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

This Question is Closed

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

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

DanielM_113
 2 years ago
Best ResponseYou've already chosen the best response.0Hmm, Im having trouble remembering this.

DanielM_113
 2 years ago
Best 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? \[\int\frac{dt}{t} = \int \frac{dy}{2y+1}\quad\mbox{?} \]

DanielM_113
 2 years ago
Best ResponseYou've already chosen the best response.0I need to visualize integration by substitution better.

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

DanielM_113
 2 years ago
Best 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.

DanielM_113
 2 years ago
Best 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.

DanielM_113
 2 years ago
Best ResponseYou've already chosen the best response.0I forgot the 1 on the previous one.

amistre64
 2 years ago
Best 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

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

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

Mendicant_Bias
 10 months ago
Best ResponseYou've already chosen the best response.0Nevermind, answered my own question, I believe.

ganeshie8
 10 months ago
Best ResponseYou've already chosen the best response.0:) `1/2C` is as arbitrary as `C`

ganeshie8
 10 months ago
Best ResponseYou've already chosen the best response.0`e^C` is as arbitrary as `C` is too

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

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

Mendicant_Bias
 10 months ago
Best ResponseYou've already chosen the best response.0Yeah, 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!

Mendicant_Bias
 10 months ago
Best ResponseYou've already chosen the best response.0\[y_h = c_1y_1+ c_2y_2\]That's what I was thinking of.

ganeshie8
 10 months ago
Best 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

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

ganeshie8
 10 months ago
Best ResponseYou've already chosen the best response.0so we get : \[\large y_h = \sum c_iy_i\]

ganeshie8
 10 months ago
Best ResponseYou've already chosen the best response.0\(\large y_i\)'s are the independent solutions to the homogeneous part
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