Question

Differential Equations: Find the solution for the given initial value problem.
y''+y'-2y=2t y(0)=0 y'(0)=0

Answer 1

I have the general solution as \[y=c _{1}e ^{-2t}+c _{2}e ^{t}\]

Answer 2

Do you have the particular solution?

Answer 3

no thats what i need ot get and im not sure how

Answer 4

Use the method of undetermined coefficients

Answer 5

It'll likely be related to the right side of the equal sign.

Answer 6

i know that im just not sure what to define as \[Y(t)\]

Answer 7

See http://tutorial.math.lamar.edu/Classes/DE/UndeterminedCoefficients.aspx for some good notes and examples.

Answer 8

Try Y=At+B

Answer 9

@CliffSedge yes

Answer 10

The basic method goes:
\[y''+y'-2y = 2t\]
we assume that y = at+b (in this case);
\[0+a-2(at+b) = 2t\]

Answer 11

Find A and B by taking first and second derivatives of At+B and plug into original equation.

Answer 12

well i got
Y(t)= 2At
Y'(t)=2A
Y''(t)=2A?

Answer 13

No

Answer 14

hold on

Answer 15

you have:
\[-2at = 2t\]
and
\[-2b + a = 0\]

Answer 16

you can cancel out the t's in the first equation:
\[-2a = 2\]
then solve for a and b

Answer 17

ok i understand but how did you get -2at=2t and -2b+a=0

Answer 18

Y = At+B
Y' = A
Y'' = 0

Answer 19

so lets go back to the equation we had previously:
\[a -2(at+b) = 2t\]
so:
\[(-2at)+(a-2b) =(2t)+(0)\]

Answer 20

oh i see now ok i got it from here thank you very much

Answer 21

the coefficients of t^1 must equal, and the coefficients of t^0 must equal

Answer 22

you are welcome