Given the function y(x)C_1e^x+C_2e^x \cos xa) Prove that it is

Wanda Kane 2021-12-15 Answered

Given the function y(x)C1ex+C2excosx
a) Prove that it is a solution to the equation y3y+4y2y=0
b) Show that the initial value problem y(0)=1; y(0)=0; y(0)=0 has no solution.

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Expert Answer

accimaroyalde
Answered 2021-12-16 Author has 29 answers
y(x)C1ex+C2excosx
y=C1ex+C2[exsinx+excosx]
y=C1ex+C2[excosxexsinxexsinx+excosx]
y=C1ex+C2[exsinxexcosxexcosxexsinxexsinxexcosxexsinx+excosx]
y=C1ex+C2[excosxexsinx]
yC1ex+C2[2exsinx]
y=C1ex+C2[2excosx2exsinx]
Substitute in y3y+4y2y
C1ex2C2excosx2C2exsinx3C1ex+6C2exsinx+4C1ex+4C2excosx4C2exsinx2C1ex2C2excosx
=0
Since all terms cancel out. Hence given y is solution.
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Terry Ray
Answered 2021-12-17 Author has 50 answers
y(x)=C1ex+C2excosx
y(0)=1C1+C2=1
y(0)=0C1=0
y(0)=0C1+C2[2]=0
c1=2C2, since C1=0C2=0
But C1+C21
Hence no solution is possible.
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