Variation of Parameters (Diffy Equ.)

[EV33]

Homework Statement

t²y"-t(t+2)y'+(t+2)y= 2t³

y1(t)=t

y2(t)=te^t

t>0

Homework Equations

w(t)=y1*y2' - y1*y2
g=2t
y=-y1∫(gy2)/w + y2∫(gy1)/w

The Attempt at a Solution

y1=t
y1'=1
y2=te^t
y2'=e^(t)+ te^(t)

w(t)=te^(t)+t²e^(t)-te^(t)=t²e(t)

∫(gy2)/w=∫(2t²e^(t))/(t²e^(t))=∫ 2dt=2t+c1

∫(gy1)/w=∫(2t²)/(e^(t)t²)=∫2e^(-t) dt=-2e^(-t) + c2

my final answer:
y=-2t²-tc1-2t+te^(t)c2


The book some how got rid of the C's...
The answer in the book is:

Y(t)= -2t²

I have no idea how to solve for the c1 and c2 without initial conditions.

 

[HallsofIvy]

You can't without some kind of additional conditions, either "initial values" or "boundary values", you cannot determine the constants. Reread the exercise and see if additional conditions are not given somewhere.

 

[EV33]

There is no additional information, but I have found on the oher problems that if I let c1 and c2=0 then I get the correct answer. On this problem though I have a correct solution when I drop the c's, but unlike the rest of the problems I don't get the exact answer in the book.

The solutions manual actually gets to the point where I ended...

-2t²-2t

but it says" However since t is a solution of the homogeneous D.E. we can choose as our particular solution Y(t)= -2t²


I am not too sure what they mean by this,If this helps anyone understand how they arrived at that answer, I would greatly appreciate an explanation.

 

[HallsofIvy]

You did not state the problem correctly then. "$-2t^2$" is NOT the general solution to the equation, nor is it a solution to the equation with specific additional conditions. It is a particular solution to the entire equation- which you would then add to the general solution to the associated homogeneous equation to get the general solution to the entire equation.

Yes, if the general solution to the associated homogeneous equation is "$C_1e^{ax}+ C_2e^{bx}$" and a particular solution is $y_p(x)$, then the general solution to the entire equation is $C_1e^{ax}+ C_2e^{bx}+ y_p(x)$ and, obviously, taking $C_1$ and $C_2$ equal to 0 gives back $y_p(x)$. IF part of the $y_p(x)$ that you get happens to be one of the solutions to the homogeneous equation, then you can add that to the homogeneous equation just changing the constant multiplying it.

Here two independent solution to the associated homogeneous equation are t and $te^t$ so the general solution to the homogeneous equation is $y= C_1t+ C_2te^t$.

You say you got, as specific solution to the entire equation $y_p= -2t^2- 2t$. Okay, that means your general solution to the entire equation can be written as $y(x)= C_1t+ C_2te^t- 2t^2- 2t$. But that is the same as $y(x)= (C_1- 2)t+ C_2te^t- 2t^2$ and that could be rewritten as $y(x)= B_1t+ B_2te^t- 2t^2$ where $B_1= C_1- 2$ and $B_2= C_2$.