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

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

Notice that

$\left(\dfrac yx\right)_y=\dfrac1x$

$\left(\ln x-\ln y-1\right)_x=\dfrac1x$

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

$\Psi_x=\dfrac yx$

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

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

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

$y\ln\dfrac xy=-e$

$\boxed{y\ln\dfrac yx=e}$