If a circuit is activated by the electromotive force $E=V_0\sin(nt)$ where $t$ and $V_0$ are constants, then the current I at time t satisfies the differential equation $$L\frac{d^2I}{dt^2} + R\frac{dI}{dt}+\frac{1}{C} I=\frac{dE}{dt}$$ where $C$, $L$ and $R$ are constants and positive.
1) Describe how I varies with t when $CR^2\gt4L$
2) Show that after some time it settles to a regular oscillation given by $I=a\sin(nt) + b\cos(nt)$, where a and b are constants, and hence show that this can be written in the form $I=K\sin(nt+\alpha)$, where $K$ and $\alpha$ are constants.
3) Hence ultimately express $K$ in terms of $V_0$, $C,$ $L,$ $R,$ and $n.$
My approach is as follows: $$\frac{dE}{dt}=nV_0\cos(nt)$$ $$L\frac{d^2I}{dt^2} + R\frac{dI}{dt}+\frac{1}{C}I=nV_0\cos(nt)$$ Then I tried to find the complementary function: $$L\lambda^2+R\lambda+\frac{1}{C}=0$$ $$\lambda=-\frac{R\pm\sqrt{(CR^2-4L)\frac{1}{C}}}{2L}$$ Next I tried to find the particular solution by letting $I=a\sin(nt)+b\cos(nt)$ and substitute into $L\frac{d^2I}{dt^2} + R\frac{dI}{dt}+\frac{1}{C}I=nV_0\cos(nt)$ which ultimately brought me to the result $a=b=\frac{V_0}{2R}$ and this gives $K$ to be $\sqrt{2}\frac{V_0}{2R}$ which is of course wrong.
K is supposedly $$\frac{V_0n}{\sqrt{(\frac{1}{C}-Ln^2)^2+R^2n^2}}$$
Help would be much appreciated, at the level of an A-level student please.
With your substitution: $$ \begin{array}{ccrlcrl} \frac{1}{C}I&=&\frac{a}{C}&\sin(nt)&+&\frac{b}{C}&\cos(nt) \\ R\frac{dI}{dt}&=&-Rnb&\sin(nt)&+&Rna&\cos(nt) \\ L\frac{d^2I}{dt^2}&=&-Ln^2a&\sin(nt)&+&-Ln^2b&\cos(nt) \\ \hline nV_0\cos(nt)&=&\left(\left(\frac{1}{C}-Ln^2\right)a-Rnb\right)&\sin(nt)&+&\left(\left(\frac{1}{C}-Ln^2\right)b-Rna\right)&\cos(nt) \end{array} $$ Matching the coefficients gives the system $$ \left\{ \begin{array}{rlcrll} \left(\frac{1}{C}-Ln^2\right)&a&+&-Rn&b&=0 \\ -Rn&a&+&\left(\frac{1}{C}-Ln^2\right)&b&=nV_0 \end{array} \right. $$ It should be clear from this why your $a$ and $b$ don't work, and this is a linear system (albeit a complicated one) so it shouldn't give you too much difficulty.