In(y) - 1]dy = 0 and y(1) = e for In(e) = 1.

Answer 1

Assume a solution of the form
`\Psi(x,y)=C`. Differentiating both sides gives

`\Psi_x\,\mathrm dx+\Psi_y\,\mathrm dy=0`

with
`\Psi_x=y` and
`\Psi_y=x(\ln x-\ln y-1)`.

Divide both sides by
`x` and we have

`\frac yx\,\mathrm dx+(\ln x-\ln y-1)\,\mathrm dy=0`

Notice that

`\left(\frac yx\right)_y=\frac1x`

`\left(\ln x-\ln y-1\right)_x=\frac1x`

so the ODE is exact. Now we can look for a solution
`\Psi` with

`\Psi_x=\frac yx`

`\Psi_y=\ln x-\ln y-1`

Integrating the first PDE with respect to
`x` gives

`\Psi(x,y)=y\ln x+f(y)`

and differentiating this with respect to
`y` gives

`\Psi_y=\ln x+f'(y)=\ln x-\ln y-1\implies f'(y)=-\ln y-1\implies f(y)=-y\ln y+C`

So this ODE has general solution

`y\ln x-y\ln y=C`

Given that
`y(1)=e`, we have

`e\ln1-e\ln e=C\implies C=-e`

so the particular solution is

`y(\ln x-\ln y)=-e`

`y\ln\frac xy=-e`

`\boxed{y\ln\frac yx=e}`