Question

If \(\theta=t^n e^\frac{-r^2}{2t}\), find the value of n that satisfies the equation, \(\frac{\partial \theta}{\partial t}=\frac{1}{r^2}\frac{\partial}{\partial r}(r^2 \frac{\partial \theta}{\partial r})\).

(a) 0

(b) -1

(c) 1

(d) 3

This question was addressed to me during an interview.

My question is taken from First Order PDE in section Partial Differential Equations of Engineering Mathematics

Answer 1

The correct answer is (b) -1

The explanation is: Here we use partial differentiation.

\(\frac{\partial \theta}{\partial t}=nt^{n-1}×e^\frac{-r^2}{2t}+t^n×e^\frac{-r^2}{2t}×\frac{r^2}{2t^2}\)

\(\frac{\partial \theta}{\partial t}=nt^{n-1}×e^\frac{-r^2}{2t}+t^n×e^\frac{-r^2}{2t}×\frac{r^2}{2t^2}\)

Substituting from the question

\(\frac{\partial \theta}{\partial t}=\frac{n\theta}{t}+\frac{r^2 \theta}{2t^2}\)

\(\frac{\partial \theta}{\partial t}=(\frac{n}{t}+\frac{r^2}{2t^2})\theta \)

\(\frac{\partial \theta}{\partial r}=t^n×e^\frac{-r^2}{2t}×\frac{-r}{t}\)

Substituting from the question

\(\frac{\partial \theta}{\partial r}=\frac{-r\theta}{t}\)

\(r^2 \frac{\partial \theta}{\partial r}=\frac{-r^3 \theta}{t}\)

Now substituting the values of \(\frac{\partial \theta}{\partial r}\) and \(\frac{\partial \theta}{\partial t}\) in the original equation,

\(\left ( \frac{n}{t}+\frac{r^2}{2t^2} \right )\theta = \frac{1}{r^2} \frac{\partial}{\partial r}(\frac{-r^3 \theta}{t})\)

\(\frac{n}{t}=\frac{-1}{t}\)

n=-1.