Question

Solve the initial-value problem

3y′ + (1 + 8x^7)y = (e^(-x^8))/y^2, y(0) = 1.
Note that this is a Bernoulli differential equation.
Write as a function of x.

Answer 1

Explanation:

To solve this initial value problem, we can use the Bernoulli differential equation technique:

Divide both sides of the equation by y^3 to obtain:

3y^(-2) y' + (1 + 8x^7)/y^3 = e^(-x^8)/y^5

Let z = y^(-1), then we can rewrite the equation as:

-3z' + (1 + 8x^7)z = e^(-x^8)

This is now a linear differential equation with an integrating factor of:

μ(x) = e^(∫(1 + 8x^7) dx) = e^(x + 4x^8)

Multiply both sides of the equation by the integrating factor to obtain:

-3(e^(x + 4x^8) z)' = e^(x - x^8)

Integrate both sides with respect to x to get:

-e^(x + 4x^8) z = ∫e^(x - x^8) dx + C

where C is a constant of integration.

Solve for z to get:

z = -1/e^(x + 4x^8) ∫e^(x - x^8) dx - C/e^(x + 4x^8)

Substituting z = y^(-1), we get:

y(x) = (C - ∫e^(x - x^8) dx)/e^(x + 4x^8)

Using the initial condition y(0) = 1, we can solve for C:

y(0) = (C - ∫e^(0 - 0^8) dx)/e^(0 + 4(0^8)) = C/1 = 1

So, C = 1.

Substituting C = 1, we get the solution:

y(x) = (1 - ∫e^(x - x^8) dx)/e^(x + 4x^8)

Therefore, the solution to the initial-value problem is:

y(x) = (1 - ∫e^(x - x^8) dx)/e^(x + 4x^8)

As an explicit function of x, the integral term cannot be simplified any further.