Find the general solution of t^2y"-3ty'+4y=0 on (0, infinity sign, given that Y1=t^2 is a solution
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To find the general solution of the differential equation t^2y" - 3ty' + 4y = 0, we can use the method of reduction of order since we already have one solution, Y1 = t^2.
Let's assume the second solution can be written in the form Y2 = v(t)Y1, where v(t) is a function to be determined.
First, let's find the derivatives of Y1:
Y1 = t^2
Y1' = 2t
Y1" = 2
Now we can substitute Y1, Y1', and Y1" into the differential equation:
t^2(2) - 3t(2t) + 4(t^2) = 0
2t^2 - 6t^2 + 4t^2 = 0
0 = 0
Since the equation reduces to 0 = 0, we have confirmed that Y1 = t^2 is indeed a solution.
Now, let's find Y2 by substituting Y1 and its derivatives into Y2 = v(t)Y1:
Y2 = v(t)(t^2)
Y2' = v'(t)(t^2) + v(t)(2t)
Y2" = v''(t)(t^2) + 2v'(t)(t^2) + 2v(t)(2t)
Substitute Y2, Y2', and Y2" into the original differential equation:
t^2(v''(t)(t^2) + 2v'(t)(t^2) + 2v(t)(2t)) - 3t(v'(t)(t^2) + v(t)(2t)) + 4v(t)(t^2) = 0
Simplifying and collecting like terms, we get:
t^4v''(t) + 2t^3v''(t) + 4t^3v'(t) + 4t^3v(t) - 3t^3v'(t) - 6t^2v(t) + 4t^2v(t) = 0
Combine like terms:
t^4v''(t) + 2t^3v''(t) + t^3v'(t) + 2t^3v(t) = 0
Dividing through by t^3, we obtain:
t(v''(t) + 2v'(t) + v(t)) = 0
Now, divide both sides by t:
v''(t) + 2v'(t) + v(t) = 0
This is a second-order linear homogeneous differential equation. Let's solve it using the auxiliary equation method.
Assume a solution in the form of v(t) = e^(rt):
v'(t) = re^(rt)
v''(t) = r^2e^(rt)
Substitute these into the differential equation:
r^2e^(rt) + 2re^(rt) + e^(rt) = 0
Factor out e^(rt):
e^(rt) (r^2 + 2r + 1) = 0
For e^(rt) to be equal to zero, we must have r^2 + 2r + 1 = 0.
The characteristic equation r^2 + 2r + 1 = 0 can be factored as (r + 1)(r + 1) = 0.
The repeated root is r = -1.
Thus, the general solution for v(t) is:
v(t) = C1e^(-t) + C2te^(-t)
Therefore, the general solution for the given differential equation is:
y(t) = t^2(C1e^(-t) + C2te^(-t))
where C1 and C2 are arbitrary constants.