Question

Solve the boundary-value problem, if possible.

y’’8y+17y=0,y(0)=3,y(π)=2y’’-8y’+17y=0, y(0)=3, y(π)=2

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y8y+17y=0y''-8 y' +17y =0

The auxiliary equation is

r28r+17=0r^2-8r+17=0

Subtract 1 from both sides

r28r+16=1r^2-8r+16=-1

(r4)2=1(r-4)^2=-1

Take square root of both sides, To get

r4=±ir-4 = \pm i

There are two solutions, r1=4ir_1 = 4-i and r2=4+ir_2 = 4+i

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