Question

Solve this differential equations

Answer 1

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

Answer 2

y not solve using seperation

Answer 3

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

Answer 4

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

Answer 5

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

Answer 6

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 :)

Answer 7

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

Answer 8

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

Answer 9

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

Answer 10

first start off by dividing by \[xe^2x\]

Answer 11

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

Answer 12

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

Answer 13

now find the integration factor

\[e^{\int{p(x)dx}}=e^{\int{\frac{2x+1}{x}dx}}\]

Answer 14

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

Answer 15

\[e^{2x-\frac{1}{x^2}}\]

Answer 16

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

Answer 17

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???

Answer 18

hbu @nubeer ???

Answer 19

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

Answer 20

yes., of course