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(A) Given the differential equation: $(x \log x) \frac{dy}{dx} + y = 2x \log x$. Divide both sides by $(x \log x)$ to get the linear form $\frac{dy}{d

Vedclass · Accepted Answer

$2$

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(A) Given the differential equation: $(x \log x) \frac{dy}{dx} + y = 2x \log x$. Divide both sides by $(x \log x)$ to get the linear form $\frac{dy}{dx} + P(x)y = Q(x)$: $\frac{dy}{dx} + \frac{1}{x \log x} y = 2$. Here,$P(x) = \frac{1}{x \log x}$ and $Q(x) = 2$. The Integrating Factor ($I$.$F$.) is given by: $I.F. = e^{\int P(x) dx} = e^{\int \frac{1}{x \log x} dx} = e^{\log(\log x)} = \log x$. The general solution is $y \cdot (I.F.) = \int Q(x) \cdot (I.F.) dx + C$: $y \log x = \int 2 \log x dx + C$. Using integration by parts,$\int \log x dx = x \log x - x$: $y \log x = 2(x \log x - x) + C$. Since the equation is defined for $x \geq 1$,at $x=1$,$\log(1)=0$. Substituting $x=1$ into the equation: $y(1) \cdot 0 = 2(1 \cdot 0 - 1) + C \implies 0 = -2 + C \implies C = 2$. Thus,the solution is $y \log x = 2x \log x - 2x + 2$. To find $y(e)$,substitute $x=e$: $y(e) \log(e) = 2(e) \log(e) - 2(e) + 2$. Since $\log(e) = 1$: $y(e) \cdot 1 = 2e - 2e + 2 = 2$. Therefore,$y(e) = 2$.

Let $y=y(x)$ be the solution of the differential equation $(x \log x) \frac{dy}{dx} + y = 2x \log x$ for $x \geq 1$. Then $y(e)$ is equal to:

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