Let y=y(x) be the solution curve of the diffe | vedclass

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(B) The given differential equation is a linear differential equation of the form $\frac{dy}{dx} + P(x)y = Q(x)$,where $P(x) = \frac{2x^2+11x+13}{(x+1

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$\frac{3}{2}$

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(B) The given differential equation is a linear differential equation of the form $\frac{dy}{dx} + P(x)y = Q(x)$,where $P(x) = \frac{2x^2+11x+13}{(x+1)(x+2)(x+3)}$ and $Q(x) = \frac{x+3}{x+1}$.Using partial fractions for $P(x)$:$\frac{2x^2+11x+13}{(x+1)(x+2)(x+3)} = \frac{A}{x+1} + \frac{B}{x+2} + \frac{C}{x+3} = \frac{2}{x+1} + \frac{1}{x+2} - \frac{1}{x+3}$.Integrating $P(x)$:$\int P(x) dx = 2\ln(x+1) + \ln(x+2) - \ln(x+3) = \ln\left(\frac{(x+1)^2(x+2)}{x+3}\right)$.The Integrating Factor ($I$.$F$.) is $e^{\int P(x) dx} = \frac{(x+1)^2(x+2)}{x+3}$.The general solution is $y \cdot (I.F.) = \int Q(x) \cdot (I.F.) dx + C$.$y \cdot \frac{(x+1)^2(x+2)}{x+3} = \int \left(\frac{x+3}{x+1}\right) \cdot \frac{(x+1)^2(x+2)}{x+3} dx = \int (x+1)(x+2) dx = \int (x^2+3x+2) dx$.$y \cdot \frac{(x+1)^2(x+2)}{x+3} = \frac{x^3}{3} + \frac{3x^2}{2} + 2x + C$.Since the curve passes through $(0,1)$:$1 \cdot \frac{(1)^2(2)}{3} = 0 + 0 + 0 + C \implies C = \frac{2}{3}$.For $x=1$:$y(1) \cdot \frac{(2)^2(3)}{4} = \frac{1}{3} + \frac{3}{2} + 2 + \frac{2}{3} = 1 + \frac{3}{2} + 2 = \frac{9}{2}$.$y(1) \cdot 3 = \frac{9}{2} \implies y(1) = \frac{3}{2}$.

Let $y=y(x)$ be the solution curve of the differential equation $\frac{dy}{dx} + \left(\frac{2x^2+11x+13}{x^3+6x^2+11x+6}\right)y = \frac{x+3}{x+1}$,where $x > -1$,which passes through the point $(0,1)$. Then $y(1)$ is equal to:

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