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gerryliyana Group TitleBest ResponseYou've already chosen the best response.2
Solve \[xe^{2x} \frac{ dy }{ dx } + e^{2x} (2x+1) y \]
 2 years ago

Outkast3r09 Group TitleBest ResponseYou've already chosen the best response.1
y not solve using seperation
 2 years ago

gerryliyana Group TitleBest ResponseYou've already chosen the best response.2
@Outkast3r09 Yes, of course, i really know.., it is use linear equations method, isn't it?
 2 years ago

Meepi Group TitleBest ResponseYou've already chosen the best response.1
It's a separable equation, which means it can be written in the form N(y) dy = M(x) dx Rewrite the equation to get it in this form, and integrate both sides
 2 years ago

Skaematik Group TitleBest ResponseYou've already chosen the best response.0
Im learning too, is this right? \[\frac{ dx }{ xe ^{2x} }= e ^{2x}y dy (2x+1)\]
 2 years ago

Meepi Group TitleBest ResponseYou've already chosen the best response.1
What you want do do is have only y's on the side of dy, and only x's on the side of dx. Here are some example problems http://tutorial.math.lamar.edu/Classes/DE/Separable.aspx which will give you a good general idea of how to tackle the problem :) \[x e^{2x} \frac{dy}{dx} + e^{2x}(2x + 1)y = 0\] \[x e^{2x} dy =  e^{2x}(2x + 1)y dx\] \[\frac{1}{y}dy = \frac{e^{2x}(2x + 1)}{xe^{2x}}dx\] Then integrate both sides :)
 2 years ago

Meepi Group TitleBest ResponseYou've already chosen the best response.1
Btw \[\frac{e^{2x}(2x + 1)}{x e^{2x}}\] can be simplified to: \[\frac{1}{x}  2\]
 2 years ago

Meepi Group TitleBest ResponseYou've already chosen the best response.1
So what you get is: \[\int\limits_{}^{} \frac{1}{y} dy = \int\limits_{}^{} ( \frac{1}{x}  2) dx\] I'll leave it to you to solve it
 2 years ago

Outkast3r09 Group TitleBest ResponseYou've already chosen the best response.1
You can probably use linear equations to solve it also... The whole point of Diff. Equations is to find multiple ways to solve the same problem
 2 years ago

Outkast3r09 Group TitleBest ResponseYou've already chosen the best response.1
first start off by dividing by \[xe^2x\]
 2 years ago

Outkast3r09 Group TitleBest ResponseYou've already chosen the best response.1
\[\frac{dy}{dx}+\frac{e^{2x}(2x+1)}{xe^{2x}}=0\]
 2 years ago

Outkast3r09 Group TitleBest ResponseYou've already chosen the best response.1
\[ \frac{dy}{dx}+\frac{(2x+1)}{x}=0\]
 2 years ago

Outkast3r09 Group TitleBest ResponseYou've already chosen the best response.1
now find the integration factor \[e^{\int{p(x)dx}}=e^{\int{\frac{2x+1}{x}dx}}\]
 2 years ago

Outkast3r09 Group TitleBest ResponseYou've already chosen the best response.1
\[e^{\int{2+\frac{1}{x}}dx}\]
 2 years ago

Outkast3r09 Group TitleBest ResponseYou've already chosen the best response.1
\[e^{2x\frac{1}{x^2}}\]
 2 years ago

Meepi Group TitleBest ResponseYou've already chosen the best response.1
Isn't it\[\int_{}^{} (2 + \frac{1}{x}) dx = 2x + \ln x \]
 2 years ago

gerryliyana Group TitleBest ResponseYou've already chosen the best response.2
Nice @Outkast3r09 and @Meepi., I think.., in order to satisfy the equation: \[\frac{ dy }{ dx } + p(x)y = g(y)\] then the original equation becomes: \[x^{e^{2x}} \frac{ dy }{ dx } + e^{2x} 2xy+e^{2x} = 0 \] \[x^{e^{2x}} \frac{ dy }{ dx } + e^{2x} 2xy = e^{2x}\] \[\frac{ dy }{ dx } + \frac{ e^{2x} 2xy }{ x^{e^{2x}} } = \frac{ e^{2x} y }{ x^{e^{2x}} }\] So P(x) = \(\frac{ e^{2x} 2x}{ x^{e^{2x}}}\) and g(y) =\(\frac{ e^{2x}y }{ x^{e^{2x}}} \) Hbu???
 2 years ago

gerryliyana Group TitleBest ResponseYou've already chosen the best response.2
hbu @nubeer ???
 2 years ago

nubeer Group TitleBest ResponseYou've already chosen the best response.0
xe^2x (dy/dx) + e^2x ( 2x+1)y = 0 right is this the question? i might take e^2x common first .. and then divide the whole equation by it
 2 years ago

