Differential equation: $(x+2y-1)dx-(x+2y-5)dy=0$ Submitted by Helpme on Wed, 08/21/2019 - 20:41 (x+2y-1)dx-(x+2y-5)dy=0 Tags nonlinear DE Log in to post comments What do you need to do with Jhun Vert Fri, 08/23/2019 - 07:52 What do you need to do with this equation? You just write the equation but there is no instruction. Log in to post comments Can i solve this using homo? Helpme Sat, 08/24/2019 - 11:20 In reply to What do you need to do with by Jhun VertCan i solve this using homo? Our professor only gave that equation and nothing else. Log in to post comments Nope use suggested amaziahbryceherrera Wed, 08/28/2019 - 16:00 In reply to Can i solve this using homo? by HelpmeNope use suggested substitution method. since the linear equation is parallel, then use u = x + 2y Log in to post comments $$(x+2y-1)dx-(x+2y-5)dy=0$$ Infinitesimal Mon, 03/02/2020 - 18:54 $$(x+2y-1)dx-(x+2y-5)dy=0$$ 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 Log in to post comments
What do you need to do with Jhun Vert Fri, 08/23/2019 - 07:52 What do you need to do with this equation? You just write the equation but there is no instruction. Log in to post comments
Can i solve this using homo? Helpme Sat, 08/24/2019 - 11:20 In reply to What do you need to do with by Jhun VertCan i solve this using homo? Our professor only gave that equation and nothing else. Log in to post comments
Nope use suggested amaziahbryceherrera Wed, 08/28/2019 - 16:00 In reply to Can i solve this using homo? by HelpmeNope use suggested substitution method. since the linear equation is parallel, then use u = x + 2y Log in to post comments
$$(x+2y-1)dx-(x+2y-5)dy=0$$ Infinitesimal Mon, 03/02/2020 - 18:54 $$(x+2y-1)dx-(x+2y-5)dy=0$$ 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 Log in to post comments
What do you need to do with
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?
In reply to What do you need to do with by Jhun Vert
Can i solve this using homo? Our professor only gave that equation and nothing else.
Nope use suggested
In reply to Can i solve this using homo? by Helpme
Nope use suggested substitution method. since the linear equation is parallel, then use u = x + 2y
$$(x+2y-1)dx-(x+2y-5)dy=0$$
$$(x+2y-1)dx-(x+2y-5)dy=0$$
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