Question

Calc I

Answer 1

if Y=A sen (ln x)+B cos (ln X) being A and B constants show that is \[x ^{2}y ^{"}+xy'+y=0\]

Answer 2

@Libniz

Answer 3

he can help?

Answer 4

yes........ im doing another person right now...... thats why if he doesnt come help let me know in like 10 min :)

Answer 5

Find the first and second derivatives of y, then plug them into the DE. Show that every thing cancels out and you get 0.

If \( \large y= A\sin(\ln x)+B\cos(\ln x)\)

Then \(\large y' = A\cos(\ln x)\cdot \frac{1}{x} -B\sin(\ln x)\cdot \frac{1}{x}\)
\( = \large \frac{1}{x}\left( A\cos(\ln x) -B\sin(\ln x)\right)\)

And \(\large y'' = -\frac{1}{x^2} \left( A\cos(\ln x) -B\sin(\ln x)\right) \)
\(\large+ \frac{1}{x} \left(-A\sin(\ln x)\cdot \frac{1}{x} - B\cos(\ln x) \cdot \frac{1}{x} \right)\)
\(\large = -\frac{1}{x^2} \left( A\cos(\ln x) -B\sin(\ln x)\right) \)
\(\large- \frac{1}{x^2} \left(A\sin(\ln x) + B\cos(\ln x)\right)\)
\(\large = -\frac{1}{x^2} \left( A\cos(\ln x) -B\sin(\ln x) +A\sin(\ln x) + B\cos(\ln x)\right)\)

Answer 6

Note that \(\large x^2y'' = x^2\cdot-\frac{1}{x^2} \left( A\cos(\ln x) -B\sin(\ln x) +A\sin(\ln x) + B\cos(\ln x)\right)\)
\( \large = - A\cos(\ln x) +B\sin(\ln x) -A\sin(\ln x) + B\cos(\ln x)\)

And \(\large xy' = x \cdot\frac{1}{x}\left( A\cos(\ln x) -B\sin(\ln x)\right)\)
\( \large = A\cos(\ln x) -B\sin(\ln x) \)

So \(\large x^2y'' + xy' + y = - A\cos(\ln x) + B\sin(\ln x) -A\sin(\ln x) \)
\(\large + B\cos(\ln x) + A\cos(\ln x) -B\sin(\ln x) + A\sin(\ln x)+B\cos(\ln x)\)

And everything cancels :)