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

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(B) The given linear differential equation is $(x + 1)f'(x) - 2x(x + 1)f(x) = \frac{e^{x^2}}{x + 1}$. Dividing by $(x + 1)$,we get $f'(x) - 2xf(x) = \

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

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(B) The given linear differential equation is $(x + 1)f'(x) - 2x(x + 1)f(x) = \frac{e^{x^2}}{x + 1}$. Dividing by $(x + 1)$,we get $f'(x) - 2xf(x) = \frac{e^{x^2}}{(x + 1)^2}$. The integrating factor $I.F. = e^{\int -2x \, dx} = e^{-x^2}$. Multiplying both sides by $I.F.$,we have $\frac{d}{dx} [f(x) e^{-x^2}] = \frac{e^{x^2}}{(x + 1)^2} e^{-x^2} = \frac{1}{(x + 1)^2}$. Integrating both sides,$f(x) e^{-x^2} = \int \frac{1}{(x + 1)^2} \, dx = -\frac{1}{x + 1} + C$. Given $f(0) = 5$,we have $5(e^0) = -\frac{1}{0 + 1} + C \Rightarrow 5 = -1 + C \Rightarrow C = 6$. Thus,$f(x) e^{-x^2} = 6 - \frac{1}{x + 1} = \frac{6x + 6 - 1}{x + 1} = \frac{6x + 5}{x + 1}$. Therefore,$f(x) = \left( \frac{6x + 5}{x + 1} \right) e^{x^2}$.

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

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