Given the general solution to #t^2y'' - 4ty' + 4y = 0# is #y= c_1t + c_2t^4#, how do I solve the problem #t^2y'' - 4ty' + 4y = -2t^2 ,   y(1) = 2,    y'(1) =0#?

1 Answer
Feb 2, 2015

The answer is: #y=-t^4+t^2+2t#.

If it is known a solution of the homogenous equation, now a particular solution of the non-homogenous equation has to be find.

Since then in the second member there is #-2t^2#, for the similarity principle the particular solution is a square equation like this:

#beta(t)=at^2+bt+c#.

Now we calculate:

#beta'(t)=2at+b#, and

#beta''(t)=2a#.

Now we have to find #a,b,c#.

#beta(t)# has to satisfy the equation, so:

#t^2(2a)-4t(2at+b)+4at^2+4bt+4c=-2t^2#
#2at^2-8at^2-4bt+4at^2+4bt+4c=-2t^2#
#-2at^2+4c=-2t^2#

Then, for the principle of the identity between polynomials:

#-2a=-2rArra=1#, and
#4c=0rArrc=0#.
#b# is not important, so we can assume that is 0.

So:

#beta(t)=t^2#.

The solution of the non-homogeneous is:

#y=c_1t+c_2t^4+t^2#

and its derivative is:

#y'=c_1+4c_2t^3+2t#

Now we have to find #c_1# and #c_2# using the conditions:

#y(1)=2# and #y'(1)=0#.

#y'=c_1+4c_2t^3+2t#.

So:

#2=c_1+c_2+1rArrc_1+c_2=1# and

#0=c_1+4c_2+2rArrc_1+4c_2=-2#.

We can substitute #c_1=1-c_2# from the first equation to the second:

#1-c_2+4c_2=-2rArr3c_2=-3rArrc_2=-1#

And:

#c_1=1-(-1)rArrc_1=2#.

So, finally:

#y=-t^4+t^2+2t#.