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 is Linux Kernel Developer & SAN Architect and is passionate about competency developments in these areas. He lives in Bangalore and delivers focused training sessions to IT professionals in Linux Kernel, Linux Debugging, Linux Device Drivers, Linux Networking, Linux Storage, Advanced C Programming, SAN Storage Technologies, SCSI Internals & Storage Protocols such as iSCSI & Fiber Channel. Stay connected with him @ LinkedIn | Youtube | Instagram | Facebook | Twitter