Determine the general solution to

2t²y'' + ty' - 3y = 0

It given that y (t) = t -1 is a solution.

 Solution

Reduction of order needs that a solution already be identified.  Without this identified solution we won't be capable to do reduction of order.

Once we have this first solution we will after that assumes a second solution will have the form as

y₂ (t) = v (t ) y1 (t )   ..................(1)

 For a suitable choice of v(t). To find out the good choice, we plug the guess in the differential equation and find a new differential equation which can be solved for v(t).

Therefore, let's do that for this problem.  Now there is the form of the second solution as well as the derivatives that we'll require.

y₂ (t) = t^-1 v,      y₂'(t) = -t² v + t^-1 v',      y₂''(t) = 2t^-3 v -2t^-2 v' + t^-1 v''

Plugging these in the differential equation provides,

2t² (2t ^-3v - 2t ^-2v′ + t ^-1v′′)+ t(-t^-2v + t^-1v′) - 3(t^-1v) = 0

Rearranging and simplifying gives

2tv′′ + ( -4 + 1) v′ + (4t^-1 - t^-1 - 3t^-1 ) v = 0

2tv′′ - 3v′ = 0

Remember that upon simplifying the simple terms remaining are those including the derivatives of v. The term including v drops out. If you've done all of your work properly this should always occur. Sometimes, as in the repeated roots case, the first derivative term will as well drop out.

Therefore, in order for (1) to be a solution after that v must satisfy,

2tv'' - 3v' = 0  .............................(2)

It appears to be a problem. So as to find a solution to a second order non-constant, coefficient differential equation we have to to solve a different second order non-constant coefficient differential equation.

Though, this isn't the problem that this appears to be. Since the term including the v drops out we can in fact solve (2) and we can do this with the knowledge which we already have at this point. We will solve it by making the subsequent change of variable.

 w = v′ ⇒         w′ = v′′

Along with this change of variable (2) becomes

 2tw′ - 3w = 0

And it is a linear; first order differential equation which we can solve. This also illustrates the name of this method. We've managed to decrease a second order differential equation down to a first order differential equation.

This is a quite simple first order differential equation thus I'll leave the details of the solving to you. If you require a refresher on solving linear, first order differential equations return to the second section and check out such section. The solution to this differential equation is,

w(t) = ct^3/2

Here, this is not fairly what we were after.  We are after a solution to (2).  Though, we can now get this.  Recall our change of variable.

v′ = w

With that we can simply solve for v(t).

v(t) = ∫w dt = ∫ ct^3/2 dt = 2/5  ct^5/2+ k

It is the most general possible v(t) which we can use to find a second solution. Therefore, just as we did in the repeated roots section, we can select the constants to be anything we want so select them to clear out all the extraneous constants. Under this case we can utilize

 c = 5/2, k = 0,

By using these gives the subsequent for v(t) and for the second solution.

v(t) = t^5/2 ⇒ y₂(t) = t^-1 (t^5/2) = t^3/2

After that general solution will be,

y(t) = c₁t^-1 +  c₂t^3/2

If we had been specified initial conditions we could after that differentiate, apply the initial conditions and resolve for the constants.

Reduction of order, the method utilized in the previous illustration can be used to get second solutions to differential equations. Though, this does need that we already have a solution and frequently finding that first solution is a very tough task and frequently in the process of finding the first solution you will also find the second solution without needing to resort to reduction of order.  Therefore, for those cases while we do have a first solution it is a nice method for finding a second solution.