datavorous/entrance-exam-dataset huggingface dataset:
What is the solution of the differential equation
$$
a\left(x \frac{d y}{d x}+2 y\right)=x y \frac{d y}{d x} ?
$$

Options:
A $x^{2}=k y e^{\frac{y}{a}}$
B $y x^{2}=k y e^{\frac{y}{a}}$
C $y^{2} x^{2}=k y e^{\frac{y^{2}}{a}}$
D None of the above

The correct answer is:
None of the above

Solution:
Given differential equation is $a\left(x \frac{d y}{d x}+2 y\right)=x y \frac{d y}{d x}$
$\Rightarrow \quad a x \frac{d y}{d x}-x y \frac{d y}{d x}=-2 a y$
$\Rightarrow(x y-a x) \frac{d y}{d x}=2 a y$
$\Rightarrow \quad x(y-a) \frac{d y}{d x}=2 a y$
$\Rightarrow x(y-a) d y=2 a y d x$
$\Rightarrow \frac{(y-a)}{y} d y=\frac{2 a}{x} d x$
$\Rightarrow\left(1-\frac{a}{y}\right) d y=\frac{2 a}{x} d x$
$d y-\frac{a}{y} d y=\frac{2 a}{x} d x$
Integrate on both side
$\int d y-a \int \frac{1}{y} d y=2 a \int \frac{1}{x} d x$
$y-a \log y=2 a \log x+\log c$
$\Rightarrow \quad y=a \log x^{2} y c$
$\Rightarrow x^{2} y=k e^{y / a} \quad(\because c=k=$ constant $)$

Topic: Mathematics, Differential Equations, NDA, NDA 2010 (Phase 1)
What is the solution of the differential equation $-\operatorname{cosec}^{2}(x+y) d y=d x ? \quad$

Options:
A $y-c=\sin (x+y)$
B $x-c=\sin (x+y)$
C $y-c=\tan (x+y)$
D None of the above

The correct answer is:
None of the above

Solution:
$-\operatorname{cosec}^{2}(x+y) d y=d x$
$\Rightarrow \frac{d y}{d x}=-\sin ^{2}(x+y)$
Put $x+y=t$
$\Rightarrow \quad 1+\frac{d y}{d x}=\frac{d t}{d x} \mathrm{p} \frac{d y}{d x}=\frac{d t}{d x}-1$
$\therefore \quad \frac{d t}{d x}-1=-\sin ^{2}(t)$
$\Rightarrow \frac{\mathrm{dt}}{\mathrm{d} \mathrm{x}}=1-\sin ^{2} \mathrm{t} \Rightarrow \frac{d t}{d x}=\cos ^{2} t=\frac{1}{\sec ^{2} t}$
$\Rightarrow \int \sec ^{2} t d t=\int d x \Rightarrow \tan t=x-c$

Topic: Mathematics, Differential Equations, NDA, NDA 2008 (Phase 2)
What is the solution of the differential equation $\frac{d y}{d x}=e^{x-y}\left(e^{y-x}-e^{y}\right) ? \quad$

Options:
A $y=x-e^{x}+c$
B $y=x+e^{x}+c$
C $y=e^{x-y}-e^{y}+c$
D None of these

The correct answer is:
$y=x-e^{x}+c$

Solution:
Given differential equation is
$$
\frac{d y}{d x}=e^{x-y}\left(e^{y-x}-e^{y}\right)=e^{-y} \cdot e^{y}\left(e^{x} \cdot e^{-x}-e^{x}\right)
$$
$$
\begin{array}{l}
\Rightarrow \quad \int 1 d y=\int\left(1-e^{x}\right) d x \\
\Rightarrow \quad y=x-e^{x}+c
\end{array}
$$

Topic: Mathematics, Differential Equations, NDA, NDA 2009 (Phase 1)
What is the solution of the differential equation
$\frac{d y}{d x}=\frac{y}{\left(x+2 y^{3}\right)} ?$

Options:
A $\mathrm{y}(1-\mathrm{xy})=\mathrm{cx}$
B $y^{3}-x=c y$
C $x(1-x y)=c y$
D $\mathrm{x}(1+\mathrm{xy})=\mathrm{cy} \quad$

The correct answer is:
$y^{3}-x=c y$

Solution:
$y^{3}-x=c y$
$\Rightarrow 3 y^{2} \frac{d y}{d x}-1=c \frac{d y}{d x}$
$\Rightarrow \frac{d y}{d x}\left(3 y^{2}-c\right)=1$
$\Rightarrow \frac{d y}{d x}\left(3 y^{2}-\frac{y^{3}-x}{y}\right)=1$
$\Rightarrow \frac{d y}{d x}\left(\frac{3 y^{3}-y^{3}+x}{y}\right)=1$
$\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{y}}{\mathrm{x}+2 \mathrm{y}^{3}}$

Topic: Mathematics, Differential Equations, NDA, NDA 2006 (Phase 1)
What is the solution of the differential equation
$\frac{d x}{d y}+\frac{x}{y}-y^{2}=0 ?$

Options:
A $x y=x^{4}+c$
B $x y=y^{4}+c$
C $4 x y=y^{4}+c$
D $3 x y=y^{3}+c$
where $c$ is an arbitrary constant.

