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

MIT 18.03SC Differential Equations
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\[\int \frac{dy}{2y+1}=\int dx \\ \ 2 \ln y = x \]
What are the steps to the integration of the equation? \[\int \frac{dy}{2y+1} \\ \frac{dt}{dy} = 2y+1 \\ \int dt = \int\\ \int \frac{dt}{t} = \ln t \\ \]
Hmm, Im having trouble remembering this.

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Other answers:

\[t=2y+1 \\ \frac{1}{t}=\frac{1}{2y+1} \\ \frac{dt}{dy}=2 \\\] Is this true? \[\int\frac{dt}{t} = \int \frac{dy}{2y+1}\quad\mbox{?} \]
I need to visualize integration by substitution better.
if I set \[u=2y+1\\ \frac{\mathrm du}{\mathrm dy}= 2\\ \frac{1}{u}\frac{\mathrm du}{\mathrm dy}=\frac{1}{2y+1}2 \\ \frac{1}{2}\int \frac{\mathrm du}{u}=\int\frac{1}{2y+1}\mathrm dy \\ \int \frac{\mathrm du}{u} = \ln u = \ln (2y +1)\\ \int\frac{1}{2y+1}\mathrm dy =\frac{1}{2}\int \frac{\mathrm du}{u}=\frac{1}{2}\ln (2y +1) \]
\[\int\frac{1}{2y+1}\mathrm dy = \int \mathrm dx \\ \frac{1}{2}\ln (2y +1) + c_1= x + c_2\\ \ln (2y +1) = 2x + 2c_2 -c_1\\ 2y+1=e^{2x+2c_2-c_1}\\ 2y+1=e^{2x+c_3}\\ y=\frac{1}{2}e^{c_3}e^{2x}\\ y=\frac{1}{2}Ce^{2x} \] This looks right for the first question.
\[\int\frac{1}{2y+1}\mathrm dy = \int \mathrm dx \\ \frac{1}{2}\ln (2y +1) + c_1= x + c_2\\ \ln (2y +1) = 2x + 2c_2 -c_1\\ 2y+1=e^{2x+2c_2-c_1}\\ 2y+1=e^{2x+c_3}\\ y=\frac{1}{2}(e^{c_3}e^{2x} - 1)\\ y=\frac{1}{2}(Ce^{2x} - 1) \] This looks right for the first question.
I forgot the -1 on the previous one.
\[ dy/dx = 2y+1 \] 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
y' + xy = x dy/dx = x(1-y) dy/(1-y) = x dx that can be seperated as well
The first one checks with my answer, thank you! And the second one looks great too.
Nevermind, answered my own question, I believe.
:) `1/2C` is as arbitrary as `C`
`e^C` is as arbitrary as `C` is too
so yeah we can lump `1/2e^C` by a new name `C`
im not sure if thats the question though, as you have deleted it before i completely reading...
Yeah, I understand that, my mistake was that I was thinking of the root-solving methods involving homogeneous higher-order 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 higher-order DE, which it clearly *is not*, so I was multiplying it by c_1 in my mind. Totally different technique. Thanks, nonetheless!
\[y_h = c_1y_1+ c_2y_2\]That's what I was thinking of.
I see..
\(y = \frac {1}{2}e^{2x+C}-\frac{1}{2}\) 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
whle cooking up homogeneous solution for higher order DEs, we take all the linear combinations of all the independent solutions
so we get : \[\large y_h = \sum c_iy_i\]
\(\large y_i\)'s are the independent solutions to the homogeneous part

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