Find the general solution for the following differential equation:
$$f''(t) + f(t) = 3 \sin(2t) + 1, \ f(0) = 3, f′ (0) = 0$$
So, I understand how to find the homogeneous solution to this equation. Given the form $f''(t) + f(t)$, you can deduce that it is in the form of $A\cos(t)+B\sin(t)$. However, I am having difficulty in finding how the specific solution to the $3\sin(2t)$ term comes into play. I know the answer is $2\cos(t) + 2\sin(t) - \sin(2t) + 1$, but I don't know how to find the specific solution with regard to the initial conditions. Can someone please please help me?
$$f′′(t) + f(t) = 3 sin(2t) + 1$$ the characteristic polynomial is
$$r^2+1=0 \implies r= \pm i \implies f(t)= c_1\cos(t)+c_2 \sin(t)$$ for the particular solution try
$$y_p= A\sin(2t)+B\cos(2t) \text { or simply } y_p=A\sin(2t)$$ $$f′′(t) + f(t) = 3 sin(2t)$$ $$(A\sin(2t))'' +A\sin(2t) = 3 sin(2t) $$ $$-4A\sin(2t)+A\sin(2t)=3\sin(2t) \implies A=-1$$ $$ \implies y_p=-\sin(2t)$$ For the constanttry $f_p=C$ so that $$f′′(t) + f(t) = 1$$ $$(C)'' + C = 1 \implies C=1$$ $$y_p=-\sin(2t)+1$$ Therefore, $$f(t)= c_1\cos(t)+c_2 \sin(t) -\sin(2t)+1$$ Try to find the coefficients with the conditions you are given.. $$ \begin{cases} f(0)=3 \\ f'(0)=0 \end{cases} \implies \begin{cases} f(0)=3 \implies f(0)= c_1\cos(0)+c_2 \sin(0) -\sin(0)+1=3 \\ f'(0)=0 \implies f'(0)= -c_1\sin(0)+c_2 \cos(0) -2\cos(0)+1=0 \end{cases} $$ It's just a linear system that's easy to solve...