Behaviour of a function in differential equation

Question

I have the following question, which is says that: $$y' - (1/3)y = e^{-t} ,~ y(0) = l ,$$ and it says that $y$ goes to zero as $t$ goes to infinity if $l = -3/4$. How is that?

Answer 1

The following Mathematica code will solve your differential equation problem:

DSolve[{y'[t] - (1/3)*y[t] == Exp[-t], y[0] == l}, y[t], t]

The result is:

{{y[t] -> 1/4 E^-t (-3 + 3 E^(4 t/3) + 4 E^(4 t/3) l)}}

Let's look at the value of the limit of this solution to the differential equation as t goes to infinity with the following code:

Limit[1/4 E^-t (-3 + 3 E^(4 t/3) + 4 E^(4 t/3) l), t -> \[Infinity]]

The result is:

(3 + 4 l) \[Infinity]

Set the above limit equal to zero and solve for l. l=-3/4 satisfies the condition.

The back slashes refer to multiplication in the symbolic language when it is copied and pasted into this window.

Answer 2

This is a first-order linear ordinary differential equation in the form $$ y'(t)+P(t)y(t)=Q(t). $$ In our ODE, we have $P(t)=-\dfrac{1}{3}$ and $Q(t)=e^{-t}$. Multiply both sides by $$ e^{\int P(t)\ dx}=e^{-\int \frac{1}{3}\ dt}=\text{C}e^{-\frac{1}{3}t}, $$ yields $$ \begin{align} y'\text{C}e^{-\frac{1}{3}t}-\frac{1}{3}\text{C}e^{-\frac{1}{3}t}y&=\text{C}e^{-\frac{4}{3}t}\\ \frac{d}{dt}\left(e^{-\frac{1}{3}t}y\right)&=e^{-\frac{4}{3}t}\\ e^{-\frac{1}{3}t}y&=\int e^{-\frac{4}{3}t}\ dt\\ &=-\frac{3}{4}e^{-\frac{4}{3}t}+\text{K}\\ y(t)&=-\frac{3}{4}e^{-t}+\text{K}e^{\frac{1}{3}t}. \end{align} $$ As $t\to\infty$, $y\to0$ then $\text{K}=0$. Thus, the solution of the ODE is $$ \boxed{\Large\color{blue}{y(t)=-\frac{3}{4}e^{-t}}} $$ As $t=0$, then $y(0)=-\dfrac{3}{4}$.

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$$\large\color{blue}{\text{# }\mathbb{Q.E.D.}\text{ #}}$$