Solve Diff. Eq. w/ Initial Conditions: y' + 2y = te^-2t | Homework Help

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SUMMARY

The differential equation y' + 2y = te^-2t with initial condition y(1) = 0 was analyzed, revealing a mistake in the calculation of the constant c. The correct integration factor is e^2t, leading to the equation e^2t(y) = (t^2)/2 + c. Upon applying the initial condition, the correct value of c is determined to be -1, resulting in the solution y = (t^2 - 1)e^-2t / 2, which matches the book's answer.

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dwilmer
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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
 
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Look for a couple of arithmetic mistakes after you used the initial condition y(1)=0.
 


dwilmer said:

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
Then you reasoned wrong. c is not equal to 1.

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
 

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