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[tex]y'''-6y''+8y'=9e^x[/tex]

[tex]y(0)=13, y'(0)=17, y''(0)=16[/tex]

solve for y(x)=

I first solved for [tex]y'''-6y''+8y'=0[/tex]

I got the general solution:

[tex]y=A0.5e^{2x}+B0.25e^{4x}+C[/tex]

I was just wondering if I should plug in the initial conditions now or should I solve the 9e^x and then solve for the initial conditions? Also, if I solved for 9e^x now, would the costant C just be zero?