Question

1. which of the following is the solution to the differential equation \\(\frac{dy}{dx} = y^2 - (xy)^2\\) with the initial condition \\(y(3) = 1\\)?

a. \\(y = \frac{5}{6} - \frac{3}{3x - x^3}\\)
b. \\(y = -\frac{3}{3x - x^3 + 15}\\)
c. \\(y = (3x - x^3 + 19)^{\frac{1}{3}}\\)
d. \\(y = \left(e^{x - \frac{1}{3}x^3 + 6}\
ight)^{\frac{1}{2}}\\)

Explanation:

Step1: Simplify the differential equation

The given differential equation is \(\frac{dy}{dx}=y^{2}-(xy)^{2}\). Factor the right - hand side: \(\frac{dy}{dx}=y^{2}(1 - x^{2})\). This is a separable differential equation. We can rewrite it as \(\frac{dy}{y^{2}}=(1 - x^{2})dx\) (assuming \(y
eq0\)).

Step2: Integrate both sides

Integrate the left - hand side with respect to \(y\) and the right - hand side with respect to \(x\).
For the left - hand side: \(\int\frac{dy}{y^{2}}=\int y^{-2}dy=\frac{y^{-2 + 1}}{-2+1}+C_1=-\frac{1}{y}+C_1\)
For the right - hand side: \(\int(1 - x^{2})dx=\int1dx-\int x^{2}dx=x-\frac{x^{3}}{3}+C_2\)

Combining the two integrals (we can combine the constants \(C = C_2 - C_1\)): \(-\frac{1}{y}=x-\frac{x^{3}}{3}+C\)

Step3: Use the initial condition \(y(3) = 1\)

Substitute \(x = 3\) and \(y = 1\) into the equation \(-\frac{1}{y}=x-\frac{x^{3}}{3}+C\).
When \(x = 3\) and \(y = 1\), we have \(-\frac{1}{1}=3-\frac{3^{3}}{3}+C\)
Simplify the right - hand side: \(3 - 9+C=-6 + C\)
So, \(-1=-6 + C\), which gives \(C = 5\)

Step4: Solve for \(y\)

From \(-\frac{1}{y}=x-\frac{x^{3}}{3}+5\), we can rewrite it as \(\frac{1}{y}=-x+\frac{x^{3}}{3}-5=\frac{x^{3}-3x - 15}{3}\)
Then \(y=\frac{3}{x^{3}-3x - 15}=-\frac{3}{3x - x^{3}+15}\) (by factoring out a negative sign from the denominator)

Answer:

B. \(y = -\frac{3}{3x - x^{3}+15}\)