A function y = f(x) satisfies (x + 1) f'(x) - | vedclass

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(B) Given the differential equation: $(x + 1) f'(x) - 2x(x + 1) f(x) = \frac{e^{x^2}}{x + 1}$.Dividing by $(x + 1)$ (since $x > -1$): $f'(x) - 2x f(x)

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$\left( \frac{6x + 5}{x + 1} \right) e^{x^2}$

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(B) Given the differential equation: $(x + 1) f'(x) - 2x(x + 1) f(x) = \frac{e^{x^2}}{x + 1}$.Dividing by $(x + 1)$ (since $x > -1$): $f'(x) - 2x f(x) = \frac{e^{x^2}}{(x + 1)^2}$.This is a linear differential equation of the form $f'(x) + P(x) f(x) = Q(x)$,where $P(x) = -2x$ and $Q(x) = \frac{e^{x^2}}{(x + 1)^2}$.The integrating factor $IF = e^{\int P(x) dx} = e^{\int -2x dx} = e^{-x^2}$.Multiplying the equation by $IF$: $e^{-x^2} f'(x) - 2x e^{-x^2} f(x) = \frac{1}{(x + 1)^2}$.This simplifies to: $\frac{d}{dx} [f(x) e^{-x^2}] = \frac{1}{(x + 1)^2}$.Integrating both sides: $f(x) e^{-x^2} = \int \frac{1}{(x + 1)^2} dx = -\frac{1}{x + 1} + C$.Given $f(0) = 5$: $5 e^0 = -\frac{1}{0 + 1} + C \implies 5 = -1 + C \implies C = 6$.Thus,$f(x) e^{-x^2} = 6 - \frac{1}{x + 1} = \frac{6(x + 1) - 1}{x + 1} = \frac{6x + 5}{x + 1}$.Therefore,$f(x) = \left( \frac{6x + 5}{x + 1} \right) e^{x^2}$.

$A$ function $y = f(x)$ satisfies $(x + 1) f'(x) - 2(x^2 + x) f(x) = \frac{e^{x^2}}{x + 1}$ for all $x > -1$. If $f(0) = 5$,then $f(x)$ is:

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