Question

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

Answer 1

Given the differential equation:
$\frac{dx}{dy} = \frac{1 + x - y^2}{y}$
with the initial condition:
$x(1) = 1$

Step 1: Rearranging the Equation
Rearranging the differential equation:
$y \frac{dx}{dy} = 1 + x - y^2$
Rewriting:
$\frac{dx}{dy} - \frac{x}{y} = \frac{1 - y^2}{y}$
This is a linear first-order differential equation in the form:
$\frac{dx}{dy} + P(y)x = Q(y)$
where:
$P(y) = -\frac{1}{y}, \quad Q(y) = \frac{1 - y^2}{y}$

Step 2: Finding the Integrating Factor
The integrating factor (IF) is given by:
$\text{IF} = e^{\int P(y) \, dy} = e^{\int -\frac{1}{y} \, dy} = e^{-\ln |y|} = \frac{1}{y}$

Step 3: Multiplying by the Integrating Factor
Multiplying the entire equation by the integrating factor:
$\frac{1}{y} \frac{dx}{dy} - \frac{x}{y^2} = \frac{1 - y^2}{y^2}$
This simplifies to:
$\frac{d}{dy} \left( \frac{x}{y} \right) = \frac{1 - y^2}{y^2}$

Step 4: Integrating Both Sides
Integrating both sides with respect to $y$:
$\frac{x}{y} = \int \frac{1 - y^2}{y^2} \, dy = \int \left( \frac{1}{y^2} - 1 \right) \, dy$
$\frac{x}{y} = \int y^{-2} \, dy - \int 1 \, dy = -y^{-1} - y + C$
Multiplying by $y$:
$x = -1 - y^2 + Cy$

Step 5: Applying the Initial Condition
Given $x(1) = 1$:
$1 = -1 - 1^2 + C \times 1$
$C = 3$
Thus, the solution is:
$x = -1 - y^2 + 3y$

Step 6: Evaluating $5x(2)$
Substituting $y = 2$:
$x(2) = -1 - 2^2 + 3 \times 2 = -1 - 4 + 6 = 1$
$5x(2) = 5 \times 1 = 5$
Conclusion: $5x(2) = 5$.