I'm having a hard time solving this problem.
$$x''+4x=t$$ $a)$Derivative the corresponding Cauchy fn.
$b)$Find the solution of the given equation to the I.C., $x(0)=0$, $x'(0)=0$
So the char. eq. is $r^2+4=0$, $r_1=2i$ and $r_2=-2i$. How can I write $x(t)$ as a function when the roots are complex?
For $b)$ I have no idea to solve it.
Any advice/solution would be appreciated.
$$x''+4x=t$$ When the roots are complex like $ r=a\pm bi$ then :$$x(t)=c_1e^{at}\cos (bt)+c_2e^{at}\sin (bt)$$ So for your solution ($r=\pm 2i$) you have to write: $$x(t)=c_1\cos (2t)+c_2\sin (2t)$$
For $b$ part of the question, you have to first solve the inhomogeneous DE. Try Undetermined coefficients method; $$x_p(t)=at$$ Plug this in your DE and find the constant $a$. Then apply initial conditions you 're given. $$x(t)=c_1\cos (2t)+c_2\sin (2t)+at$$
Apply intial conditions: $$x(0)=c_1\cos (2*0)+c_2\sin (2*0)+a*0$$ $$0=c_1$$ Apply the second intial condition $$x(t)=c_2\sin (2t)+at$$ Differentiate the apply initial condition.