Partial Differential Equations Questions and Answers – First Order PDE

This set of Partial Differential Equations Multiple Choice Questions & Answers (MCQs) focuses on “First Order PDE”.

1. Find \(\frac{\partial z}{\partial x}\) where \(z=ax^2+2by^2+2bxy\).
a) 3by
b) 2ax
c) 3(ax+by)
d) 2(ax+by)
View Answer

Answer: d
Explanation: Here we use partial differentiation.
\(\frac{\partial z}{\partial x}\)=a(2x)+0+2by
\(\frac{\partial z}{\partial x}\)=2ax+2by
\(\frac{\partial z}{\partial x}\)=2(ax+by).

2. Find \(\frac{\partial z}{\partial x}\) where \(z=sin⁡x^2×cos⁡y^2\).
a) 2xsin⁡x2
b) x sin2x
c) 2xsin⁡x2 cos⁡y2
d) 6xsin⁡x2 cos⁡y2
View Answer

Answer: c
Explanation: Here we use partial differentiation.
\(\frac{\partial z}{\partial x}=(cos⁡x^2×2x)× cos⁡y^2\)
\(\frac{\partial z}{\partial x}=2xsin⁡x^2 cos⁡y^2\).

3. Find \(\frac{\partial u}{\partial x}\) where \(u=cos⁡(\sqrt x+\sqrt y)\).
a) \(\frac{-1}{2\sqrt x}×tan⁡(\sqrt x+\sqrt y)\)
b) \(\frac{-1}{2\sqrt x}×cos⁡(\sqrt x+\sqrt y)\)
c) \(\frac{-1}{2\sqrt x}×sin⁡(\sqrt x+\sqrt y)\)
d) \(\frac{-1}{\sqrt x}×sin⁡(\sqrt x+\sqrt y)\)
View Answer

Answer: c
Explanation: Here we use partial differentiation.
\(u=cos⁡(\sqrt x+\sqrt y)\)
\(\frac{\partial u}{\partial x}= -sin⁡(\sqrt x+\sqrt y)×\frac{1}{2\sqrt x}\)
\(\frac{\partial z}{\partial x}=\frac{-1}{2\sqrt x}×sin⁡(\sqrt x+\sqrt y)\).

4. If \(u=\frac{e^{x+y}}{e^x-e^y}\), what is \(\frac{\partial u}{\partial x}+\frac{\partial u}{\partial y}\)?
a) \(\frac{2((e^x-e^y)×e^{x+y})-(e^{x+y}) (e^x+e^y)}{(e^x-e^y)^2} \)
b) \(\frac{2((e^x-e^y)×e^{x+y})-(e^{x+y}) (e^x+e^y)}{(e^x+e^y)^2} \)
c) \(\frac{2((e^x-e^y)×e^{x+y})-(e^{x+y}) (e^x-e^y)}{(e^x-e^y)^2} \)
d) u
View Answer

Answer: a
Explanation: Here we use partial differentiation.
\(\frac{\partial u}{\partial x}=\frac{(e^x-e^y)×e^{x+y}-(e^{x+y})(e^x)}{(e^x-e^y)^2}\)
\(\frac{\partial u}{\partial y}=\frac{(e^x-e^y)×e^{x+y}-(e^{x+y})(e^y)}{(e^x-e^y)^2}\)
\(\frac{\partial u}{\partial x}+\frac{\partial u}{\partial y}=\frac{(e^x-e^y)×e^{x+y}-(e^{x+y})(e^x)}{(e^x-e^y)^2} + \frac{(e^x-e^y)×e^{x+y}-(e^(x+y))(e^y)}{(e^x-e^y)^2} \)
\(\frac{\partial u}{\partial x}+\frac{\partial u}{\partial y}=\frac{2((e^x-e^y)×e^{x+y})-(e^{x+y}) (e^x+e^y)}{(e^x-e^y)^2}\).

5. 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
View Answer

Answer: b
Explanation: 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,
\(\left ( \frac{n}{t}+\frac{r^2}{2t^2} \right )\theta = \frac{1}{r^2} \frac{\partial}{\partial r}(\frac{-r^3 \theta}{t})\)
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Manish Bhojasia, a technology veteran with 20+ years @ Cisco & Wipro, is Founder and CTO at Sanfoundry. He lives in Bangalore, and focuses on development of Linux Kernel, SAN Technologies, Advanced C, Data Structures & Alogrithms. Stay connected with him at LinkedIn.

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