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?

1 Answer
NickTheTurtle
Jul 6, 2018

#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|))#

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