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gerryliyana

  • 3 years ago

Solve this differential equations

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  1. gerryliyana
    • 3 years ago
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    Solve \[xe^{2x} \frac{ dy }{ dx } + e^{2x} (2x+1) y \]

  2. Outkast3r09
    • 3 years ago
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    y not solve using seperation

  3. gerryliyana
    • 3 years ago
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    @Outkast3r09 Yes, of course, i really know.., it is use linear equations method, isn't it?

  4. Meepi
    • 3 years ago
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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

  5. Skaematik
    • 3 years ago
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    Im learning too, is this right? \[\frac{ dx }{ xe ^{2x} }= e ^{2x}y dy (2x+1)\]

  6. Meepi
    • 3 years ago
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    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 :)

  7. Meepi
    • 3 years ago
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    Btw \[\frac{-e^{2x}(2x + 1)}{x e^{2x}}\] can be simplified to: \[-\frac{1}{x} - 2\]

  8. Meepi
    • 3 years ago
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    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

  9. Outkast3r09
    • 3 years ago
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    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

  10. Outkast3r09
    • 3 years ago
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    first start off by dividing by \[xe^2x\]

  11. Outkast3r09
    • 3 years ago
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    \[\frac{dy}{dx}+\frac{e^{2x}(2x+1)}{xe^{2x}}=0\]

  12. Outkast3r09
    • 3 years ago
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    \[ \frac{dy}{dx}+\frac{(2x+1)}{x}=0\]

  13. Outkast3r09
    • 3 years ago
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    now find the integration factor \[e^{\int{p(x)dx}}=e^{\int{\frac{2x+1}{x}dx}}\]

  14. Outkast3r09
    • 3 years ago
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    \[e^{\int{2+\frac{1}{x}}dx}\]

  15. Outkast3r09
    • 3 years ago
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    \[e^{2x-\frac{1}{x^2}}\]

  16. Meepi
    • 3 years ago
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    Isn't it\[\int_{}^{} (2 + \frac{1}{x}) dx = 2x + \ln |x| \]

  17. gerryliyana
    • 3 years ago
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    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???

  18. gerryliyana
    • 3 years ago
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    hbu @nubeer ???

  19. nubeer
    • 3 years ago
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    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

  20. gerryliyana
    • 3 years ago
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    yes., of course

  21. nubeer
    • 3 years ago
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    |dw:1353775372293:dw|

  22. nubeer
    • 3 years ago
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    so answer should be lny = -2x -lnx +c

  23. gerryliyana
    • 3 years ago
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    ok.., it's separable equations method, right? what if "linear equations method"? because its from the questions bank of Linear equations method., :)

  24. nubeer
    • 3 years ago
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    honestly i dont know most methods by name.. but quite dont remember about linear equation method.

  25. gerryliyana
    • 3 years ago
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    ok np., thank u so much @nubeer :)

  26. nubeer
    • 3 years ago
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    you are most welcome :)

  27. UnkleRhaukus
    • 3 years ago
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    \[xe^{2x} \frac{ \text dy }{\text dx } + e^{2x} (2x+1) y=0\]\[xe^{2x} \text dy + e^{2x} (2x+1) y{\text dx }=0\] 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 \[\int M\text dy=\int xe^{2x} \text dy =yxe^{2x}+c_1\] \[\int N\text dx=\int e^{2x} (2x+1) y\text dx\]\[\qquad\qquad=y\int e^{2x} \text dx+2y\int xe^{2x}\text dx\]\[\qquad\qquad=\frac{ye^{2x}}2+c_2+2y\left[\left.\frac{xe^{2x}}2\right| -\int \frac{e^{2x} }2\text dx\right]\]\[\qquad\qquad=\frac{ye^{2x}}2+2y\left[\frac{xe^{2x}}2 -\frac{e^{2x} }4\right]+c_3 \]\[\qquad\qquad= y{xe^{2x}}+c_3\] \[f(x,y)=y{xe^{2x}}=c\]

  28. Outkast3r09
    • 3 years ago
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    your first way was wrong

  29. Outkast3r09
    • 3 years ago
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    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

  30. Outkast3r09
    • 3 years ago
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    Rhaukaus answer is the general solution. He used exact method though. you should get the same answer using linear

  31. gerryliyana
    • 3 years ago
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    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????

  32. Outkast3r09
    • 3 years ago
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    that's right

  33. UnkleRhaukus
    • 3 years ago
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    great work @gerryliyana \[\color{red}\checkmark\color{red}\checkmark\color{red}\checkmark\]

  34. gerryliyana
    • 3 years ago
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    nice.., I thank all of you :)

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