(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

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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