A. $y'=(3\sin x)'=3\cos x$ . Differentiate again to get $y''=(3\cos x)'=-3\sin x$ , then $y''+y=-3\sin x+3\sin x=0\neq 3\sin x$ so this choice of y doesn't solve the equation.

B. $y'=3\sin x+3x\cos x-5\cosx +5x\sin x=(5x+3)\sin x+(3x-5)\cos x$ and $y''=5\sin x+(5x+3)\cos x+3\cos x-(3x-5)\sin x=(10-3x)\sin x+(5x+6)\cos x$ , then $y''+y=10\sin x+6\cos x\neq 3\sin x$ so y is not a solution

C. $y'=\frac{-3}{2}\cos x+\frac{3}{2}x\sin x$ hence $y''=\frac{3}{2}\sin x+\frac{3}{2}\sin x+\frac{3}{2}x\cos x=3\sin x+\frac{3}{2}x\cos x$ . Then $y''+y=3\sin x+\frac{3}{2}x\cos x-\frac{3}{2}x\cos x=3\sin x$ so y is a solution.

D. $y'=-3\sin x$ and $y''=-3\cos x$ , then $y''+y=0$ thus y isn't a solution

E. $y'=\frac{3}{2}\sin x+\frac{3}{2}x\cos x$ hence $y''=\frac{3}{2}\cos x+\frac{3}{2}\cos x-\frac{3}{2}x\sin x=3\cos x-\frac{3}{2}x\cos x$ . Then $y''+y=3\cos x-\frac{3}{2}x\cos x-\frac{3}{2}x\cos x=3\cos x\neq 3\sin x$ then y is not a solution.

3 0

3 years ago

What are the 5 perfect prime numbers between 50 and 140

grin007 [14]

Answer:

61 67 71 73 79

There you go:)

7 0

3 years ago

Read 2 more answers

Which shapes are congruent to shape I

Natasha2012 [34]

Answer: all of them except III

Step-by-step explanation:

8 0

3 years ago

Read 2 more answers

A woman deposit $100 into her son saving account on his first birthday on his second birthday she deposits $125 and $150 on his

OleMash [197]

This is just showing what I did.
100=1 125=2 150=3 175=4 200=5 225=6 250=7 275=8 300=9 325=10 350=11 375=12 400=13 425=14 450=15 475=16 500=17. So on the son's 17th birthday his mom deposited $500 in her son's saving account. Hope I helped!

5 0

3 years ago