Question

Can someone please tell me what I am doing wrong?
the question is Solve the following initial-value problem

1y''+1y=sin(x);y(0)=9y(0)=−6

y=

Answer 1

My work
r^2+1=0-->r==i,-i
y_c=C1cosx+c2sinx
f(x)=sinx
y_p(x)=Asinx
y'p(x)=Acosx
y''p(x)=-sinx

y''+y=sinx
-Asinx+Asinx=sinx

Answer 2

That cannot be true because I'm not supposed to get zero for the left hand side

Answer 3

oh and btw its supposed to read y(0)=9 and y'(0)=-6

Answer 4

You suppose: yp=Asinx
then yp''=Asinx
so:
yp''+yp=2Asinx
Must be equal to: sinx
so:
2Asinx=sinx
2A=1
A=1/2
So: yp=(1/2)*sinx (this is the particular solution)

But now we have to find the complementary one.
we set:
yh=c1*cos(x)+c2*sin(x)
So now the solution that we are looking for is like this:

y=c1*cosx+c2*sinx+(1/2)*sin(x)

And now we have to use the initial conditions.Ok?
Do you remember how do continue?

Answer 5

ok thanks

Answer 6

:)

Answer 7

wait, im sorry but I'm confused on what you did! y''=-Asinx so that would be -Asinx+Asinx=sinx but because there's two of the same it becomes -Asinx+Asinx=2sinx but theres a negative, does that mean we get rid of the negative

Answer 8

but then when you plug the values in, it doesnt make sense because I get yp=(1/2)sinx
y'=(1/2)cosx, if I were to get rid of the negative

Answer 9

but then when you put y(0)=9 and y'(0)=-6, you get an answer that doesnt make any sense

Answer 10

\[y_h=c_1\cos(x)+c_2\sin(x)\]
\[y_p≠A\sin(x)\]

Answer 11

I'm sorry but I think I am misunderstanding what it is were supposed to be doing

Answer 12

so we derive y_h?

Answer 13

if the homogenous solution already has the form of the 'guess' of the y_p form in undetermined coefficients , there is resonance so multiply by x
\[y_p=x\left(A\sin (x)+B\cos (x)\right)\]

Answer 14

wait i think I got it! the general solution is yx=yh+yp and e found the particular solution to be (1/2)sin x so plus that in and hat we found to be the complimentary solution and we derive the general solution and then find the values of c1 and c2 using the values given. right?

Answer 15

so youd getc1=9 and c2=2.5

Answer 16

so the solution would be y(x)=9cosx+2.5sinx+.5sinx

Answer 17

Be careful, the particular is not Asin(x), it's (-1/2)*x*cos(x)
You have to use what UnkleRhaukus said
After you have the particular, you add it to the complementary.
And then use the initial conditions.
Makes sense?

Answer 18

so yp=x(asinx+Bcosx

Answer 19

but why?

Answer 20

how does that ebcome (-.5)xcosx?

Answer 21

become*

Answer 22

you can use wroniskian, or use the yc parts in a variation of parameters is my thought

Answer 23

Thats what I would use to but the directions say otherwise! Thanks though!

Answer 24

I understand everything but the particular solution part

Answer 25

\[y_c=c_1cosx+c_2sinx\]

\[y_p=A(x)cosx+B(x)sinx\]
\[y'_p=A'(x)cosx+B'(x)sinx\]\[\hspace{3em}-A(x)sinx+B(x)cosx\]

the A'B' parts = 0 to root out the homogenous part

\[y'_p=-A(x)sinx+B(x)cosx\]
\[ y''_p=-A'(x)sinx+B'(x)cosx\]\[\hspace{3em} -A(x)cosx-B(x)sinx\]

y''p+ yp = sinx

\[\ \ \ y_p=\cancel{A(x)cosx+B(x)sinx}\]\[ y''_p=-A'(x)sinx+B'(x)cosx\]\[\hspace{3em} \cancel{-A(x)cosx-B(x)sinx}\]--------------------------- \[ \hspace{3em} -A'(x)cosx +B'(x)sinx=sinx\]

we have a system of 2 equations in 2 unknowns: A' and B'

\[-A'(x)sinx+B'(x)cosx=0\]\[\ \ A'(x)cosx +B'(x)sinx=sinx\]

determine the values of A' and B' then integrate them back into A and B

the Wronskian just goes straight to the last part

Answer 26

ok thanks sooo much!!

Answer 27

youre welcome, i find the longer method to be more informative since the shorter methods suppose that you already have a working understanding