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anonymous
 4 years ago
Solve this differential equations
anonymous
 4 years ago
Solve this differential equations

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anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Solve \[xe^{2x} \frac{ dy }{ dx } + e^{2x} (2x+1) y \]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0y not solve using seperation

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0@Outkast3r09 Yes, of course, i really know.., it is use linear equations method, isn't it?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0It'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

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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0What 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 :)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Btw \[\frac{e^{2x}(2x + 1)}{x e^{2x}}\] can be simplified to: \[\frac{1}{x}  2\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0So 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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0You 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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0first start off by dividing by \[xe^2x\]

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

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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0now find the integration factor \[e^{\int{p(x)dx}}=e^{\int{\frac{2x+1}{x}dx}}\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[e^{\int{2+\frac{1}{x}}dx}\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[e^{2x\frac{1}{x^2}}\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Isn't it\[\int_{}^{} (2 + \frac{1}{x}) dx = 2x + \ln x \]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Nice @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???

Nubeer
 4 years ago
Best ResponseYou've already chosen the best response.0xe^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

Nubeer
 4 years ago
Best ResponseYou've already chosen the best response.0so answer should be lny = 2x lnx +c

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0ok.., it's separable equations method, right? what if "linear equations method"? because its from the questions bank of Linear equations method., :)

Nubeer
 4 years ago
Best ResponseYou've already chosen the best response.0honestly i dont know most methods by name.. but quite dont remember about linear equation method.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0ok np., thank u so much @nubeer :)

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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0your first way was wrong

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0the 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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Rhaukaus answer is the general solution. He used exact method though. you should get the same answer using linear

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Thank 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????

UnkleRhaukus
 4 years ago
Best ResponseYou've already chosen the best response.1great work @gerryliyana \[\color{red}\checkmark\color{red}\checkmark\color{red}\checkmark\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0nice.., I thank all of you :)
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