## gerryliyana Group Title Solve this differential equations one year ago one year ago

1. gerryliyana Group Title

Solve $xe^{2x} \frac{ dy }{ dx } + e^{2x} (2x+1) y$

2. Outkast3r09 Group Title

y not solve using seperation

3. gerryliyana Group Title

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

4. Meepi Group Title

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 Group Title

Im learning too, is this right? $\frac{ dx }{ xe ^{2x} }= e ^{2x}y dy (2x+1)$

6. Meepi Group Title

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 Group Title

Btw $\frac{-e^{2x}(2x + 1)}{x e^{2x}}$ can be simplified to: $-\frac{1}{x} - 2$

8. Meepi Group Title

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 Group Title

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 Group Title

first start off by dividing by $xe^2x$

11. Outkast3r09 Group Title

$\frac{dy}{dx}+\frac{e^{2x}(2x+1)}{xe^{2x}}=0$

12. Outkast3r09 Group Title

$\frac{dy}{dx}+\frac{(2x+1)}{x}=0$

13. Outkast3r09 Group Title

now find the integration factor $e^{\int{p(x)dx}}=e^{\int{\frac{2x+1}{x}dx}}$

14. Outkast3r09 Group Title

$e^{\int{2+\frac{1}{x}}dx}$

15. Outkast3r09 Group Title

$e^{2x-\frac{1}{x^2}}$

16. Meepi Group Title

Isn't it$\int_{}^{} (2 + \frac{1}{x}) dx = 2x + \ln |x|$

17. gerryliyana Group Title

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 Group Title

hbu @nubeer ???

19. nubeer Group Title

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 Group Title

yes., of course

21. nubeer Group Title

|dw:1353775372293:dw|

22. nubeer Group Title

so answer should be lny = -2x -lnx +c

23. gerryliyana Group Title

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

24. nubeer Group Title

honestly i dont know most methods by name.. but quite dont remember about linear equation method.

25. gerryliyana Group Title

ok np., thank u so much @nubeer :)

26. nubeer Group Title

you are most welcome :)

27. UnkleRhaukus Group Title

$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 Group Title

29. Outkast3r09 Group Title

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 Group Title

Rhaukaus answer is the general solution. He used exact method though. you should get the same answer using linear

31. gerryliyana Group Title

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 Group Title

that's right

33. UnkleRhaukus Group Title

great work @gerryliyana $\color{red}\checkmark\color{red}\checkmark\color{red}\checkmark$

34. gerryliyana Group Title

nice.., I thank all of you :)