Question

\(\large y=e^{ax} \sin(bx)\) prove that \(\large y_2-2ay_1+(a^2+b^2)y=0 \)

Answer 1

y2 here is \(\large \rm y^2\) and y1 is just \(\large \rm y\)?

Answer 2

and is the equation written like this \(\large \rm y=e^{ax}sin(bx)\)

Answer 3

in one side \(y^2=e^{2ax}sin^2(bx)\)

Answer 4

you need to clear what is y1? here

Answer 5

you wrote y1 and y? are they different?

Answer 6

i think he means first derivative and second derivative

Answer 7

let me edit the q

Answer 8

eh i thought about that as well but i wasn't sure lol

Answer 9

refresh and see if it looks okay

Answer 10

@Gobikrishna

Answer 11

that looks good^_^

Answer 12

one way is to just find first and 2nd derivative of y
and evaluate LHS to prove it =0

Answer 13

first take the first derivative! once that is done
find the 2nd one

Answer 14

@Gobikrishna do you know how to find first derivative?

Answer 15

coming nobody will help ya of you set there watching lol
say something!!

Answer 16

ok First derivative we need product rule first yes?

Answer 17

let me write that down so you know what happening
\(y'=(e^{ax})'sin(bx)+e^{ax}(sin(bx))'\)

Answer 18

Now we need chain rule?

Answer 19

i will do the first derivative for you but you need to finish off the work
\(y'=ae^{ax}sin(bx)+be^{ax}cos(bx)=e^{ax}(asin(bx)+bcos(bx))\)

Answer 20

carry on...

Answer 21

I able to do until the second differentiate