Question

what is the integral of e^2y/x dx

Answer 1

\[e ^{2(\frac{ y }{ x })} dx\]

Answer 2

is it e^2lnx

Answer 3

Looks wrong. You cannot integrate y with respect to x w/o plugging in the expression for y

Answer 4

May I know how you arrived at that integral ?

Answer 5

ah im applying it in linear equation

Answer 6

May I see your work ?
if possible take a screenshot of your work and attach here

Answer 7

okay

Answer 8

here @ganeshie8

Answer 9

nice
there is a mistake in third line

Answer 10

you have flipped the sign of 2y/x

Answer 11

yes becauseof the formula

Answer 12

What formula

Answer 13

There are no formulas in math that allow you to flip the signs magically

Answer 14

here in linear

Answer 15

Given eqn :
```
xdy/dx - 2y = x^3cos4x
```

Answer 16

dividing x through out gives


```
dy/dx - 2y/x = x^2cos4x
```

Answer 17

Notice that

`P(x) = -2/x`

Answer 18

yes, but the px is -2 y/x

Answer 19

No
P(x) is simply -2/x

Answer 20

huh, how come. the equation says it is `2y/x

Answer 21

I want you take some time and compare above equation with the general linear form

Answer 22

ah got it the y will be cancel

Answer 23

Compare below two eqns :


```
dy/dx - 2y/x = x^2cos4x
```

and

```
dy/dx + P(x)y = Q(x)
```

Answer 24

yes got it

Answer 25

P(x) = -2/x


you don't need to wry about Q(x)

Answer 26

ah yes, the proceed to I.F

Answer 27

so the integrating factor is just
\[IF = e^{\int -2/x\,dx}\]

Answer 28

YES so e^2lnx

Answer 29

careful with the sings

Answer 30

with negative

Answer 31

its e^(-2lnx)

Answer 32

yes so it is now -2x

Answer 33

How ?

Answer 34

e^ln is 1 so then -2x

Answer 35

No

Answer 36

huh

Answer 37

\[e^{-2\ln x} = \left(e^{\ln x}\right)^{-2} = (x)^{-2} = \dfrac{1}{x^2}\]

Answer 38

ah, i know now. then in solution

Answer 39

recall that \((x^m)^n = x^{mn}\)

Answer 40

(I,F)y = int Q(x) (I.F) dx +c

Answer 41

\[(\frac{ 1 }{ x ^{2} } )y = \int\limits x ^{2} \cos 4x (x)(\frac{ 1 }{ x ^{2} })\]

Answer 42

\[(\frac{ 1 }{ x ^{2} } )y = \int\limits x ^{2} \cos (4x)(\frac{ 1 }{ x ^{2} })\]

Answer 43

with plus c

Answer 44

\[(\frac{ 1 }{ x ^{2} } )y = \int\limits x ^{2} \cos (4x)(\frac{ 1 }{ x ^{2} })\,dx\]

Answer 45

you can put +c after evaluating the integral

Answer 46

ah okay \[(\frac{ 1 }{ x ^{2} }) y = \int\limits \frac{ x ^{3} }{ 3 } \sin 4x \]

Answer 47

Don't put the integral sign after evaluating the integral

Answer 48

also that is wrong

Answer 49

oops sorry

Answer 50

\[(\frac{ 1 }{ x ^{2} } )y = \int\limits x ^{2} \cos (4x)(\frac{ 1 }{ x ^{2} })\,dx\]


right hand, cancel x^2 top and bottom first

Answer 51

ah okay so \[\frac{ 1 }{ x^2 } y = \sin4x + c\]

Answer 52

then multiply both side by x^2

Answer 53

Still wrong

Answer 54

whats the integral of cos(4x) ?

Answer 55

1/4 sin 4x

Answer 56

Yes, 1/4 is missing in your answer

Answer 57

yep i forgot tho rewrite \[\frac{ 1 }{ x^2 } y = \frac{ 1 }{ 4 } \sin 4x + c\]

Answer 58

then multiply both side by 4x^2

Answer 59

so it become \[4y = x^2 \sin 4x + 4cx^2\]

Answer 60

Looks perfect!

Answer 61

You might enjoy watching this video on linear des

http://ocw.mit.edu/courses/mathematics/18-03-differential-equations-spring-2010/video-lectures/lecture-3-solving-first-order-linear-odes/

Answer 62

thanks for the help. need to know more about linear.

Answer 63

okay thanks i looked for it

Answer 64

np :)