# DE involving limits

## Homework Statement

Find the general solution of the given differential equation, and use it to determine how
solutions behave as t→∞.

a) y' − 2y = t2e2t

DE

## The Attempt at a Solution

After doing the linear DE steps I end up getting y(t)= t3e2t/3 + ce2t. Now I have a problem in getting the limit. The book says the 'It is
evident that all solutions increase at an exponential rate.' Now, my doubt is how would it increase. Or what makes us sure that it will increase? The first part indeed goes to infinity but the second part has a c which can either be positive or negative. So we can end up getting infinity - infinity. Can someone help me?

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UltrafastPED
Gold Member
y(t)= t3e2t/3 + ce2t. Now I have a problem in getting the limit. The book says the 'It is evident that all solutions increase at an exponential rate.'
Assuming that you have found the general solution, note that both terms have exponentials with positive exponents for t>0 - the exponential dominates the t^3, so the solution increases exponentially as t-> infinity.

Excuse me sir, but I found this an irrelevant reply. I kept on mentioning the constant C. With examples, consider C to be -5 the limit would then be NEGATIVE INFINITY as t gets larger. But in cases such as C being 6 the answer goes to POSITIVE INFINITY. WHy is that the book mentions only the shooting UP!

UltrafastPED
Gold Member
y(t)= t3e2t/3 + ce2t.
Spend some time looking at the expression that you have ...

y(t)=(t^3/3 +C)exp(2t)

Does the value of C matter at all?

Well, if C is NEGATIVE, then yes it will matter. The first half definitely goes to positive infinity. But the second half would go to Negative inifinity since exp(2t) shoots up, this negative C can simply reflect this on the x axis and it is negative infinity???

UltrafastPED
Gold Member
What is the limit as t-> infinity for this expression: (t^3/3 +C)?

In this case it is infinity because C comes after a plus sign unsupported with any exponential or something. What would be the limit then for the expression (t^3/3 + Ct^3/3). Consider the case where C is negative.

UltrafastPED
Gold Member
Your assertion is incorrect - are you clear how limits work?

Ray Vickson
Homework Helper
Dearly Missed
Excuse me sir, but I found this an irrelevant reply. I kept on mentioning the constant C. With examples, consider C to be -5 the limit would then be NEGATIVE INFINITY as t gets larger. But in cases such as C being 6 the answer goes to POSITIVE INFINITY. WHy is that the book mentions only the shooting UP!
His answer is not irrelevant; it is 100% accurate and gives you exactly all you need. Perhaps you do not understand the issues, but that is a separate matter, and one that YOU need to address.

For example, suppose C = -5. Then $C + t^3/3 = t^3/3 - 5 \geq 1$ if $t \geq 18^{1/3} \doteq 2.6207$; that is, if $t \geq 2.6207$, $y(t) \geq e^{2t}$. Does this go to $+\infty$ as $t \to + \infty$?

Suppose C = -5,000,000. Then $C + t^3/3 = t^3/3 - 5000000 \geq 1$ if $t \geq 100 \times 15^{1/3} \doteq 246.6212$. So, if $t \geq 246.6212$, $y(t) \geq e^{2t}$. Does this go to $+\infty$ as $t \to + \infty$?

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