# 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)

Explanation: Here we use partial differentiation.
$$z=ax^2+2by^2+2bxy$$.
$$\frac{\partial z}{\partial x}$$=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

Explanation: Here we use partial differentiation.
z=sin⁡x2×cos⁡y2
$$\frac{\partial z}{\partial x}=(cos⁡x^2×2x)× cos⁡y^2$$
$$\frac{\partial z}{\partial x}=2xsin⁡x^2 cos⁡y^2$$.

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) $$\frac{-1}{\sqrt x}×sin⁡(\sqrt x+\sqrt y)$$

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

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

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

Sanfoundry Global Education & Learning Series – Partial Differential Equations.

To practice all areas of Partial Differential Equations, here is complete set of 1000+ Multiple Choice Questions and Answers.