Q.$y=e^{ax}\cos(bx+c)$

Let $y=e^{ax}\cos(bx+c)$.

Step 1 — differentiate by the product rule: $y_1=ae^{ax}\cos(bx+c)-be^{ax}\sin(bx+c)=e^{ax}\left[a\cos(bx+c)-b\sin(bx+c)\right]$.

Step 2 — write $a\cos\theta-b\sin\theta$ (with $\theta=bx+c$) in the auxiliary-angle form $R\cos(\theta+\alpha)$: expanding $R\cos(\theta+\alpha)=R\cos\alpha\cos\theta-R\sin\alpha\sin\theta$ and matching coefficients gives $R\cos\alpha=a,\,R\sin\alpha=b$, so $R=\sqrt{a^2+b^2}$ and $\alpha=\tan^{-1}(b/a)$.

So $a\cos\theta-b\sin\theta=\sqrt{a^2+b^2}\,\cos(\theta+\alpha)$, giving $y_1=\sqrt{a^2+b^2}\,e^{ax}\cos(bx+c+\alpha)$. …

Differentiate once, re-express $a\cos\theta-b\sin\theta$ as $R\cos(\theta+\alpha)$ via the auxiliary-angle (R-method), showing each differentiation reproduces the same form with an ext …

Getting the auxiliary angle wrong — for $a\cos\theta-b\sin\theta$ it is $R\cos(\theta+\alpha)$ with $\tan\alpha=b/a$ (this is a different pairing from t …