Question

Show that a particular solution of y"+2ay' + b²y = Asin(wx) is given by,
$y=\frac{A\sin(\omega x-\alpha)}{\sqrt{(b^2-\omega^2)^2+4\omega^2a^2}}$
Where,

$\alpha=\tan^{ - 1} (\frac{2a\omega}{b^2-\omega^2})$

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Answer 1

Answer:

$\huge{\fbox{\fbox{\bigstar{\mathfrak{\red{Answer}}}}}}$

My hand calculation solution of

y′′+192y=0,y(0)=1/6,y′(0)=−1y″+192y=0,y(0)=1/6,y′(0)=−1

is

y=16cos83–√t−3–√24sin83–√ty=16cos⁡83t−324sin⁡83t

Now a few more hand calculations provides this equivalent solution:

y=19−−√24cos(83–√t−tan−1(−3–√/4))y=1924cos⁡(83t−tan−1⁡(−3/4))

Now, when we solve the equation using

Answer 2

Answer:

$\huge\underline\mathtt\pink{AnSweR:}$

check the above attachment!

Hope it will be helpful :)