Undetermined Coefficients Method

Question

$$Y'''+4y'=t$$ My attempt: For t I took my guess to be in the form $y_p= At+B$, however, this method fails to give the correct answer, is it because 'y' is missing from the original ODE? Annihilator method works though to give the correct answer

Answer 1

It works fine, whatever method you use annihilator or else

$$Y'''+4y'=t \implies r^3+4r=0 \implies r(r^2+4)=0 \implies r=0,-2i,2i$$ $$y_h=c_1+c_2\cos(2x)+c_3\sin(2x)$$ Try $y_p=at^2+bt+c $ instead of $y_p=at+b$ $$\implies 4(2at+b)=t \implies a=\frac18 \,; b=0 $$ $$\implies y_p=\frac {t^2}8+c$$ Therefore $$y(x)=c_1+c_2\cos(2x)+c_3\sin(2x)+\frac {t^2}8$$

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Or you can integrate the equation directly $$y'''+4y'=t $$ $$y''+4y=\frac {t^2}2+K$$ And apply the variation of constant...