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(C) The given differential equation is a linear differential equation of the form $\frac{dy}{dx} + P(x)y = Q(x)$,where $P(x) = -1$ and $Q(x) = 1 + 4 \

Vedclass · Accepted Answer

$7$

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(C) The given differential equation is a linear differential equation of the form $\frac{dy}{dx} + P(x)y = Q(x)$,where $P(x) = -1$ and $Q(x) = 1 + 4 \sin x$.The integrating factor $(IF)$ is $e^{\int P(x) dx} = e^{\int -1 dx} = e^{-x}$.Multiplying both sides by the $IF$,we get: $e^{-x} \frac{dy}{dx} - e^{-x} y = e^{-x}(1 + 4 \sin x)$,which simplifies to $\frac{d}{dx}(y e^{-x}) = e^{-x} + 4 e^{-x} \sin x$.Integrating both sides: $y e^{-x} = \int e^{-x} dx + 4 \int e^{-x} \sin x dx$.Using the formula $\int e^{ax} \sin(bx) dx = \frac{e^{ax}}{a^2+b^2}(a \sin(bx) - b \cos(bx))$,we get $\int e^{-x} \sin x dx = \frac{e^{-x}}{(-1)^2+1^2}(-1 \sin x - 1 \cos x) = -\frac{e^{-x}}{2}(\sin x + \cos x)$.Thus,$y e^{-x} = -e^{-x} + 4 \left( -\frac{e^{-x}}{2}(\sin x + \cos x) \right) + C = -e^{-x} - 2e^{-x}(\sin x + \cos x) + C$.Dividing by $e^{-x}$,we get $y = -1 - 2(\sin x + \cos x) + C e^x$.Given $y(\pi) = 1$: $1 = -1 - 2(\sin \pi + \cos \pi) + C e^{\pi} \Rightarrow 1 = -1 - 2(0 - 1) + C e^{\pi} \Rightarrow 1 = -1 + 2 + C e^{\pi} \Rightarrow 1 = 1 + C e^{\pi} \Rightarrow C = 0$.So,$y(x) = -1 - 2(\sin x + \cos x)$.Then $y\left(\frac{\pi}{2}\right) = -1 - 2(\sin \frac{\pi}{2} + \cos \frac{\pi}{2}) = -1 - 2(1 + 0) = -3$.Finally,$y\left(\frac{\pi}{2}\right) + 10 = -3 + 10 = 7$.

Let the solution $y=y(x)$ of the differential equation $\frac{dy}{dx}-y=1+4 \sin x$ satisfy $y(\pi)=1$. Then $y\left(\frac{\pi}{2}\right)+10$ is equal to:

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