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

To solve the differential equation \(\frac{dx}{dy} = \frac{1 + x - y^2}{y}\) with the initial condition \(x(1) = 1\) and to find \(5x(2)\), we use the method of separation of variables and integrating factors.

First, rearrange the equation:
\[\frac{dx}{dy} - \frac{x}{y} = \frac{1 - y^2}{y}\]
This is a linear first-order 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 \(\mu(y)\) is given by:
\[\mu(y) = e^{\int -\frac{1}{y} \, dy} = e^{-\ln|y|} = \frac{1}{y}\]
Multiplying through by the integrating factor:
\[\frac{1}{y}\frac{dx}{dy} - \frac{x}{y^2} = \frac{1 - y^2}{y^2}\]
Recognize the left side as a derivative:
\[\frac{d}{dy}\left(\frac{x}{y}\right) = \frac{1}{y^2} - 1\]
Integrate both sides with respect to \(y\):
\[\int \frac{d}{dy}\left(\frac{x}{y}\right) dy = \int \left(\frac{1}{y^2} - 1\right) dy\]
\[\frac{x}{y} = \int \frac{1}{y^2} \, dy - \int 1 \, dy = -\frac{1}{y} - y + C\]
Multiply through by \(y\):
\[x = -1 - y^2 + Cy\]
Use the initial condition \(x(1) = 1\) to find \(C\):
\[1 = -1 - 1 + C(1)\]
\[C = 3\]
So the solution is:
\[x = -1 - y^2 + 3y\]
Evaluate \(x(2)\):
\[x(2) = -1 - 4 + 6 = 1\]
Finally, compute \(5x(2)\):
\[5x(2) = 5 \times 1 = 5\]
This result is within the expected range of [5, 5].

Answer 2

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