I need help with this problem. Thanks. Solve the differential equation. 1+xy=xy'

Question

How do you do this ,and what is the answer?

Answer 1

Starting from 1 + xy = xy', let's move all terms except the constant to the same side.

xy' -xy = 1

Now divide both sides by the factor of x since it will simply the first order linear differential equation.

y' - y = 1/x

This now places the equstion into the format of y' + p(x)y = q(x). We can identify p(x) = -1 and q(x) = 1/x.

Using the integratingμ factor as defined by...

μ(x) = exp(-integral(p(s)ds)) --> exp(-x)

This term when multiplied by y' and y will allow us to represent the left side as the derivative of the product rule...

μ(x)y' - μ(x)y = μ(x)/x

μ(x)y' + μ'(x)y = μ(x)/x

d/dx [ μ(x)y ] = μ(x)/x

μ(x)y = integral ((μ(x)/x) dx)

y(x) = 1/μ(x) integral ((μ(x)/x) dx)

Substitution of μ(x) with the necessary integration (using integration by parts) will provide the solution to y(x) with a constant to be determined by an initial condition.

The integral to solve represents exp(-x)/x. The easiest way to solve this integral is via power series representation for exp(-x)...

exp(-x) = sum_k=0_infinity (-x)^k/k! ( Note the alternating signs due to odd powers in infinite sum.

The first few factors would be --> 1 - x + x^2/2 - x^3/6 + ...

Dividing by x yields --> 1/x - 1 + x/2 - x^2/6 + ...

Integrating term by terms will yield --> ln(x) - x + x^2/4 - x^3/18 + ...

Now we want to find a pattern to represent the new series...

ln(x) - x( 1 - x/(2*2!) + x^2/(3*3!) - x^3/(4*4!) + ...)

ln(x) -x * sum_k=0_infinity x^k/((k+1)(k+1)!)

The bolded answer is a power series representation of the integral. Now one can divide by μ(x) = exp(-x) or attempt the division of two infinite power series...

y(x) = exp(x)*ln(x) - x*exp(x) * sum_k=0_infinity x^k/((k+1)(k+1)!)