- #1
dwilmer
- 11
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Homework Statement
Solve the differential equation: y' + 2y = te^-2t with initial conditions y(1) = 0?
Please show me what I am doing wrong... (or right).. also please don't show me shortcut i need to know where I am going wrong thanks.
Homework Equations
The Attempt at a Solution
first i multiplied whole equation by integrating factor e^2t (e^integral(2) = e^2t)
after rearranging this gave me
e^2t(y) = integ te^-2t * e^2t
i then simplified RHS to get:
e^2t(y) = integ t
then i integrated RHS to get
e^2t(y) = (t^2)/2 + c
then i isolated y to get:
y = ((t^2)e^-2t) / 2 + ce^-2t
then i applied the initial condition y(1) = 0
this gave me
0 = e^-2 / 2 + ce^-2
then i multiplied equation by 2
0 = e^-2 + ce^-2
then i factored out e^-2
0 = e^-2 (1 + c)
then i reasoned that c must = 1 and put this back into the explicit equation
y = ((t^2)e^-2t) /2 + (1) e^-2t
is this correct? the book answer says y = (t^2 -1) e^-2t /2
THANKS FOR ANY HELP