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Chapter 4: Q4.3-14E (page 172)

Find a general solution. y''+2y'+5y=0

Short Answer

Expert verified

The general solution of the given equationy''+2y'+5y=0isy(t)=e-t(c1cos(2t)+c2sin(2t)).

Step by step solution

01

Complex conjugate roots.

If the auxiliary equation has complex conjugate roots 伪卤i尾, then the general solution is given as:

y(t)=c1e伪tcos尾t+c2e伪tsin尾t.

02

Finding roots of the auxiliary equation.

Given differential equation isy''+2y'+5y=0.

Then the auxiliary equation isr2+2r+5=0.

The roots of the auxiliary equation are:

role="math" localid="1654070409803" r=-2卤22-4脳1脳52脳1r=-2卤4-202r=-2卤-162r=-2卤4i2r=-1卤2i

03

Final answer.

Therefore, the general solution is:

y(t)=e-1脳t(c1cos(2t)+c2sin(2t))=e-t(c1cos(2t)+c2sin(2t))

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