If frac{dx}{dy} = frac{1+x-y^2}{y} and x(1) = | vedclass

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(A) Given the differential equation: $\frac{dx}{dy} = \frac{1+x-y^2}{y}$.Rearranging the terms,we get: $\frac{dx}{dy} - \frac{x}{y} = \frac{1-y^2}{y}$

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$5$

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(A) Given the differential equation: $\frac{dx}{dy} = \frac{1+x-y^2}{y}$.Rearranging the terms,we get: $\frac{dx}{dy} - \frac{x}{y} = \frac{1-y^2}{y}$.This is a linear differential equation of the form $\frac{dx}{dy} + P(y)x = Q(y)$,where $P(y) = -\frac{1}{y}$ and $Q(y) = \frac{1-y^2}{y}$.The integrating factor $(IF)$ is given by $IF = e^{\int P(y) dy} = e^{\int -\frac{1}{y} dy} = e^{-\ln|y|} = \frac{1}{y}$.The general solution is $x \cdot IF = \int Q(y) \cdot IF dy + C$.Substituting the values: $x \cdot \frac{1}{y} = \int \left(\frac{1-y^2}{y}\right) \cdot \frac{1}{y} dy + C$.$x \cdot \frac{1}{y} = \int \left(\frac{1}{y^2} - 1\right) dy + C$.$x \cdot \frac{1}{y} = -\frac{1}{y} - y + C$.Multiplying by $y$: $x = -1 - y^2 + Cy$.Given the condition $x(1) = 1$,we substitute $y=1$ and $x=1$: $1 = -1 - (1)^2 + C(1) \Rightarrow 1 = -2 + C \Rightarrow C = 3$.Thus,the particular solution is $x = -1 - y^2 + 3y$.To find $5x(2)$,we substitute $y=2$: $x(2) = -1 - (2)^2 + 3(2) = -1 - 4 + 6 = 1$.Therefore,$5x(2) = 5(1) = 5$.

If $\frac{dx}{dy} = \frac{1+x-y^2}{y}$ and $x(1) = 1$,then $5x(2)$ is equal to:

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