1. Given differential equation

We have

$$\frac{dx}{dy}=\frac{1+x-y^2}{y}, \qquad x(1)=1.$$

We need to find $x(2)$ and then compute $5x(2)$.

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2. Rewrite in linear form

Write the equation as

$$\frac{dx}{dy}=\frac{1+x-y^2}{y}=\frac{1}{y}+\frac{x}{y}-y.$$

So,

$$\frac{dx}{dy}-\frac{1}{y}x=\frac{1}{y}-y.$$

This is a linear differential equation in $x$ as a function of $y$.

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3. Find the integrating factor

For

$$\frac{dx}{dy}+P(y)x=Q(y),$$

we have

$$P(y)=-\frac{1}{y}.$$

Hence the integrating factor is

$$\text{I.F.}=e^{\int -\frac{1}{y}\,dy}=e^{-\ln y}=\frac{1}{y}$$

(taking $y>0$ since the condition is given at $y=1$).

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4. Multiply throughout by the integrating factor

Multiplying the equation

$$\frac{dx}{dy}-\frac{x}{y}=\frac{1}{y}-y$$

by $\frac{1}{y}$, we get

$$\frac{1}{y}\frac{dx}{dy}-\frac{x}{y^2}=\frac{1}{y^2}-1.$$

Notice that

$$\frac{d}{dy}\left(\frac{x}{y}\right)=\frac{1}{y}\frac{dx}{dy}-\frac{x}{y^2}.$$

Therefore,

$$\frac{d}{dy}\left(\frac{x}{y}\right)=\frac{1}{y^2}-1.$$

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5. Integrate

Integrating both sides,

$$\frac{x}{y}=\int \left(\frac{1}{y^2}-1\right)dy =\int y^{-2}dy-\int 1\,dy.$$

Thus,

$$\frac{x}{y}=-\frac{1}{y}-y+C.$$

Multiply by $y$:

$$x=-1-y^2+Cy.$$

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6. Use the initial condition

Given $x(1)=1$,

$$1=-1-1+C(1).$$

So,

$$1=-2+C \implies C=3.$$

Hence,

$$x(y)=-1-y^2+3y.$$

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7. Find $x(2)$

$$x(2)=-1-4+6=1.$$

Therefore,

$$5x(2)=5\cdot 1=5.$$

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8. Comparison with stored answer

Our derived answer is $5$, which matches the stored correct answer.