2.1.2 Find the general solution of y'−2y=t^{2}e^{2t}

  • #1
karush
Gold Member
MHB
3,269
5
Find the general solution of

and use it to determine how solutions behave as


ok presume the first thing to do is to find
 
Last edited:
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  • #2
is the DE
?

if so, then the integrating factor is







 
  • #3
do you mean



if so we have



The integrating factor is given by



 
  • #4
y- 2y? That's equal to -y! Do you mean ?
You say "the first thing to do is find ux". That makes no sense because there is no "u" or "x" in the problem!

I presume you mean you are looking for an "integrating factor". That would be a function, u(t), such that u(t)y'- 2u(t)y= (u(t)y)'. Since (u(t)y)'= u(t)y'+ u'(t)y we must have u'(t)y= -2u(t)y so that u'(t)= -2u(t). Then u'(t)/u(t)= -2. Integrating, ln(u)= -2t+ C and where .

Multiplying both sides of by (since we can use any integrating factor we can take C'= 1) we get .

Integrating both sides so that

.
 
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  • #5
romsek said:
do you mean



if so we have



The integrating factor is given by



yes my bad #2

Screenshot 2021-08-20 2.34.38 PM.png
 

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