Answer:
[tex]y=-e^{-t}(t+1)+1+\frac{2}{e}[/tex]
Step-by-step explanation:
The given differential equation is
[tex]y'(t)=te^{-t}[/tex]
It can be written as
[tex]\frac{dy}{dt}=te^{-t}[/tex]
[tex]dy=te^{-t}dt[/tex]
Integrate both sides.
[tex]\int dy=\int te^{-t}dt[/tex]
Apply ILATE rule on right side. Here, t is first function and [tex]e^{-t}[/tex] is the second function.
[tex]y=t\int e^{-t}-\int (\frac{d}{dt}t\int e^{-t})[/tex]
[tex]y=-te^{-t}-\int (1\times (-e^{-t}))[/tex] [tex]\int e^{-x}=-e^{-x}+C[/tex]
[tex]y=-te^{-t}+\int e^{-t}[/tex]
[tex]y=-te^{-t}-e^{-t}+C[/tex] .... (1)
Initial condition is y(1) = 1. It means at t=1 the value of y is 1.
[tex]1=-(1)e^{-t}-e^{-(1)}+C[/tex]
[tex]1=-e^{-1}-e^{-1}+C[/tex]
[tex]1=-2e^{-1}+C[/tex]
[tex]1=-\frac{2}{e}+C[/tex]
Add [tex]\frac{2}{e}[/tex] on both sides.
[tex]1+\frac{2}{e}=C[/tex]
Substitute the value of C in equation (1).
[tex]y=-te^{-t}-e^{-t}+1+\frac{2}{e}[/tex]
[tex]y=-e^{-t}(t+1)+1+\frac{2}{e}[/tex]
Therefore, the solution of given initial value problem is [tex]y=-e^{-t}(t+1)+1+\frac{2}{e}[/tex].
