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Bas_tet [7]
4 years ago
9

Ap calc please help (question 4)

Mathematics
1 answer:
sineoko [7]4 years ago
5 0
A)

y(t)=5\sin(t)-\cos(t)\\\\y'(t)=5\cos(t)+\sin(t)\\\\y''(t)=-5\sin(t)+\cos(t)\\\\\\7y(t)-3y''(t)=7(5\sin(t)-\cos(t))-3(-5\sin(t)+\cos(t))=\\\\=35\sin(t)-7\cos(t)+15\sin(t)-3\cos(t)=\\\\=50\sin(t)-10\cos(t)=5\big(10\sin(t)-2\cos(t)\big)\neq5\cos(t)

Not a solution.

b)

y(t)=\sin(t)\\\\y'(t)=\cos(t)\\\\y''(t)=-\sin(t)\\\\\\7y(t)-3y''(t)=7\sin(t)+3\sin(t)=10\sin(t)\neq5\cos(t)

Not a solution.

c)

y(t)=\frac{1}{2}\sin(t)\\\\y'(t)=\frac{1}{2}\cos(t)\\\\y''(t)=-\frac{1}{2}\sin(t)\\\\\\7y(t)-3y''(t)=\frac{7}{2}\sin(t)+\frac{3}{2}\sin(t)=\frac{10}{2}\sin(t)=5\sin(t)\neq5\cos(t)

Not a solution.

d)

y(t)=\frac{1}{2}\cos(t)\\\\y'(t)=-\frac{1}{2}\sin(t)\\\\y''(t)=-\frac{1}{2}\cos(t)\\\\\\7y(t)-3y''(t)=\frac{7}{2}\cos(t)+\frac{3}{2}\cos(t)=\frac{10}{2}\cos(t)=\boxed{5\cos(t)\quad\checkmark}

e)

y(t)=\cos(t)\\\\y'(t)=-\sin(t)\\\\y''(t)=-\cos(t)\\\\\\7y(t)-3y''(t)=7\cos(t)+3\cos(t)=10\cos(t)\neq5\cos(t)

Not a solution.

Correct answer is d)
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