- #1

dwilmer

- 11

- 0

## 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