Question

(t^2)y''-4ty'+6y=0 t>0 y1(t)=t^2 How do i find the 2nd solution using the method of reduction of order?

Answer 1

here is a good reference:
http://tutorial.math.lamar.edu/Classes/DE/ReductionofOrder.aspx

Answer 2

\[y_2(t) = vt^{2}\]
\[y'_2(t) = v' t^{2} + 2vt\]
\[y''_2(t) = v''t^{2}+4v' t+2v\]
plug that into diff equ
simplify (a lot of terms cancel)
\[v''t^{4}=0\]
\[v'' = 0\]
\[v = c_1 t\]
so
\[y_2(t) = c_1t^{3}\]
and the general solution is:
\[y(t) = c_1t^{3} +c_2t^{2}\]

Answer 3

yours is \[t^2y"-4ty'+6y =0\] in which, P(t)= 4t. Let \(y_2=Vy_1\) So, the formula for reduction of order is
\[y_1V"+(2y_1'+P(t)y_1)V'=0\]
Plug them in, you have
\[t^2 V"+(2t^2 +4t*t^2)V'=0\\t^2V"+(2t^2 +4t^3)V'=0\]
Let W=V' (*) so, yours is
\[t^2 W' +(2t^2+4t^3)W=0\\ t^2W' = -(2t^2+4t^3)W\\\frac{W'}{W}=-\frac{2t^2+4t^3}{t^2}=-4t-2\]
Now \[\frac{dW}{W}=(-4t-2)dt\]integral both sides, lnW = -2t^2 -2t
\[W = e^{(-2t^2 -2t)}\text{Plug back to W = V' so,} V'=e^{(-2t^2-2t)}\]
Again, take integral both sides to get V, after you have V, plug back to\(y_2 = V y_1\) to get the second solution.
That's what my prof taught me. I don't know why it doesn't look like what dumbcow did.