If y=e^{ax}(cos bx+sin bx) satisfies the equa | vedclass

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(B) Given $y=e^{ax}(\cos bx+\sin bx)$.First derivative: $\frac{dy}{dx}=ae^{ax}(\cos bx+\sin bx)+e^{ax}(-b\sin bx+b\cos bx) = ay+be^{ax}(\cos bx-\sin b

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$(a+b)^2$

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(B) Given $y=e^{ax}(\cos bx+\sin bx)$.First derivative: $\frac{dy}{dx}=ae^{ax}(\cos bx+\sin bx)+e^{ax}(-b\sin bx+b\cos bx) = ay+be^{ax}(\cos bx-\sin bx)$.Rearranging gives $be^{ax}(\cos bx-\sin bx) = \frac{dy}{dx}-ay$.Second derivative: $\frac{d^2y}{dx^2} = a\frac{dy}{dx} + b[ae^{ax}(\cos bx-\sin bx) + e^{ax}(-b\sin bx-b\cos bx)]$.$\frac{d^2y}{dx^2} = a\frac{dy}{dx} + a[be^{ax}(\cos bx-\sin bx)] - b^2e^{ax}(\sin bx+\cos bx)$.Substituting $be^{ax}(\cos bx-\sin bx) = \frac{dy}{dx}-ay$ and $e^{ax}(\cos bx+\sin bx) = y$:$\frac{d^2y}{dx^2} = a\frac{dy}{dx} + a(\frac{dy}{dx}-ay) - b^2y$.$\frac{d^2y}{dx^2} - 2a\frac{dy}{dx} + (a^2+b^2)y = 0$.Comparing with $\frac{d^2y}{dx^2}-K\frac{dy}{dx}+Ly=0$, we get $K=2a$ and $L=a^2+b^2$.Thus, $L+bK = a^2+b^2+b(2a) = a^2+b^2+2ab = (a+b)^2$.

If $y=e^{ax}(\cos bx+\sin bx)$ satisfies the equation $\frac{d^2y}{dx^2}-K\frac{dy}{dx}+Ly=0$, then $L+bK=$

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