Variation of Parameters for Solving Differential Equations

[ascheras]

I can't seem to properly solve this problem:

ty" - (1+t) y' + y= t^2 e^(2t) , t> 0; ysub1= 1+t

any help would be appreciated

 

[HallsofIvy]

I wonder if you are even clear on what the problem is since you didn't state it completely.

The problem is, I would guess, "Find the general solution to ty" - (1+t) y' + y= t^2 e^(2t) for t> 0, given that y₁= 1+t is a solution to the associated homogeneous differential equation".

You did title this thread "help with reduction method" so you should understand what to do now. Since the equation is a non-homogeneous linear second order d.e., we need to find another solution to the associated homogeneous equation.

Using "reduction of order", we look for a solution of the form
y(t)= (1+t)u(t): the know solution multiplied by an unknown function u. Now y'= (1+t)u'+ u and y"= (1+t)u"+ 2u'. Putting those into the original equation:
ty"= t(1+t)u"+ 2tu'
-(1+t)y'= -(1+t)²u'- (1+t)u
y = (1+t)u
so ty"-(1+t)y'+ y= t(1+t)u"+(2t- 1- 2t- t²)u'=0
= t(1+t)u"-(t²+1)u'= 0
The fact that 1+t was itself a solution to the d.e. meant that while we have u" and u' we do not have u explicitely in the equation. Let
v= u' and the equation is t(1+t)v'- (t²+1)v= 0, a first order equation: we have reduced the order.

In fact, t(1+t)v'- (t²+1)u'= 0 is a separable equation: it can be integrated directly. Then replace v with u' and integrate again. That will give the general solution to the associated homogeneous equation. Finally, you can use "variation of parameters", which is very similar, to find a specific solution to the entire equation.

 

[ascheras]

reply

i'm glad you decided to help, but your belittling was unwelcome. i figured this problem out already using the variation of parameters. thank you anyway. and i did fully understand the problem and it was stated that way. it is understood that y1 is the solution to the homogenous equation. where else would it have come from? i do appreciate your help.