# Construct ODE that approaches an asymptote

## 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.

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pasmith
Homework Helper

## 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.

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
Homework Helper
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 $x = y - (2-t)$.

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.