\(\displaystyle{y}{''}+{16}{y}=-{\frac{{{2}}}{{{\sin{{\left({4}{x}\right)}}}}}}\) I try to solve this ode using

sexoagotadorogyr

sexoagotadorogyr

Answered question

2022-03-30

y +16y=2sin(4x)
I try to solve this ode using the variation of parameters theorem.
The characteristic polynomial of the homogenous equation is r2+16=0.
Then u1(x)=sin(4x),u2(x)=cos(4x)
y(x)=c1(x)sin(4x)+c2(x)cos(4x)
c1(x)sin(4x)+c2cos(4x)=0
II c1(x)cos(4x)c2sin(4x)=2sin(4x)
Multiply I by cos(4x) and II by sin(4x) and subtract.
c1(sin(4x)cos(4x)cos(4x)sin(4x))+c2(cos2(4x)+sin2(4x))=2
We get c2=2c2=2x,
c1'=2cos(4x)sin(4x)=2cot(4x)c1=12ln|sin(4x)|.
I don't get why it incorrect, where am I wrong?

Answer & Explanation

Ali Price

Ali Price

Beginner2022-03-31Added 6 answers

Explanation:
Your y(x) should be y(x)=c1(x)cos(4x)+c2(x)sin(4x).
Using this, you will get two equations which are different from yours. So your second equation is wrong.
Update: First
y(x)=c1(x)cos(4x)+c2(x)sin(4x)4c1(x)sin(4x)+4c2(x)cos(4x).
Let c1(x)cos(4x)+c2(x)sin(4x)=0.
and then y(x)=4c1(x)sin(4x)+4c2(x)cos(4x).
Now you can get y''(x).
Chad Robinson

Chad Robinson

Beginner2022-04-01Added 6 answers

Step 1
It seems like you are not paying enough attention to your differentiation. If you have u1,u2 as your homogeneous solution and look for
y(x)=c1(x)u1(x)+c2(x)u2(x),
then just taking one derivative gives:
y=(c1u1)+(c2u2)=c1u1+c1u1+c2u2+c2u2.
Step 2
The usual assumption (your equation I) makes is it so that
y=c1u1+c2u2.
The way you have written this right now, it reads like
y=d2sin4x
which is your equation II and this is wrong. You take another derivative of y' to find y'' and while you are doing so, you are going to use equation I again to obtain a simple expression. Then, you plug y and y'' to your original DE to obtain two sets of equations for c1 and c2.

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