Question

A solution of IVP [1-t^(2)]d^2y/dt^2+2tdy/dt-2y=0,y(0)=3,y'(0)=-4 is y_1=t.Use the method of reduction of order to find a general solution of IVP on interval -1<t<1?

Answer 1

`c_1t+c_2t(sqrt(1-t^2)+1/2ln((sqrt(1-t^2)+1)/|t|))`

Explanation:

The differential equation is of the form
`(1-t^2)y''+2ty'-2y=0`

Since we want the general solution, we can ignore the initial conditions.

It is given to us that one of the solutions to the differential equation is `y_1=t`.

Now, using reduction of order, we assume that there is another solution `y_2=tv(t)`, where `v(t)` is some function of `t`. The first and second derivatives are, respectively, `y_2'=tv'(t)+v(t)` and `y_2''=tv''(t)+2v'(t)`

Substitute this new solution into the differential equation:
`(1-t^2)y_2''+2ty_2'-2y_2=0`
`(1-t^2)(tv''(t)+2v'(t))+2t(tv'(t)+v(t))-2tv(t)=0`

Now, the rationale of the reduction of order is that this solution will simplify to a first-order equation:
`(1-t^2)tv''(t)+2v'(t)=0`

This indeed does by substituting `w(t)=v'(t)` and `w'(t)=v''(t)`:
`(1-t^2)tw'(t)+2w(t)=0`

Now, this is just a separable equation that can be solved easily to find that
`w(t)=sqrt(1-t^2)/t` (Note that the domain is set in the problem as `-1 < t < 1`)

Then,
`v(t)=int\ w(t)\ dt=sqrt(1-t^2)+1/2ln((sqrt(1-t^2)+1)/|t|)+C` (Again noting the domain of `t`)

Thus,
`y_2=tv(t)=tsqrt(1-t^2)+t/2ln((sqrt(1-t^2)+1)/|t|)+Ct`

And the general solution is thus
`c_1y_1+c_2y_2`

`=c_1t+c_2t(sqrt(1-t^2)+1/2ln((sqrt(1-t^2)+1)/|t|))`