The differential equation whose general solut | vedclass

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(B) Given the general solution: $y = c_1 \cos(x + c_2) - c_3 e^{-x + c_4} + c_5 \sin x$ Using trigonometric identity $\cos(A+B) = \cos A \cos B - \sin

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$\frac{d^3y}{dx^3} + \frac{d^2y}{dx^2} + \frac{dy}{dx} + y = 0$

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(B) Given the general solution: $y = c_1 \cos(x + c_2) - c_3 e^{-x + c_4} + c_5 \sin x$ Using trigonometric identity $\cos(A+B) = \cos A \cos B - \sin A \sin B$ and property of exponents $e^{A+B} = e^A e^B$: $y = c_1(\cos x \cos c_2 - \sin x \sin c_2) - (c_3 e^{c_4}) e^{-x} + c_5 \sin x$ Let $A = c_1 \cos c_2$,$B = (c_1 \sin c_2 - c_5)$,and $C = c_3 e^{c_4}$. Then $y = A \cos x - B \sin x - C e^{-x} \dots (i)$ Differentiating with respect to $x$: $y' = -A \sin x - B \cos x + C e^{-x} \dots (ii)$ $y'' = -A \cos x + B \sin x - C e^{-x} \dots (iii)$ $y''' = A \sin x + B \cos x + C e^{-x} \dots (iv)$ Adding $(i)$ and $(iii)$: $y'' + y = -2C e^{-x} \dots (v)$ Adding $(ii)$ and $(iv)$: $y''' + y' = 2C e^{-x} \dots (vi)$ Adding $(v)$ and $(vi)$: $y''' + y'' + y' + y = 0$ Thus,the differential equation is $\frac{d^3y}{dx^3} + \frac{d^2y}{dx^2} + \frac{dy}{dx} + y = 0$.

The differential equation whose general solution is given by $y = (c_1 \cos(x + c_2)) - (c_3 e^{(-x + c_4)}) + (c_5 \sin x)$,where $c_1, c_2, c_3, c_4, c_5$ are arbitrary constants,is

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