If the differential equation having y=Ae^x+B | vedclass

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(D) Given the general solution $y = Ae^x + B \sin x$. Step $1$: Differentiate with respect to $x$: $\frac{dy}{dx} = Ae^x + B \cos x$. Step $2$: Differ

Vedclass · Accepted Answer

$\cos x - \sin x$

Vedclass · Answer

(D) Given the general solution $y = Ae^x + B \sin x$. Step $1$: Differentiate with respect to $x$: $\frac{dy}{dx} = Ae^x + B \cos x$. Step $2$: Differentiate again: $\frac{d^2y}{dx^2} = Ae^x - B \sin x$. Step $3$: Eliminate constants $A$ and $B$. From the original equation, $Ae^x = y - B \sin x$. Substitute into the second derivative: $\frac{d^2y}{dx^2} = (y - B \sin x) - B \sin x = y - 2B \sin x$. From the first derivative: $B \cos x = \frac{dy}{dx} - Ae^x = \frac{dy}{dx} - (y - B \sin x) = \frac{dy}{dx} - y + B \sin x$. This approach is complex. Let's use the operator method. The solution $y = Ae^x + B \sin x$ implies $y$ is a linear combination of $e^x$ and $\sin x$. The characteristic roots are $1, i, -i$. The differential operator is $(D-1)(D^2+1)y = 0$. Expanding this: $(D^3 - D^2 + D - 1)y = 0$. This gives $\frac{d^3y}{dx^3} - \frac{d^2y}{dx^2} + \frac{dy}{dx} - y = 0$. However, the question specifies a second-order equation $f(x) \frac{d^2y}{dx^2} + g(x) \frac{dy}{dx} + h(x) y = 0$. For $y = Ae^x + B \sin x$, we have $y' = Ae^x + B \cos x$ and $y'' = Ae^x - B \sin x$. $y'' - y = -B \sin x - B \sin x = -2B \sin x$. $y' - y = B \cos x - B \sin x$. By eliminating $A$ and $B$, the resulting equation is $(\cos x + \sin x) y'' - (\cos x - \sin x) y' - (\cos x + \sin x) y = 0$. Thus $f(x) = \cos x + \sin x$, $g(x) = -\cos x + \sin x$, $h(x) = -\cos x - \sin x$. Summing these: $f(x) + g(x) + h(x) = (\cos x + \sin x) + (-\cos x + \sin x) + (-\cos x - \sin x) = -\cos x + \sin x$.

If the differential equation having $y=Ae^x+B \sin x$ as its general solution is $f(x) \frac{d^2 y}{d x^2}+g(x) \frac{d y}{d x}+h(x) y=0$, then $f(x)+g(x)+h(x)=$

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