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 .

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

If there is a solution find it. It could happen that there is a one parameter family of solutions, i.e., infinitely many solutions determined by a single parameter which we denote by , e.g., for example an answer could look like y(x)=ce^x (sin(x)+cos(x)). If there is no solution then enter .

Explanation / Answer

y'' - 2y' + 2y = 0

given general solution is y=c1e^xcos(x)+c2e^xsin(x)

A) connditions: y(0)=1, y(pi/2)=0

y=c1e^xcos(x)+c2e^xsin(x)

put y = 1, x = 0

1 = C1 *1*1 + 0 ==> C1 = 1

y=c1e^xcos(x)+c2e^xsin(x)

put y = 0, x = pi/2

0 = 0 + C2*e^(pi/2) * 1

=> C2 = 0

solution is

y = e^xcos(x)

B) conditions y(0)=1, y(pi)=0

y=c1e^xcos(x)+c2e^xsin(x)

put y = 1, x = 0

1 = C1 *1*1 + 0 ==> C1 = 1

y=c1e^xcos(x)+c2e^xsin(x)

put y = 0, x = pi

0 = C1*1*-1 + 0 => C1 = 0

here we have c1 = 0 and c1 = 1

SO, no solution

111

C) y(-pi)=1, y(pi/2)=1

y=c1e^xcos(x)+c2e^xsin(x)

put y = 1, x = -pi

1 = c1 e^(-pi) * 1 + 0

=> C1 = e^pi

y=c1e^xcos(x)+c2e^xsin(x)

put y = 1, x = pi/2

1 = 0 + C2 * e^(pi/2) * 1

=> C2 = e^(-pi/2)

solutionis

y = e^pi * e^x cos(x) + e^(-pi/2) * e^x sin(x)

D)

conditions y(-pi/2)=0, y(pi/2)=0

y=c1e^xcos(x)+c2e^xsin(x)

put y = 0, x = -pi/2

0 = 0 - c2 * e^(-pi/2)

=> C2 = 0

y=c1e^xcos(x)+c2e^xsin(x)

put y = 0, x = pi/2

0 = 0 + C2 *e^(pi/2)

=> C2 = 0

solution is

y = c1e^xcos(x)

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