gerryliyana Group TitleBest ResponseYou've already chosen the best response.2
yes., of course
 2 years ago

nubeer Group TitleBest ResponseYou've already chosen the best response.0
dw:1353775372293:dw
 2 years ago

nubeer Group TitleBest ResponseYou've already chosen the best response.0
so answer should be lny = 2x lnx +c
 2 years ago

gerryliyana Group TitleBest ResponseYou've already chosen the best response.2
ok.., it's separable equations method, right? what if "linear equations method"? because its from the questions bank of Linear equations method., :)
 2 years ago

nubeer Group TitleBest ResponseYou've already chosen the best response.0
honestly i dont know most methods by name.. but quite dont remember about linear equation method.
 2 years ago

gerryliyana Group TitleBest ResponseYou've already chosen the best response.2
ok np., thank u so much @nubeer :)
 2 years ago

nubeer Group TitleBest ResponseYou've already chosen the best response.0
you are most welcome :)
 2 years ago

UnkleRhaukus Group TitleBest ResponseYou've already chosen the best response.1
\[xe^{2x} \frac{ \text dy }{\text dx } + e^{2x} (2x+1) y=0\]\[xe^{2x} \text dy + e^{2x} (2x+1) y{\text dx }=0\] is of the form \(M\text dy+N\text dx=0\) \[\frac{\partial M}{\partial x}=e^{2x}+2xe^{2x}=(1+2x)e^{2x}\]\[\frac{\partial N}{\partial y}=(2x+1)e^{2x}\] \(\frac{\partial M}{\partial x}=\frac{\partial N}{\partial y}\) the equation is exact \[\int M\text dy=\int xe^{2x} \text dy =yxe^{2x}+c_1\] \[\int N\text dx=\int e^{2x} (2x+1) y\text dx\]\[\qquad\qquad=y\int e^{2x} \text dx+2y\int xe^{2x}\text dx\]\[\qquad\qquad=\frac{ye^{2x}}2+c_2+2y\left[\left.\frac{xe^{2x}}2\right \int \frac{e^{2x} }2\text dx\right]\]\[\qquad\qquad=\frac{ye^{2x}}2+2y\left[\frac{xe^{2x}}2 \frac{e^{2x} }4\right]+c_3 \]\[\qquad\qquad= y{xe^{2x}}+c_3\] \[f(x,y)=y{xe^{2x}}=c\]
 2 years ago

Outkast3r09 Group TitleBest ResponseYou've already chosen the best response.1
your first way was wrong
 2 years ago

Outkast3r09 Group TitleBest ResponseYou've already chosen the best response.1
the y is distributed to both of the latter terms... so you can't distribute... you don't need a constant in order to satisfy the linear equation method. you only need a coefficient of one for the derivative term , a p(x) and a g(x) which can equal 0
 2 years ago

Outkast3r09 Group TitleBest ResponseYou've already chosen the best response.1
Rhaukaus answer is the general solution. He used exact method though. you should get the same answer using linear
 2 years ago

gerryliyana Group TitleBest ResponseYou've already chosen the best response.2
Thank u @UnkleRhaukus and @Outkast3r09 how about this \[xe^{2x} \frac{ dy }{ dx } + e^{2x} (2x+1)y = 0\] \[\frac{ dy }{ dx }+\frac{ e^{2x}(2x+1) }{ xe^{2x} }y = 0\] \[\frac{ dy }{ dx }+\frac{(2x+1) }{x }y = 0\] so , \(p(x) = \frac{ (2x+1) }{ x} \) and g(x) = 0 for integration factor: \[\mu(x) = e^{\int\limits p(x) dx}\] \[\mu(x) = \exp(\int\limits_{0}^{x} \frac{ 2x+1 }{ x } dx)\] \[\mu(x) = \exp(2x+\ln x) = e^{2x} e^{\ln x}\] \[\mu(x) = xe^{2x}\] \[y \mu(x) = \int\limits_{0}^{x} \mu(x) g(x) dx\] finally, for the solution : \[y = \frac{ 1 }{ \mu(x) } \int\limits \mu(x) g(x) dx \] \[y \mu(x) = \int\limits_{0}^{x} \mu(x) g(x) dx\] \[yxe^{2x} = \int\limits_{0}^{x} xe^{2x} (0) dx\] \[yxe^{2x} = 0\] \[xye^{2x} = C\] am i right or wrong????
 one year ago

Outkast3r09 Group TitleBest ResponseYou've already chosen the best response.1
that's right
 one year ago

UnkleRhaukus Group TitleBest ResponseYou've already chosen the best response.1
great work @gerryliyana \[\color{red}\checkmark\color{red}\checkmark\color{red}\checkmark\]
 one year ago

gerryliyana Group TitleBest ResponseYou've already chosen the best response.2
nice.., I thank all of you :)
 one year ago
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