solve the differential equation

Question

x²y"(x)+xy'(x)-y(x)=0 x is greater than 0; subject to y(1)=1 and y'(1)=1/2

a. solve the d.e
b. determine the values of the parameters from the initial condition

Answer 1

Solving this differential equation amounts to determining if it is analytic in the domain in question. We can attempt to use a power series representation or use Euler's method if it is analytic...

x² y''(x) + x y'(x) - y(x) = 0 = y''(x) + 1/x * y'(x) - 1/x² * y(x)

define P(x) as coefficient of y''(x)

define Q(x) as coefficient of y'(x)

define R(x) as coefficient of y(x)

Analyticity demands that undefined points based on "reduction" have finite values if it is a continuous or regular singular point. Irregular singular points require power series representations with other conditions...

Therefore...

lim (x --> 0) P(x) and lim (x --> 0) xQ(x) and lim (x --> 0) x²R(x) must all be defined...

lim (x --> 0) P(x) = 1

lim (x --> 0) xQ(x) = lim (x --> 0) x * 1/x = 1

lim (x --> 0) x²R(x) = lim (x --> 0) x² * 1/x² = 1

We can use Euler's method. Assume a solution to the problem can be cast as y(x) = x^r

y'(x) = r x^(r-1)

y''(x) = r * (r-1) x^(r-2)

Therefore the differential equation is now...

r * (r-1) x^(r-2) + r x^(r-2) - x^(r-2) = 0

Since we can factor this as --> (r*(r-1) + r - 1) * (x^(r-2)) = 0, we know that x^(r-2) cannot be zero unless x = 0 ( the trivial solution), we are interested in what values of r can also be made to make this differential equation to be zero with our ansatz.

r² - r + r - 1 = 0 --> r² -1 = 0

r = ±1

Therefore, the two independent solutions for y(x) are...

y₁(x) = x and y₂(x) = 1/x

y(x) = C₁y₁(x) + C₂y₂(x) = C₁x + C₂/x

y'(x) = C₁y₁'(x) + C₂y₂'(x) = C1 - C2/x²

Using the initial conditions will allow is to solve for C₁ and C₂.

y(1) = 1 = C1 + C2

y'(1) = 1/2 = C1 - C2

3/2 = 2C₁ --> C₁ = 3/4 --> C₂ = 1/4

Solution --> 3x/4 + 1/(4x)