Topic: Engineering Mathematics
The value of y as t→∞ for an initial value of y(1)=0 for the differential equation $(4t^{2}+l)\frac{dy}{dt}+8yt-t=0,$ is
- A. 1
- B. 1/2
- C. 1/4
- D. 1/8
Answer: D. 1/8
Explanation: A linear differential equation is of the form $\frac{dy}{dt}+Py=Q$, where P and Q are functions of t. Arranging the given equation in this form gives $\frac{dy}{dt}+\frac{8t}{4t^{2}+1}y=\frac{t}{4t^{2}+1}$. The solution is $y(e^{\int Pdt}) = \int Qe^{\int Pdt}dt + C$. The integrating factor is $e^{\int Pdt} = e^{\int\frac{8t}{4t^{2}+1}dt} = e^{\ln(4t^{2}+1)} = 4t^{2}+1$. Substituting this into the solution equation yields $y(4t^{2}+1) = \int\frac{t}{4t^{2}+1}(4t^{2}+1)dt + C$, which simplifies to $y(4t^{2}+1) = \frac{t^{2}}{2} + C$. Given that at t=1, y=0, we find $C = -\frac{1}{2}$. The equation becomes $y(4t^{2}+1) = \frac{t^{2}}{2} - \frac{1}{2}$. Dividing by $t^{2}$ gives $4y+\frac{y}{t^{2}} = \frac{1}{2}-\frac{1}{2t^{2}}$. For $t \rightarrow \infty$, $\frac{1}{t^{2}} \rightarrow 0$, leaving $4y = \frac{1}{2}$, which results in $y = \frac{1}{8}$.