(x+2y-1)dx-(x+2y-5)dy=0

What do you need to do with this equation? You just write the equation but there is no instruction.

Can i solve this using homo? Our professor only gave that equation and nothing else.

Nope use suggested substitution method. since the linear equation is parallel, then use u = x + 2y

$$(x+2y-1)dx-(x+2y-5)dy=0$$

Let $u = x + 2y - 1 \rightarrow du = dx + 2dy$ which gives $dx = du - 2dy$ Substituting to the equation $$\begin{eqnarray} u(du - 2dy) - (u - 4)dy &=& 0\\ u du - (3u - 4)dy &=& 0\\ \dfrac{udu}{3u-4} - dy &=& 0\\ \dfrac{1}{3} \left(1 - \dfrac{4}{3u - 4}\right)du - dy &=& 0\\ \displaystyle{\int \dfrac{1}{3} \left(1 - \dfrac{4}{3u - 4}\right)}du - \int dy &=& C\\ \dfrac{1}{3}u - \dfrac{4}{3} \ln|3u - 4| - y &=& C\\ u - 4 \ln|3u - 4| - 3y &=& C\\ 4 \ln|3u - 4| &=& u - 3y + C\\ 4 \ln|3(x + 2y - 1) - 4| &=& x + 2y - 3y + C\\ 4 \ln|3x + 6y - 7| &=& x - y + C\\ 3x + 6y - 7 &=& Ce^{\frac{x-y}{4}} (ANSWER) \end{eqnarray}$$

P.S. There can also be other forms of the general solution

MATHalino

What do you need to do with this equation? You just write the equation but there is no instruction.

Can i solve this using homo? Our professor only gave that equation and nothing else.

Nope use suggested substitution method. since the linear equation is parallel, then use u = x + 2y

$$(x+2y-1)dx-(x+2y-5)dy=0$$

Let $u = x + 2y - 1 \rightarrow du = dx + 2dy$ which gives $dx = du - 2dy$

Substituting to the equation

$$\begin{eqnarray}

u(du - 2dy) - (u - 4)dy &=& 0\\

u du - (3u - 4)dy &=& 0\\

\dfrac{udu}{3u-4} - dy &=& 0\\

\dfrac{1}{3} \left(1 - \dfrac{4}{3u - 4}\right)du - dy &=& 0\\

\displaystyle{\int \dfrac{1}{3} \left(1 - \dfrac{4}{3u - 4}\right)}du - \int dy &=& C\\

\dfrac{1}{3}u - \dfrac{4}{3} \ln|3u - 4| - y &=& C\\

u - 4 \ln|3u - 4| - 3y &=& C\\

4 \ln|3u - 4| &=& u - 3y + C\\

4 \ln|3(x + 2y - 1) - 4| &=& x + 2y - 3y + C\\

4 \ln|3x + 6y - 7| &=& x - y + C\\

3x + 6y - 7 &=& Ce^{\frac{x-y}{4}} (ANSWER)

\end{eqnarray}$$

P.S. There can also be other forms of the general solution