We have, $y=e^{a x}(\cos b x+\sin b x)$
$\frac{d y}{d x}=e^{a x}(-\sin b x \cdot b+\cos b x \cdot b)$ $+a \cdot e^{a x}(\cos b x+\sin b x)$
$\frac{d y}{d x}=b e^{a x}(\cos b x-\sin b x)+a y$ $\ldots(\mathrm{i})$
$\Rightarrow \quad \frac{d^2 y}{d x^2}=b\left[e^{a x}(-\sin b x \cdot b-\cos b x \cdot b)\right.$ $\left.+a \cdot e^{a x}(\cos b x-\sin b x)\right]+a \frac{d y}{d x}$
$\Rightarrow \quad \frac{d^2 y}{d x^2}=b$
$\left[-b e^{a x}(\sin b x+\cos b x)+a e^{a x}(\cos b x-\sin b x)\right]$ $+a \frac{d y}{d x}=0$
$\Rightarrow \quad \frac{d^2 y}{d x^2}=b\left[(-b)(y)+a e^{a x}(\cos b x-\sin b x)\right]+a \frac{d y}{d x}$
$\Rightarrow \frac{d^2 y}{d x^2}-a \frac{d y}{d x}+b^2 y-a b e^{a x}(\cos b x-\sin b x)=0$
$\Rightarrow \frac{d^2 y}{d x^2}-a \frac{d y}{d x}+b^2 y-a\left(\frac{d y}{d x}-a y\right)=0$
$\Rightarrow \frac{d^2 y}{d x^2}-2 a \frac{d y}{d x}+\left(b^2+a^2\right) y=0$
On comparing with $\frac{d^2 y}{d x^2}-k \frac{d y}{d x}+L y=0$
$\therefore \quad K=2 a, L=b^2+a^2$
$\therefore \quad L+b k=b^2+a^2+2 a b=(a+b)^2$