Answer to Question #199551 in Differential Equations for Raj kumar

Question

Find a continuous solution of the Intial Value Problem

dy/dx + y = g(t), y(0) = 0

where g(t) = {2, when 0<=t<=1

and g(t) = {0, when t>1

Answer 1

The given ODE is an inhomogeneous linear ordinary differential equation. We can apply the method of variation of parameters.

At first we solve the homogeneous equation:

dy/dx + y = 0

$dy/y=-dx$

$d\log y=-dx$

$y=ce^{-x}$

If c is an arbitrary constant, the last equation is a general solution of the homogeneous ODE. In order to solve inhomogeneous ODE, consider c as a function on x. Then

$dy/dx + y =(c'e^{-x}-ce^{-x}) + ce^{-x}=c'e^{-x}=g(x)$

$c'=g(x)e^x$

$c(x)=c_0+\int\limits_{1}^{x}g(t)e^tdt$

If $0\leq x\leq1$ then

$c(x)=c_0+\int\limits_{1}^{x}2e^tdt=c_0+2e^x-2e$

$y(x)=c(x)e^{-x}=2+(c_0-2e)e^{-x}$

$y(0)=2+c_0-2e=0$

$c_0=2e-2$

$y(x)=2+(c_0-2e)e^{-x}=2-2e^{-x}$

If x>1 then

$c(x)=c_0+\int\limits_{1}^{x}0e^tdt=2e-2$

$y(x)=c(x)e^{-x}=(2e-2)e^{-x}=2e^{1-x}-2e^{-x}$

Answer.

If $0\leq x\leq1$ then $y(x)=2-2e^{-x}$

If x>1 then $y(x)=2e^{1-x}-2e^{-x}$

It is clear that this function is continuous.