What is a suitable choice of form for $Y(t)$ for the following nonhomogeneous differential equation?
$(a). y''-8y'+20y=7t^3e^{4t}\sin (2t)-t^2e^{4t}-9\\ (b). y''-8y'+16 y=7t^3e^{4t}\sin (2t)-t^2e^{4t}-9$
My idea
for (a) the corresponding homogeneous equation is $y''-8y'+20y=0$
So the axuilary equaton is $m^2-8m+20=0 \Rightarrow m= 4\pm2i$
then $y_c=e^{4t}(c_1\cos (2t)+c_2 \sin (2t))$ how to we find $y_p$ using $y_c$
thank you...
Little hint for b
$$(b). y''-8y'+16 y=7t^3e^{4t}\sin (2t)-t^2e^{4t}-9$$ $$(b). (y''-4y') -4(y'-4y) =7t^3e^{4t}\sin (2t)-t^2e^{4t}-9$$ Substitute $z=y'-4y$ $$z' -4z =7t^3e^{4t}\sin (2t)-t^2e^{4t}-9$$ $$(ze^{-4t})' =e^{-4t}(7t^3e^{4t}\sin (2t)-t^2e^{4t}-9)$$ $$(ze^{-4t})' =-9e^{-4t}+7t^3\sin (2t)-t^2$$ $$ze^{-4t} =\int -9e^{-4t}+7t^3\sin (2t)-t^2dt $$ $$y'-4y=e^{4t}\int -9e^{-4t}+7t^3\sin (2t)-t^2dt $$ $$(ye^{-4t})'=\int -9e^{-4t}+7t^3\sin (2t)-t^2dt $$ $$(ye^{-4t})'=\frac 94 e^{-4t}- \frac {t^3} 3+7\int t^3\sin (2t)dt+K_1 $$ Evaluate the last integral and you are almost done... $$I=\int t^3\sin (2t)dt= ? $$ $$............$$