I want to guess for the particular solution to the given differential equation without finding the coefficients.
$4y"+y=e^{-2t}\sin{(\frac{t}{2}})$+ 6t $\cos(\frac{t}{2}).$
The guess for the form of the particular solution is
$Y_p(t)=e^{-2t}\big(A\cos{(\frac{t}{2})}+B\sin{(\frac{t}{2})}\big)+t(Ct+D)\cos{(\frac{t}{2})}+t(Et+f)\sin{(\frac{t}{2})}$
But the complementary solution to this differential equation is
$y_c(t)=c_1\cos{(\frac{t}{2})}+c_2\sin{(\frac{t}{2})}$.
I want to know how the complementary solution is computed? If any one knows the answer, post the reply.
Best Answer
$$4y"+y=e^{-2t}\sin(t/2)+6t\cos(t/2)$$ The characteristic equation is $$4r^2+1=0 \implies r^2-\frac {i^2}4=0 $$$$\implies (r-i/2)(r+i/2)=0 \implies r=\pm\frac i2$$ Therefore $$y(t)=c_1\cos(t/2)+c_2\sin(t/2)$$
Your guess is correct for the particular solution $$4y"+y=e^{-2t}\sin(t/2)$$ $$ \implies y_p=e^{-2t}(A \cos(t/2)+B\sin(t/2))$$ $$4y"+y=6t\cos(t/2)$$ $$\implies y_p=t(Ct+D)\cos(t/2)+t(Et+F)\sin(t/2)$$