Question

ssolve y"+2y'+1=3te^-t y(0)=4 y'(0)=0

Answer 1

Step-by-step explanation:

I had to go to sleep now so I hope this helps you

Answer 2

The solution is

$= > y' + 3y + - 13= - 3t {e}^{ - t} - {e}^{ - t}$

Step-by-step explanation:

Given :

$= > y''+2y' + 1 = 3t {e}^{ - t}$

• y(0) = 4
• y'(0) = 0

On integrating both side wrt t, we get

$= > \int(y''+2y' + 1 )dt= \int(3t {e}^{ - t} )dt$

$= > \int \: y''dt+ \int \: 2y' dt+ \int1 dt \: = \int(3t {e}^{ - t} )dt$

$= > y' + 2y \: + y + c = 3 \times (t \int{e}^{ - t} - \int \int {e} ^{ - t}dt \: dt )$

$= > y' + 3y + c = - 3t {e}^{ - t} - {e}^{ - t}$

for t=0,

$= > 0 + 3 \times 4 + c = 0 - 1$

$= > c = - 13$

$= > y' + 3y - 13 = - 3t {e}^{ - t} - {e}^{ - t}$