Solve the differential equation 2y"+20y'+51y=0, y(0)=2 y'(0)=0

Phoebe 2021-02-08 Answered
Solve the differential equation2y"+20y+51y=0,y(0)=2
y(0)=0
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Expert Answer

Ezra Herbert
Answered 2021-02-09 Author has 99 answers
Step 1
Let y=emt be a trial solution. Then the auxiliary equation is given by 2m2+20m+51=0asemt0, for all t
Thus the solutions of auxiliary equations are
m=5±i22
Therefore the solution of the differential equation is given by
y=e5(c1cos(t2)+c2sin(t2))
Now apply initial condition
y(0)=e5×0(c1cos(02)+c2sin(02))=1×(cos(0)c1+sin(0)c2)=c1
Thus y(0)=2 gives us c1=2
Step 2
Further by allpying the product rule of differention we get
y(x)=(e5t)(c1cos(t2)+c2sin(t2))+(c1(t2)+c2sin(t2))e5t
=(5e5t)(c1cos(t2)+c2sin(t2))+(c1sin(t2)+c2cos(t2))2e5t
=5e5t(c1cos(t2)+c2sin(t2))+(e5t(c1sin(t2)+c2cos(t2))2
Notice that
y(0)=5e50(c1cos(02)+c2sin(02))+(e50(c1sin(02)+c2cos(02))2
=5e50(cos(0)c1+sin(0)c2)+(e50(cos(0)c2sin(0)c1))2=5c1+c22
Now y'(0)=0 gives us
5c1+c22=010+c22=0 [c1=2]
c2=102
Therefore the required solution is given by
y=e5t(2cos
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