Question

Consider the differential equation y'+\lambda y=e^{-t}, when \lambda is some constant.

Answer 1

Answer:

Part a) value of $\lambda$ 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

$y'+\lambda y=e^{-t}$

This is a linear differential equation of first order of form $\frac{dy}{dt}+P(t)\cdot y=Q(t)$ whose solution is given by

$y\cdot e^{\int P(t)dt}=\int e^{\int P(t)dt}\cdot Q(t)dt$

Applying values we get

$y\cdot e^{\int \lambda dt}=\int e^{\int \lambda dt}\cdot e^{-t}dt\\\\y\cdot e^{\lambda t}=\int (e^{(\lambda -1)t})dt\\\\y\cdot e^{\lambda t}=\frac{e^{(\lambda -1)t}}{(\lambda -1)}+c\\\\\therefore y(t)=\frac{c_{1}}{\lambda -1}(e^{-t}+c_{2}e^{-\lambda t})$

here $c_{1},c_{2}$ are arbitrary constants

part 1)

For all the function to approach 0 as t approaches infinity we have

$y(t)=\lim_{t\to \infty }[\frac{c_{1}}{\lambda -1}(e^{-t}+c_{2}e^{-\lambda t})]\\\\y(\infty )=\frac{c_{1}}{\lambda -1}=0\\\\\therefore \lambda =1$

Part b)

For a particular solution to tend to 0 will depend on the boundary conditions as $c_{1},c_{2}$ are arbitrary constants