Question

The differential equation t^2y'' - t(t+2)y' + (t + 2)y = 0 has y1 = t as a solution. Applying reduction of order we set y2 = v y1 = v · t. Then (using the prime notation for the derivatives) y2' = y2'' = So, substituting y2 and its derivatives into the left side of the differential equation, and reducing, we get t^2y2'' - t(t+2)y2' + (t+2)y2 = The reduced form has a common factor of t^3 which we can divide out of the equation. Since this equation does not have any v terms in it we can make the substitution u = v' giving us the first order linear equation in u: = 0. If we use c as the constant of integration, the solution to this equation is u = Integrating to get v, and then finding y2 gives the general solution: c1y1 + c2y2 =

          The differential equation
t^2y'' - t(t+2)y' + (t + 2)y = 0
has y1 = t as a solution.
Applying reduction of order we set y2 = v y1 = v · t.
Then (using the prime notation for the derivatives)
y2' =
y2'' =
So, substituting y2 and its derivatives into the left side of the differential equation, and reducing, we get
t^2y2'' - t(t+2)y2' + (t+2)y2 =
The reduced form has a common factor of t^3 which we can divide out of the equation.
Since this equation does not have any v terms in it we can make the substitution u = v' giving us the first order linear equation in u:
= 0.
If we use c as the constant of integration, the solution to this equation is u =
Integrating to get v, and then finding y2 gives the general solution:
c1y1 + c2y2 =

The differential equation
t^2y” - t(t+2)y' + (t + 2)y = 0
has y1 = t as a solution.
Applying reduction of order we set y2 = v y1 = v · t.
Then (using the prime notation for the derivatives)
y2' =
y2” =
So, substituting y2 and its derivatives into the left side of the differential equation, and reducing, we get
t^2y2” - t(t+2)y2' + (t+2)y2 =
The reduced form has a common factor of t^3 which we can divide out of the equation.
Since this equation does not have any v terms in it we can make the substitution u = v' giving us the first order linear equation in u:
= 0.
If we use c as the constant of integration, the solution to this equation is u =
Integrating to get v, and then finding y2 gives the general solution:
c1y1 + c2y2 =

Added by Aaron W.

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