250+ TOP MCQs on First Order PDE and Answers

Partial Differential Equations Multiple Choice Questions 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.
(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
View Answer

Answer: c
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}{2sqrt x}×tan⁡(sqrt x+sqrt y))
b) (frac{-1}{2sqrt x}×cos⁡(sqrt x+sqrt y))
c) (frac{-1}{2sqrt 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}{2sqrt x})
(frac{partial z}{partial x}=frac{-1}{2sqrt 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{ntheta}{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{-rtheta}{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.

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