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DanielM_113
Group Title
What is the general solution to the ODE
/[dy/dx = 2y+1?/]
Use separation of variables.
Is y' + xy = x separable?
 one year ago
 one year ago
DanielM_113 Group Title
What is the general solution to the ODE /[dy/dx = 2y+1?/] Use separation of variables. Is y' + xy = x separable?
 one year ago
 one year ago

This Question is Closed

DanielM_113 Group TitleBest ResponseYou've already chosen the best response.0
\[\int \frac{dy}{2y+1}=\int dx \\ \ 2 \ln y = x \]
 one year ago

DanielM_113 Group TitleBest ResponseYou've already chosen the best response.0
What are the steps to the integration of the equation? \[\int \frac{dy}{2y+1} \\ \frac{dt}{dy} = 2y+1 \\ \int dt = \int\\ \int \frac{dt}{t} = \ln t \\ \]
 one year ago

DanielM_113 Group TitleBest ResponseYou've already chosen the best response.0
Hmm, Im having trouble remembering this.
 one year ago

DanielM_113 Group TitleBest ResponseYou've already chosen the best response.0
\[t=2y+1 \\ \frac{1}{t}=\frac{1}{2y+1} \\ \frac{dt}{dy}=2 \\\] Is this true? \[\int\frac{dt}{t} = \int \frac{dy}{2y+1}\quad\mbox{?} \]
 one year ago

DanielM_113 Group TitleBest ResponseYou've already chosen the best response.0
I need to visualize integration by substitution better.
 one year ago

DanielM_113 Group TitleBest ResponseYou've already chosen the best response.0
if I set \[u=2y+1\\ \frac{\mathrm du}{\mathrm dy}= 2\\ \frac{1}{u}\frac{\mathrm du}{\mathrm dy}=\frac{1}{2y+1}2 \\ \frac{1}{2}\int \frac{\mathrm du}{u}=\int\frac{1}{2y+1}\mathrm dy \\ \int \frac{\mathrm du}{u} = \ln u = \ln (2y +1)\\ \int\frac{1}{2y+1}\mathrm dy =\frac{1}{2}\int \frac{\mathrm du}{u}=\frac{1}{2}\ln (2y +1) \]
 one year ago

DanielM_113 Group TitleBest ResponseYou've already chosen the best response.0
\[\int\frac{1}{2y+1}\mathrm dy = \int \mathrm dx \\ \frac{1}{2}\ln (2y +1) + c_1= x + c_2\\ \ln (2y +1) = 2x + 2c_2 c_1\\ 2y+1=e^{2x+2c_2c_1}\\ 2y+1=e^{2x+c_3}\\ y=\frac{1}{2}e^{c_3}e^{2x}\\ y=\frac{1}{2}Ce^{2x} \] This looks right for the first question.
 one year ago

DanielM_113 Group TitleBest ResponseYou've already chosen the best response.0
\[\int\frac{1}{2y+1}\mathrm dy = \int \mathrm dx \\ \frac{1}{2}\ln (2y +1) + c_1= x + c_2\\ \ln (2y +1) = 2x + 2c_2 c_1\\ 2y+1=e^{2x+2c_2c_1}\\ 2y+1=e^{2x+c_3}\\ y=\frac{1}{2}(e^{c_3}e^{2x}  1)\\ y=\frac{1}{2}(Ce^{2x}  1) \] This looks right for the first question.
 one year ago

DanielM_113 Group TitleBest ResponseYou've already chosen the best response.0
I forgot the 1 on the previous one.
 one year ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.2
\[ dy/dx = 2y+1 \] is seperable in the sense that we can form \[g(y)~dy = f(x)~dx\] \[ dy/dx = 2y+1 \] \[ \frac{1}{2y+1}dy = dx \] \[\int \frac{1}{2y+1}dy =\int dx \] \[\frac12ln(2y+1) =x+C \] keep in mind the that +C is an all encompassing constant: therefore 2C and c1+c2 and the like are contained within it. \[ln(2y+1) =2x+C \] \[2y+1 =e^{2x+C} \] \[2y =e^{2x+C}1 \] \[y =\frac12e^{2x+C}\frac12 \] \[y =\frac12Ce^{2x}\frac12 \] so tahts fine
 one year ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.2
y' + xy = x dy/dx = x(1y) dy/(1y) = x dx that can be seperated as well
 one year ago

DanielM_113 Group TitleBest ResponseYou've already chosen the best response.0
The first one checks with my answer, thank you! And the second one looks great too.
 one year ago

Mendicant_Bias Group TitleBest ResponseYou've already chosen the best response.0
Nevermind, answered my own question, I believe.
 23 days ago

ganeshie8 Group TitleBest ResponseYou've already chosen the best response.0
:) `1/2C` is as arbitrary as `C`
 23 days ago

ganeshie8 Group TitleBest ResponseYou've already chosen the best response.0
`e^C` is as arbitrary as `C` is too
 23 days ago

ganeshie8 Group TitleBest ResponseYou've already chosen the best response.0
so yeah we can lump `1/2e^C` by a new name `C`
 23 days ago

ganeshie8 Group TitleBest ResponseYou've already chosen the best response.0
im not sure if thats the question though, as you have deleted it before i completely reading...
 23 days ago

Mendicant_Bias Group TitleBest ResponseYou've already chosen the best response.0
Yeah, I understand that, my mistake was that I was thinking of the rootsolving methods involving homogeneous higherorder DE's, and thinking of homogeneous solutions, where y_1 and y_2 are multiplied by a *different* arbitrary constant than the original ones obtained from integration, e.g. I thought that \[\frac {1}{2}e^{2x+C}\frac{1}{2}\]was itself a single solution to a homogeneous higherorder DE, which it clearly *is not*, so I was multiplying it by c_1 in my mind. Totally different technique. Thanks, nonetheless!
 23 days ago

Mendicant_Bias Group TitleBest ResponseYou've already chosen the best response.0
\[y_h = c_1y_1+ c_2y_2\]That's what I was thinking of.
 23 days ago

ganeshie8 Group TitleBest ResponseYou've already chosen the best response.0
\(y = \frac {1}{2}e^{2x+C}\frac{1}{2}\) is the general solution and represents all the family of curves that satisfy the DE in question... as you said the arbitrary constant is already there and there wont be any more solutions
 23 days ago

ganeshie8 Group TitleBest ResponseYou've already chosen the best response.0
whle cooking up homogeneous solution for higher order DEs, we take all the linear combinations of all the independent solutions
 23 days ago

ganeshie8 Group TitleBest ResponseYou've already chosen the best response.0
so we get : \[\large y_h = \sum c_iy_i\]
 23 days ago

ganeshie8 Group TitleBest ResponseYou've already chosen the best response.0
\(\large y_i\)'s are the independent solutions to the homogeneous part
 23 days ago
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