Question

how to solve this kind of problem in linear differential equation?

Answer 1

problem is \[x \frac{ dy }{ dx } = y + x^3 +3x^2 - 2x\]

Answer 2

Where are you stuck?

Answer 3

\[\frac{ dy }{ dx } = \frac{ y }{ x } + x^2 + 3x - 2\]

Answer 4

yup

Answer 5

there, now i cant identifies the Q(x) and p(x)

Answer 6

\(\dfrac{dy}{dx}-\dfrac{y}{x}=x^2+3x-2\)

Answer 7

so the p(x) is x

Answer 8

-1/x

Answer 9

ah, and the remaing terms is the q(x)

Answer 10

yup

Answer 11

so IF is \[\int\limits e ^{\frac{ -1 }{ x }dx}\]

Answer 12

so e ^ -lnx

Answer 13

IF = 1/x

Answer 14

You always make mistake at this. It is NOT \[\int e^{-1/x} dx\]

Answer 15

\[IF = e^{\int p(x) dx}\]

Answer 16

yes so the soln is now \[(\frac{ 1 }{ x })y = [x^2 +3x - 2 ](x) (\frac{ 1 }{ x }) dx + c\]

Answer 17

Again, you always make mistake at (x) from the RHS, where does it come from?
\[(\frac{ 1 }{ x })y = [x^2 +3x - 2 ]\color{red}{(x)(?)} (\frac{ 1 }{ x }) dx + c\]

Answer 18

what x

Answer 19

the red one

Answer 20

ah, i just putted it sorry its Q(X) only

Answer 21

And this is NOT correct also

Answer 22

why

Answer 23

\[\dfrac{d}{dx}(\dfrac{1}{x}*y)= Q(x)*(\dfrac{1}{x})+C\]

Answer 24

Then take integral both sides
\[\dfrac{1}{x}y = \int Q(x) *\dfrac{1}{x} dx +C\]

Answer 25

integral of the Q(x) is \[\frac{ x^3 }{ 3 } + \frac{ 3x^2 }{ 2} - 2x(\frac{ 1 }{ x }) + c\]

Answer 26

NNNNNNNNNNNNO

Answer 27

huhhhhhhh

Answer 28

You must do multiplication before taking integral.