Use the change of variable x = exp(s) to recast the differential equation
xy′′+y′+(L/x)y=0
Please HELP ME
To recast the given differential equation using the change of variable x = exp(s), we can start by expressing the derivatives of y with respect to x using the chain rule:
dy/dx = dy/ds * ds/dx = dy/ds * 1/(dx/ds) = dy/ds * 1/(d/ds(exp(s))) = dy/ds * 1/exp(s)
d^2y/dx^2 = d/dx(dy/dx) = d/ds(dy/dx) * ds/dx = d/ds(dy/dx) * 1/(dx/ds) = d/ds(dy/dx) * 1/(d/ds(exp(s))) = d/ds(dy/dx) * 1/exp(s)
Now let's substitute back into the differential equation:
x * d^2y/dx^2 + dy/dx + (L/x) * y = 0
Substituting the expressions for dy/dx and d^2y/dx^2:
exp(s) * (d/ds(dy/dx) * 1/exp(s)) + dy/ds * 1/exp(s) + (L/exp(s)) * y = 0
Simplifying:
d/ds(dy/dx) + dy/ds + (L/exp(s)) * y = 0
Now, we can rearrange the equation and factor out dy/ds to get the modified differential equation:
dy/ds * (1 + exp(s)/L) + (L/exp(s)) * y = 0
This is the recast form of the differential equation using the change of variable x = exp(s).