Differential equation: $(x+2y-1)dx-(x+2y-5)dy=0$ Member for 6 years 7 months Submitted by Helpme on Wed, 08/21/2019 - 20:41 (x+2y-1)dx-(x+2y-5)dy=0 Tags nonlinear DE Log in or register to post comments What do you need to do with Member for 17 years 6 months 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 or register to post comments Can i solve this using homo? Member for 6 years 7 months 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 or register to post comments Nope use suggested Member for 6 years 7 months 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 or register to post comments $$(x+2y-1)dx-(x+2y-5)dy=0$$ Member for 10 years 1 month 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 or register to post comments
What do you need to do with Member for 17 years 6 months 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 or register to post comments
Can i solve this using homo? Member for 6 years 7 months 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 or register to post comments
Nope use suggested Member for 6 years 7 months 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 or register to post comments
$$(x+2y-1)dx-(x+2y-5)dy=0$$ Member for 10 years 1 month 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 or register to post comments
What do you need to do with
Member for
17 years 6 monthsWhat do you need to do with this equation? You just write the equation but there is no instruction.
Can i solve this using homo?
Member for
6 years 7 monthsIn 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
Member for
6 years 7 monthsIn 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
$$(x+2y-1)dx-(x+2y-5)dy=0$$
Member for
10 years 1 month$$(x+2y-1)dx-(x+2y-5)dy=0$$
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