If x = x(y) is the solution of the differenti | vedclass

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(A) The given differential equation is $y \frac{dx}{dy} = 2x + y^{3}(y+1)e^{y}$.Dividing by $y$,we get $\frac{dx}{dy} - \frac{2}{y}x = y^{2}(y+1)e^{y}

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$e^{3}(e^{e}-1)$

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(A) The given differential equation is $y \frac{dx}{dy} = 2x + y^{3}(y+1)e^{y}$.Dividing by $y$,we get $\frac{dx}{dy} - \frac{2}{y}x = y^{2}(y+1)e^{y}$.This is a linear differential equation of the form $\frac{dx}{dy} + P(y)x = Q(y)$,where $P(y) = -\frac{2}{y}$ and $Q(y) = y^{2}(y+1)e^{y}$.The integrating factor ($I$.$F$.) is $e^{\int P(y) dy} = e^{\int -\frac{2}{y} dy} = e^{-2 \ln|y|} = \frac{1}{y^{2}}$.The solution is given by $x \cdot (I.F.) = \int Q(y) \cdot (I.F.) dy + C$.$x \cdot \frac{1}{y^{2}} = \int y^{2}(y+1)e^{y} \cdot \frac{1}{y^{2}} dy = \int (y+1)e^{y} dy$.Using integration by parts,$\int (y+1)e^{y} dy = (y+1)e^{y} - \int e^{y} dy = (y+1)e^{y} - e^{y} + C = ye^{y} + C$.So,$\frac{x}{y^{2}} = ye^{y} + C$,which implies $x = y^{3}e^{y} + Cy^{2}$.Given $x(1) = 0$,we have $0 = (1)^{3}e^{1} + C(1)^{2} \Rightarrow C = -e$.Thus,the solution is $x = y^{3}e^{y} - ey^{2}$.For $x(e)$,we substitute $y = e$: $x(e) = e^{3}e^{e} - ee^{2} = e^{3}e^{e} - e^{3} = e^{3}(e^{e}-1)$.

If $x = x(y)$ is the solution of the differential equation $y \frac{dx}{dy} = 2x + y^{3}(y+1)e^{y}$ with the initial condition $x(1) = 0$,then $x(e)$ is equal to:

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