Q: Solve t^2 y''+ty'+y=0 for t>0
A:
Let x=ln(t)
Thus, (d^2 y)/(dt^2 )=1/t^2 ((d^2 y)/(dx^2 ) - dy/dx)
dy/dt=1/t ∙ dy/dx
Plug them into the equation:
t^2 ∙ 1/t^2 ∙ ((d^2 y)/(dx^2 ) - dy/dx) + t ∙ 1/t ∙ dy/dx + y = 0
(d^2 y)/(dx^2 ) - dy/dx + dy/dx + y = 0
(d^2 y)/(dx^2 ) + y = 0
y'' (x) + y = 0
Set y''=r^2 , y=1
r^2 + 1 = 0
r = ±i
Thus, y(x) = c_1 e^0 cos(x) + c_2 e^0 sin(x)
y(x)= c_1 cos(x) + c_2 sin(x)
Substitute x = ln(t) in the solution above:
y(t) = c_1 cos(lnt) + c_2 sin(lnt)
Therefore, the general solution is y(t) = c_1 cos(lnt) + c_2 sin(lnt) for t>0
Clearer answer is shown in the picture below:
