# Please check my math on diff equation with initial conditions question (easy for you)

1. Nov 24, 2009

### dwilmer

1. The problem statement, all variables and given/known data

Solve the differential equation: y' + 2y = te^-2t with initial conditions y(1) = 0?
Please show me what im doing wrong... (or right).. also please dont show me shortcut i need to know where im going wrong thanks.

2. Relevant equations

3. 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
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Nov 24, 2009

### diazona

Re: Please check my math on diff equation with initial conditions question (easy for

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

3. Nov 24, 2009

### HallsofIvy

Staff Emeritus
Re: Please check my math on diff equation with initial conditions question (easy for

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