Construct ODE that approaches an asymptote

[usn7564]

Homework Statement

Construct a first order linear differential equation whose solutions have the required behavior as t approaches infinity. Then solve your equation and confirm that the solutions do indeed have the specified property.

All solutions are asymptotic to the line y = 2 - t as t approaches infinity.



I don't even know where to begin, could someone give me a kick in the right direction? I tried just writing a linear differential equations with two unknown functions f and g (y' + fy = g) and getting the general solution but obviously that involves a completely unknown integral and I doubt it would lead anywhere.

 

[pasmith]

Homework Statement

Construct a first order linear differential equation whose solutions have the required behavior as t approaches infinity. Then solve your equation and confirm that the solutions do indeed have the specified property.

All solutions are asymptotic to the line y = 2 - t as t approaches infinity.

Let x(t) = y(t) - (2-t).

Can you think of a linear first-order ODE with constant coefficients for which all solutions tend to zero as t \to \infty?

Now let x satisfy that ODE.

 

[usn7564]

Thanks, could use a few more pointers though. So I put x(t) = y(t) - (2-t) where x approaches zero as t approaches infinity which makes sense as the equation describes what y(t) I want.
Can't see how I relate it to the ODE though. Say I find one where all solutions tend to zero (y' + y = 0) for example, how do I connect it to x(t)?

 

[pasmith]

Thanks, could use a few more pointers though. So I put x(t) = y(t) - (2-t) where x approaches zero as t approaches infinity which makes sense as the equation describes what y(t) I want.
Can't see how I relate it to the ODE though. Say I find one where all solutions tend to zero (y' + y = 0) for example, how do I connect it to x(t)?

Take x' + x = 0 and substitute x = y - (2-t).

 

[usn7564]

Take x' + x = 0 and substitute x = y - (2-t).

Understand now, thank you. No idea why it was so tricky to wrap my head around it now, but there you go.