Question

Test for exactness and find the complete solution of y/(x-1) dx+ {ln(2x-2)+ 1/y}dy=0

Answer 1

We check the differential equation for exactness:

$$\dfrac{y}{x-1}dx+(\ln(2x-2)+\dfrac{1}{y})dy=0$$

$$M(x,y)=\dfrac{y}{x-1}, \dfrac{\partial M}{\partial y}=\dfrac{1}{x-1}$$

$$N(x, y)=\ln(2x-2)+\dfrac{1}{y}, \dfrac{\partial N}{\partial x}=\dfrac{1}{x-1}$$

$$\dfrac{\partial M}{\partial y}=\dfrac{1}{x-1}= \dfrac{\partial N}{\partial x}$$

We see that $\dfrac{\partial M}{\partial y}= \dfrac{\partial N}{\partial x},$ so that this equation is exact.

We have the following system of differential equations to find the function $u(x, y)$

$$\dfrac{\partial u}{\partial x}=\dfrac{y}{x-1}$$

$$\dfrac{\partial u}{\partial y}=\ln(2x-2)+\dfrac{1}{y}$$

Hence

$$u(x, y)=\int\dfrac{y}{x-1}dx+\varphi(y)$$

$$=y\ln(x-1)+\varphi(y)$$

Now, by differentiating this expression with respect to $y$ and equating it to $\dfrac{\partial u}{\partial y},$ we find the derivative $\varphi'(y)$

$$\ln(x-1)+\varphi'(y)=\ln(2x-2)+\dfrac{1}{y}$$

$$\varphi'(y)=\ln2+\dfrac{1}{y}$$

$$\varphi(y)=\int (\ln2+\dfrac{1}{y})dy=y\ln2+\ln y-C$$

Thus, the general solution of the differential equation is

$$y\ln(2x-2)+\ln y=C$$