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Solve this differential equations

Mathematics
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Solve \[xe^{2x} \frac{ dy }{ dx } + e^{2x} (2x+1) y \]
y not solve using seperation
@Outkast3r09 Yes, of course, i really know.., it is use linear equations method, isn't it?

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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
Im learning too, is this right? \[\frac{ dx }{ xe ^{2x} }= e ^{2x}y dy (2x+1)\]
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 :)
Btw \[\frac{-e^{2x}(2x + 1)}{x e^{2x}}\] can be simplified to: \[-\frac{1}{x} - 2\]
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
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
first start off by dividing by \[xe^2x\]
\[\frac{dy}{dx}+\frac{e^{2x}(2x+1)}{xe^{2x}}=0\]
\[ \frac{dy}{dx}+\frac{(2x+1)}{x}=0\]
now find the integration factor \[e^{\int{p(x)dx}}=e^{\int{\frac{2x+1}{x}dx}}\]
\[e^{\int{2+\frac{1}{x}}dx}\]
\[e^{2x-\frac{1}{x^2}}\]
Isn't it\[\int_{}^{} (2 + \frac{1}{x}) dx = 2x + \ln |x| \]
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???
hbu @nubeer ???
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
yes., of course
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so answer should be lny = -2x -lnx +c
ok.., it's separable equations method, right? what if "linear equations method"? because its from the questions bank of Linear equations method., :)
honestly i dont know most methods by name.. but quite dont remember about linear equation method.
ok np., thank u so much @nubeer :)
you are most welcome :)
\[xe^{2x} \frac{ \text dy }{\text dx } + e^{2x} (2x+1) y=0\]\[xe^{2x} \text dy + e^{2x} (2x+1) y{\text dx }=0\] 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 \[\int M\text dy=\int xe^{2x} \text dy =yxe^{2x}+c_1\] \[\int N\text dx=\int e^{2x} (2x+1) y\text dx\]\[\qquad\qquad=y\int e^{2x} \text dx+2y\int xe^{2x}\text dx\]\[\qquad\qquad=\frac{ye^{2x}}2+c_2+2y\left[\left.\frac{xe^{2x}}2\right| -\int \frac{e^{2x} }2\text dx\right]\]\[\qquad\qquad=\frac{ye^{2x}}2+2y\left[\frac{xe^{2x}}2 -\frac{e^{2x} }4\right]+c_3 \]\[\qquad\qquad= y{xe^{2x}}+c_3\] \[f(x,y)=y{xe^{2x}}=c\]
your first way was wrong
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
Rhaukaus answer is the general solution. He used exact method though. you should get the same answer using linear
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????
that's right
great work @gerryliyana \[\color{red}\checkmark\color{red}\checkmark\color{red}\checkmark\]
nice.., I thank all of you :)

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