## gerryliyana 3 years ago Solve this differential equations

1. gerryliyana

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

2. Outkast3r09

y not solve using seperation

3. gerryliyana

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

4. Meepi

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

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

6. Meepi

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

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

8. Meepi

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

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

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

11. Outkast3r09

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

12. Outkast3r09

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

13. Outkast3r09

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

14. Outkast3r09

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

15. Outkast3r09

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

16. Meepi

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

17. gerryliyana

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

hbu @nubeer ???

19. 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

20. gerryliyana

yes., of course

21. nubeer

|dw:1353775372293:dw|

22. nubeer

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

23. gerryliyana

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

24. nubeer

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

25. gerryliyana

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

26. nubeer

you are most welcome :)

27. UnkleRhaukus

$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

29. Outkast3r09

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

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

31. gerryliyana

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

that's right

33. UnkleRhaukus

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

34. gerryliyana

nice.., I thank all of you :)