Question

Obtain the general solution of (e2x+4)dy/dx=y

Answer 1

That is \[(e ^{2x} + 4) dy/dx = y\]

Answer 2

I always get confused by those dy/dx
if that means derivative.. then here is how it goes :)

\[ \LARGE y'=\left(e^{2x}+4\right)'\]
\[\LARGE y'=\left(e^{2x}\right)'+4'\]
\[\LARGE y'=e^{2x}\cdot (2x)'+0\]
\[\LARGE y'=e^{2x}\cdot 2\]
\[\LARGE y'=2e^{2x}\]


this rule was used :
\[\LARGE \left(e^u\right)'=e^u\cdot u'\]

Answer 3

the right side is just a y not a y'

Answer 4

\[\left( e^{2x} + 4\right)\frac{dy}{dx}=y\]separate the variable\[\frac{dy}{dx}=\frac{y}{e^{2x} + 4}\]\[\frac{dy}{y}=\frac{dx}{e^{2x} + 4}\]and then integrate both sides

Answer 5

auff... I'm wrong then :(

Answer 6

im really sorry for acting really stupid but how do you integrate the right side?

Answer 7

its all good Kreshnik, atleast you had an idea, i can't understand a thing... hahaha

Answer 8

\[\int\limits \left( \frac{1}{e^{2x} + 4} \right)dx\]\[= \int\limits \left(\frac{e^{2x}}{e^{2x}\left( e^{2x} + 4 \right)}\right)dx\]\[u = e^{2x}\]\[du = 2e^{2x} dx\]\[= \frac{1}{2} \int\limits \left( \frac{1}{u \left( u+4 \right)} \right)du\]can you continue after this ?

Answer 9

sorry, i can't... =(

Answer 10

i can't seem to get how this stuff works...

Answer 11

why don't you try wolframa?

Answer 12

use partial fractions

Answer 13

ohhh... got it...

Answer 14

yeah wolframalpha is also a great computer algebra, it can solve most integration analytically and show you the steps

Answer 15

you show the solutions better thought mr. exraven... =)

Answer 16

I can only show solutions for problems that I can solve.. and coincidentally I can solve this :), usually I got stuck when solving differential equations too

Answer 17

i get stuck with diff eqns when i forget my basics... thanks again sir... i have to go... i hope to be of help some day...