Question

write a differential equation where all solutions approach y=3 as t->infinity in the form dy/dt=ay+b

Answer 1

\[\frac{ dy }{ y+\frac{ b }{ a } }=a \times dt\]
after integration
\[\ln y+\frac{ b }{ a }=a^2 \times t +C\]

Answer 2

Solve the general differential equation\[\frac{\text{d}y}{\text{d}x}=ay+b\]\[\int\frac{\text{d}y}{ay+b}=\int\text{d}x=x+C\]Substitute \(u=ay+b\), \(\text{d}y=\frac{1}{a}\text{d}u\):\[\frac{1}{a}\int\frac{\text{d}u}{u}=x+C\]\[\ln u=ax+C\]\[u=ke^{ax}\]\[ay+b=ke^{ax}\]\[y=ke^{ax}-\frac{b}{a}\]

Answer 3

Where \(k\) is an arbitrary constant.

Answer 4

Now, if \(y\) becomes constant as \(t\) (sorry for using \(x\) in the previous posts, I'm retarded) approaches infinity, then the non-constant term (\(ke^{at}\)) must vanish (approach zero).

Does that make sense so far?

\(e^{at}\rightarrow 0\) as \(t\rightarrow\infty\) if and only if \(a<0\). Does that make sense?

So the constant term, the only non-vanishing term, must equal 3. So we can pick any negative number for \(a\), and then let \(b=-3a\).

Answer 5

ya so -b/a = 3 then \[\frac{ dt }{ dt }=ay-3a\]