Finding Wronskian of Diff. Eq. with Abel's Formula: Help Needed

[physics=world]

1. Use Abel's Formula to find the Wronskian of two solutions of the given differential equation
without solving the equation.

x²y" - x(x+2)y' + (t + 2)y = 0

2.

Abel's Formula

W(y₁, y₂)(x) = ce^-∫p(x)dx3.

I put it in the form of

y" + p(x)y' + q(x)y = 0

to find my p(x) to use for Abel's formulap(x) = - (x+2 / x)this would give:

W(y₁, y₂)(x) = ce^{-∫(-)(x+2 / x)dx}I'm not sure if I'm going in the right direction. I need some help.

 

[Mark44]

1. Use Abel's Formula to find the Wronskian of two solutions of the given differential equation
without solving the equation.

x²y" - x(x+2)y' + (t + 2)y = 0

2.

Abel's Formula

W(y₁, y₂)(x) = ce^-∫p(x)dx





3.

I put it in the form of

y" + p(x)y' + q(x)y = 0

to find my p(x) to use for Abel's formula


p(x) = - (x+2 / x)

Your use of parentheses is commendable, although you have them in the wrong place here and below.
You should have p(x) = -(x + 2)/x

What you wrote is the same as $-(x + \frac 2 x)$.

this would give:

W(y₁, y₂)(x) = ce^{-∫(-)(x+2 / x)dx}


I'm not sure if I'm going in the right direction. I need some help.

This seems OK to me (aside from the parentheses thing). Are you having trouble with the integration?

 

[physics=world]

After integrating I get:

ce^{(x) + 2ln(x)}

Is this the answer? What about the value for c?

 

[Mark44]

After integrating I get:

ce^{(x) + 2ln(x)}

You should simplify that.

Is this the answer? What about the value for c?

I believe that c is W(y₁(x), y₂(x))(x₀). IOW, the Wronskian of the two functions, evaluated at some initial x value.