Given the function y(x)C1ex+C2excos⁡xa) Prove that it is a solution to the equation y‴−3y″+4y′−2y=0b) Show that the initial value problem y(0)=1; y′(0)=0; y″(0)=0 has no solution.

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Given the function y(x)C1ex+C2excos⁡xa) Prove that it is a solution to the equation y‴−3y″+4y′−2y=0b) Show that the initial value problem y(0)=1; y′(0)=0; y″(0)=0 has no solution.

accimaroyalde · Accepted Answer

y(x)C1ex+C2excos⁡xy′=C1ex+C2[−exsin⁡x+excos⁡x]y″=C1ex+C2[−excos⁡x−exsin⁡x−exsin⁡x+excos⁡x]y‴=C1ex+C2[exsin⁡x−excos⁡x−excos⁡x−exsin⁡x−exsin⁡x−excos⁡x−exsin⁡x+excos⁡x]∴y′=C1ex+C2[excos⁡x−exsin⁡x]y″C1ex+C2[−2exsin⁡x]y‴=C1ex+C2[−2excos⁡x−2exsin⁡x]Substitute in y″−3y″+4y′−2y⇒C1ex−2C2excos⁡x−2C2exsin⁡x−3C1ex+6C2exsin⁡x+4C1ex+4C2excos⁡x−4C2exsin⁡x−2C1ex−2C2excos⁡x=0Since all terms cancel out. Hence given y is solution.

Terry Ray · Accepted Answer

y(x)=C1ex+C2excos⁡xy(0)=1⇒C1+C2=1y′(0)=0⇒C1=0y″(0)=0⇒C1+C2[−2]=0⇒c1=2C2, since C1=0⇒C2=0But C1+C2≠1Hence no solution is possible.

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

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