Question

Let $y = y(t)$ be a solution to the differential equation $y' + 2ty = t^2$, then $16 \lim_{t\to\infty} \frac{y}{t}$ is:

• A. 8
• B. 16
• C. 24
• D. 2

Answer 1

Answer: (A)

8

Answer 2

The given differential equation is a first-order linear differential equation of the form $y' + P(t)y = Q(t)$, where $P(t) = 2t$ and $Q(t) = t^2$.

Step 1: Find the integrating factor (IF). The integrating factor is given by $IF = e^{\int P(t) dt}$. $IF = e^{\int 2t dt} = e^{t^2}$.

Step 2: Write the general solution. The general solution for a linear differential equation is $y \cdot IF = \int Q(t) \cdot IF dt + C$. Substituting the values: $y e^{t^2} = \int t^2 e^{t^2} dt + C$ So, $y(t) = e^{-t^2} \int t^2 e^{t^2} dt + C e^{-t^2}$.

Step 3: Evaluate the limit $\lim_{t\to\infty} \frac{y}{t}$. We need to find $\lim_{t\to\infty} \frac{e^{-t^2} \int t^2 e^{t^2} dt + C e^{-t^2}}{t}$. We can split this into two parts: $\lim_{t\to\infty} \frac{e^{-t^2} \int t^2 e^{t^2} dt}{t} + \lim_{t\to\infty} \frac{C e^{-t^2}}{t}$

The second term: $\lim_{t\to\infty} \frac{C e^{-t^2}}{t} = 0$, because $e^{-t^2}$ approaches 0 much faster than $t$ approaches infinity.

For the first term, rewrite it as $\lim_{t\to\infty} \frac{\int t^2 e^{t^2} dt}{t e^{t^2}}$. This limit is of the indeterminate form $\frac{\infty}{\infty}$, so we can apply L'Hopital's Rule.

Let $F(t) = \int t^2 e^{t^2} dt$. By the Fundamental Theorem of Calculus, $F'(t) = t^2 e^{t^2}$. Let $G(t) = t e^{t^2}$. $G'(t) = \frac{d}{dt}(t e^{t^2}) = 1 \cdot e^{t^2} + t \cdot (e^{t^2} \cdot 2t)$ (using product rule and chain rule) $G'(t) = e^{t^2} + 2t^2 e^{t^2} = e^{t^2}(1 + 2t^2)$.

Applying L'Hopital's Rule: $\lim_{t\to\infty} \frac{F'(t)}{G'(t)} = \lim_{t\to\infty} \frac{t^2 e^{t^2}}{e^{t^2}(1 + 2t^2)}$ $= \lim_{t\to\infty} \frac{t^2}{1 + 2t^2}$

To evaluate this limit, divide the numerator and the denominator by the highest power of $t$ in the denominator, which is $t^2$: $= \lim_{t\to\infty} \frac{\frac{t^2}{t^2}}{\frac{1}{t^2} + \frac{2t^2}{t^2}}$ $= \lim_{t\to\infty} \frac{1}{\frac{1}{t^2} + 2}$ As $t \to \infty$, $\frac{1}{t^2} \to 0$. So, the limit is $\frac{1}{0 + 2} = \frac{1}{2}$.

Therefore, $\lim_{t\to\infty} \frac{y}{t} = \frac{1}{2}$.

Step 4: Calculate the final expression. We need to find $16 \lim_{t\to\infty} \frac{y}{t}$. $16 \times \frac{1}{2} = 8$.

The final answer is 8.

Explanation of the solution (minimal) The differential equation $y' + 2ty = t^2$ is a first-order linear DE. The integrating factor is $IF = e^{\int 2t dt} = e^{t^2}$. Multiplying by IF, we get $\frac{d}{dt}(y e^{t^2}) = t^2 e^{t^2}$. Integrating both sides gives $y e^{t^2} = \int t^2 e^{t^2} dt + C$. So, $y(t) = e^{-t^2} \int t^2 e^{t^2} dt + C e^{-t^2}$. We need to find $\lim_{t\to\infty} \frac{y}{t} = \lim_{t\to\infty} \frac{e^{-t^2} \int t^2 e^{t^2} dt + C e^{-t^2}}{t}$. The term $\lim_{t\to\infty} \frac{C e^{-t^2}}{t} = 0$. For the main term, $\lim_{t\to\infty} \frac{\int t^2 e^{t^2} dt}{t e^{t^2}}$, apply L'Hopital's Rule. Derivative of numerator is $t^2 e^{t^2}$. Derivative of denominator is $e^{t^2}(1 + 2t^2)$. The limit becomes $\lim_{t\to\infty} \frac{t^2 e^{t^2}}{e^{t^2}(1 + 2t^2)} = \lim_{t\to\infty} \frac{t^2}{1 + 2t^2} = \lim_{t\to\infty} \frac{1}{1/t^2 + 2} = \frac{1}{2}$. Thus, $\lim_{t\to\infty} \frac{y}{t} = \frac{1}{2}$. Finally, $16 \lim_{t\to\infty} \frac{y}{t} = 16 \times \frac{1}{2} = 8$.