Solve the ivp problem yyxex y01 y00 y01 Solve the ivp proble

Solve the ivp problem

y\'\'\'-y\'\'=x+e^(-x)

y(0)=1

y\'(0)=0

y\'\'(0)=1

Solve the ivp problem

y\'\'\'-y\'\'=x+e^(-x)

y(0)=1

y\'(0)=0

y\'\'(0)=1

y\'\'\'-y\'\'=x+e^(-x)

y(0)=1

y\'(0)=0

y\'\'(0)=1

First we solve homogeneous equation

y\'\'\'-y\'\'=0

LEt y=exp(kx)

Substituting gives

k^3-k^2=0

k=1,k^2=0

So,k=0 is a repeated root

SO general solution to homogeneous equation is

y(x)=Ae^x+e^{0*x}(B+Cx)=Ae^x+B+Cx

y(x)=Ae^x+B+Cx

Now based on hte inhomogeneous part ie

x+e^{-x} and knowing that x^2 is already solution to homogeneous equation ,we guess the particular solution to be

Dx^2+Ee^{-x}

Substituting gives

2D-Ee^{-x}-Ee^{-x}=x+e^{-x}

So no terms involving x left on Left Hand Side

So we choose particular solution

Dx^2+Ex^3+Fe^{-x}

Substituting gives

6E-Fe^{-x}-6Ex-2D-Fe^{-x}=x+e^{-x}

Comparing coefficients gives

6E=2D ie D=3E

-2F=1 ie F=-1/2

-6E=1 ie E=-1/6

D=-1/2

Hence solution to IVP is

y(x)=Ae^x+B+Cx-x^3/6-x^2/2-e^{-x}/2

$Solve the ivp problem y\'\'\'-y\'\'=x+e^(-x) y(0)=1 y\'(0)=0 y\'\'(0)=1 Solve the ivp problem y\'\'\'-y\'\'=x+e^(-x) y(0)=1 y\'(0)=0 y\'\'(0)=1 y\'\'\'-y\'\'=x$

Solve the ivp problem yyxex y01 y00 y01 Solve the ivp proble

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