Question

show that y=xsinx is a solution to: second derivative y + y =2cosx

Answer 1

Wait...is the function like:

\[\large y'' + y = 2cos(x)\]?

Answer 2

YES

Answer 3

Ahh okay

Well, from that function, we know we need 'y' and we need the second derivative of 'y'

So if:

\[\large y = xsin(x)\]
\[\large y' =?\]
\[\large y'' =?\]

Answer 4

I got for the first derivative: sinx+xcosx

Answer 5

Perfect!

And the second derivative?

Answer 6

for the second I got: cosx+cosx-xsinx

Answer 7

OMG I GOT IT RIGHT IM SO SORRY I GET IT NOW LOL

Answer 8

And great again!

So now we just plug in everything we have!

\[\large y'' + y = 2cos(x)\]
Turns into
\[\large cos(x) + cos(x) - xsin(x) + xsin(x) = 2cos(x)\]

I'll let you simplify :)

Answer 9

so the sinx will cancel then it will be 2cosx=2cosx
yup exactly

Answer 10

Lol great job!

Answer 11

now the next part is : show that y=xsinx is a solution to the equation 4th derivative + second derivative=-2cosx

Answer 12

Follow the same steps :)

Answer 13

okay so for the 3rd derivative I got -3sinx-xcosx

Answer 14

And the fourth?

Answer 15

AH I GOT IT :), for the fourth I got -4cosx+xsinx, I plugged em both in and simplified to show that -2cosx=-2cosx

Answer 16

And there we go! :)