The family of functions y=c1e^xcos(x)+c2e^xsin(x) is the general solution of y\'

Question

The family of functions y=c1e^xcos(x)+c2e^xsin(x) is the general solution of y''-2y'+2y=0

(A) Decide whether there is a member of the family that satisfies the boundary conditions y(0)=1, y(pi/2)=0 .

If there is a solution find it. If not then enter 111 .


(B) Decide whether there is a member of the family that satisfies the boundary conditions y(0)=1, y(pi)=0 .

If there is a solution find it. If not then enter 111 .


(C) Decide whether there is a member of the family that satisfies the boundary conditions y(-pi)=1, y(pi/2)=1 .

If there is a solution find it. If not then enter 111 .

Explanation / Answer

a) y(x) = aexcos(x) + bexsin(x)

y(0) = 1 = ae0cos(0) + be0sin(0) = a

a = 1

y(/2) = 0 = ae/2cos(/2) + be/2sin(/2) = be/2

b = 0

y(x) = excos(x)

b) y(x) = aexcos(x) + bexsin(x)

y(0) = 1 = ae0cos(0) + be0sin(0) = a

a = 1

y() = 0 = aecos() + besin() = ae

no solution fits these BC.

c) y(x) = aexcos(x) + bexsin(x)

y(-) = 1 = ae-cos(-) + be-sin(-) = -ae-

a = -e

y(/2) = 1 = ae/2cos(/2) + be/2sin(/2) = be/2

b = e-/2

y(x) = -eexcos(x) + e-/2exsin(x)

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.