Question

Solve the given initial-value problem. The differential equation is a Bernoulli equation. (x² racdydx - 2xy = 6y⁴), (y(1) = 1/3)

Answer 1

The solution to the given initial-value problem,
`\(x^2 (dy)/(dx) - 2xy = 6y^4\) with \(y(1) = (1)/(3)\), is \(y = (1)/(√(3x^2 + 2x))\).`

Step-by-step explanation:

To solve this Bernoulli equation, we'll first transform it to a linear differential equation through substitution. Let
`\(u = y^(-3)\), then \((du)/(dx) = -3y^(-4) (dy)/(dx)\)`. Rearranging the original equation in terms of
`\(u\) yields \(x^2 (du)/(dx) + 2u = -18\).`

This is now a linear first-order differential equation, solvable using an integrating factor
`\(\mu(x) = e^{\int{(2)/(x^2)dx}} = e^(-2/x)\).` Multiplying both sides of the equation by
`\(\mu(x)\) leads to \(e^(-2/x)x^2 (du)/(dx) + 2e^(-2/x)u = -18e^(-2/x)\).`

Recognizing the left-hand side as the derivative of
`\((u \cdot e^(-2/x)x^2)\)` with respect to x, we integrate both sides to get
`\(u \cdot e^(-2/x)x^2 = C - 18 \int{e^(-2/x)dx}\).` Solving the integral and substituting
`\(u = y^(-3)\) yields \(y = \frac{1}{\sqrt[3]{C - 18\left(-(x)/(2)\right)e^(2/x)}} = (1)/(√(3x^2 + 2x))\).`

Applying the initial condition
`\(y(1) = (1)/(3)\)` allows us to determine the value of
`\(C\) as \(C = (1)/(3)\)` and results in the final solution
`\(y = (1)/(√(3x^2 + 2x))\).`