Question

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

Answer 1

Correct Answer: 5

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: \( 5x(2) = 5 \).