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

Outkast3r09Best ResponseYou've already chosen the best response.1
y not solve using seperation
 one year ago

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

MeepiBest 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
 one year ago

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

MeepiBest 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 :)
 one year ago

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

MeepiBest 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
 one year ago

Outkast3r09Best 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
 one year ago

Outkast3r09Best ResponseYou've already chosen the best response.1
first start off by dividing by \[xe^2x\]
 one year ago

Outkast3r09Best ResponseYou've already chosen the best response.1
\[\frac{dy}{dx}+\frac{e^{2x}(2x+1)}{xe^{2x}}=0\]
 one year ago

Outkast3r09Best ResponseYou've already chosen the best response.1
\[ \frac{dy}{dx}+\frac{(2x+1)}{x}=0\]
 one year ago

Outkast3r09Best ResponseYou've already chosen the best response.1
now find the integration factor \[e^{\int{p(x)dx}}=e^{\int{\frac{2x+1}{x}dx}}\]
 one year ago

Outkast3r09Best ResponseYou've already chosen the best response.1
\[e^{\int{2+\frac{1}{x}}dx}\]
 one year ago

Outkast3r09Best ResponseYou've already chosen the best response.1
\[e^{2x\frac{1}{x^2}}\]
 one year ago

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

gerryliyanaBest 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???
 one year ago

nubeerBest 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
 one year ago

nubeerBest ResponseYou've already chosen the best response.0
so answer should be lny = 2x lnx +c
 one year ago

gerryliyanaBest 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., :)
 one year ago

nubeerBest ResponseYou've already chosen the best response.0
honestly i dont know most methods by name.. but quite dont remember about linear equation method.
 one year ago

gerryliyanaBest ResponseYou've already chosen the best response.2
ok np., thank u so much @nubeer :)
 one year ago

nubeerBest ResponseYou've already chosen the best response.0
you are most welcome :)
 one year ago

UnkleRhaukusBest 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 \[\int M\text dy=\int xe^{2x} \text dy =yxe^{2x}+c_1\] \[\int N\text dx=\int e^{2x} (2x+1) y\text dx\]\[\qquad\qquad=y\int e^{2x} \text dx+2y\int xe^{2x}\text dx\]\[\qquad\qquad=\frac{ye^{2x}}2+c_2+2y\left[\left.\frac{xe^{2x}}2\right \int \frac{e^{2x} }2\text dx\right]\]\[\qquad\qquad=\frac{ye^{2x}}2+2y\left[\frac{xe^{2x}}2 \frac{e^{2x} }4\right]+c_3 \]\[\qquad\qquad= y{xe^{2x}}+c_3\] \[f(x,y)=y{xe^{2x}}=c\]
 one year ago

Outkast3r09Best ResponseYou've already chosen the best response.1
your first way was wrong
 one year ago

Outkast3r09Best 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
 one year ago

Outkast3r09Best 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
 one year ago

gerryliyanaBest 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

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

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