Answer:
Part a) value of [tex]\lambda [/tex] such that all the solutions tend to zero equals 1.
Part b)
For a particular solution to tend to 0 will depend on the boundary conditions.
Step-by-step explanation:
The given differential equation is
[tex]y'+\lambda y=e^{-t}[/tex]
This is a linear differential equation of first order of form [tex]\frac{dy}{dt}+P(t)\cdot y=Q(t)[/tex] whose solution is given by
[tex]y\cdot e^{\int P(t)dt}=\int e^{\int P(t)dt}\cdot Q(t)dt[/tex]
Applying values we get
[tex]y\cdot e^{\int \lambda dt}=\int e^{\int \lambda dt}\cdot e^{-t}dt\\\\y\cdot e^{\lambda t}=\int (e^{(\lambda -1)t})dt\\\\y\cdot e^{\lambda t}=\frac{e^{(\lambda -1)t}}{(\lambda -1)}+c\\\\\therefore y(t)=\frac{c_{1}}{\lambda -1}(e^{-t}+c_{2}e^{-\lambda t})[/tex]
here [tex]c_{1},c_{2}[/tex] are arbitrary constants
part 1)
For all the function to approach 0 as t approaches infinity we have
[tex]y(t)=\lim_{t\to \infty }[\frac{c_{1}}{\lambda -1}(e^{-t}+c_{2}e^{-\lambda t})]\\\\y(\infty )=\frac{c_{1}}{\lambda -1}=0\\\\\therefore \lambda =1[/tex]
Part b)
For a particular solution to tend to 0 will depend on the boundary conditions as [tex]c_{1},c_{2}[/tex] are arbitrary constants