The correct answer is:
$4 x y=y^{4}+c$

Solution:
$\frac{\mathrm{dx}}{\mathrm{dy}}+\frac{\mathrm{x}}{\mathrm{y}}-\mathrm{y}^{2}=0$
$\frac{d x}{d y}+\frac{x}{y}=y^{2}$
This is a linear differential equation of the form
$\frac{\mathrm{dx}}{\mathrm{dy}}+\mathrm{P}_{1} \mathrm{x}=\mathrm{Q}_{1}$
Here, $\mathrm{P}=\frac{1}{\mathrm{y}}$ and $\mathrm{Q}=\mathrm{y}^{2}$
$\therefore \quad \mathrm{I} . \mathrm{F}=\mathrm{e}^{\int \mathrm{P} \mathrm{dy}}=\mathrm{e}^{\int \frac{1}{\mathrm{y}} \mathrm{dy}}=\mathrm{e}^{\log \mathrm{y}}=\mathrm{y}$
So, required solution is $\mathrm{x} \cdot \mathrm{y}=\int \mathrm{y}^{2} \cdot \mathrm{y} \mathrm{dy}+\mathrm{c}$
$\mathrm{xy}=\int \mathrm{y}^{3} \mathrm{dy}+\mathrm{c}$
$\mathrm{xy}=\frac{\mathrm{y}^{4}}{4}+\mathrm{c}$
$4 x y=y^{4}+c$
$\therefore$ Option (c) is correct.

Topic: Mathematics, Differential Equations, NDA, NDA 2015 (Phase 1)
What is the solution of the differential equation
$\frac{\mathrm{dy}}{\mathrm{dx}}+\sqrt{\frac{1-\mathrm{y}^{2}}{1-\mathrm{x}^{2}}}=0 ?$

Options:
A $\sin ^{-1} \mathrm{y}+\sin ^{-1} \mathrm{x}=\mathrm{C}$
B $\sin ^{-1} \mathrm{y}-\sin ^{-1} \mathrm{x}=\mathrm{C}$
C $2 \sin ^{-1} \mathrm{y}+\sin ^{-1} \mathrm{x}=\mathrm{C}$
D $2 \sin ^{-1} \mathrm{y}-\sin ^{-1} \mathrm{x}=\mathrm{C}$
Where $\mathrm{C}$ is a constant.

The correct answer is:
$\sin ^{-1} \mathrm{y}+\sin ^{-1} \mathrm{x}=\mathrm{C}$

Solution:
The differential equation is
$$
\frac{d y}{d x}+\sqrt{\frac{1-y^{2}}{1-x^{2}}}=0
$$
$\Rightarrow \frac{d y}{d x}=-\sqrt{\frac{1-y^{2}}{1-x^{2}}}$
$\Rightarrow \frac{d y}{\sqrt{1-y^{2}}}=\frac{-d x}{\sqrt{1-x^{2}}}$
$\Rightarrow \int \frac{1}{\sqrt{1-y^{2}}} d y+\int \frac{1}{\sqrt{1-x^{2}}} d x=0$
$\Rightarrow \sin ^{-1} y+\sin ^{-1} x=C$

Topic: Mathematics, Differential Equations, NDA, NDA 2011 (Phase 1)
What is the solution of the differential equation
$\frac{\mathrm{dy}}{\mathrm{dx}}=\sec (\mathrm{x}+\mathrm{y}) ?$

Options:
A $\mathrm{y}+\tan (\mathrm{x}+\mathrm{y})=\mathrm{c}$
B $\mathrm{y}-\tan \left\{\frac{(\mathrm{x}+\mathrm{y})}{2}\right\}=\mathrm{c}$
C $\mathrm{y}+\tan \left\{\frac{(\mathrm{x}+\mathrm{y})}{2}\right\}=\mathrm{c}$
D $\mathrm{y}+\tan \left\{\frac{(\mathrm{x}-\mathrm{y})}{2}\right\}=\mathrm{c}$

The correct answer is:
$\mathrm{y}-\tan \left\{\frac{(\mathrm{x}+\mathrm{y})}{2}\right\}=\mathrm{c}$

Solution:
In the equation $\frac{d y}{d x}=\sec (x+y)$
Let $x+y=v$
So, $1+\frac{d y}{d x}=\frac{d v}{d x}$ or $\frac{d y}{d x}=\frac{d v}{d x}-1$
and $\frac{\mathrm{dv}}{\mathrm{dx}}-1=\mathrm{sec} \mathrm{v}$
$\frac{d v}{d x}=1+\sec v=\frac{1+\cos v}{\cos v}$
or, $\frac{\cos \mathrm{v}}{1+\cos \mathrm{v}} \mathrm{dv}=\mathrm{dx}$
$\cos \mathrm{v}=\cos ^{2} \frac{\mathrm{v}}{2}-\sin ^{2} \frac{\mathrm{v}}{2}$
and $1+\cos v=2 \cos ^{2} \frac{\mathrm{v}}{2}$
so, $\left(\frac{2 \cos ^{2} \frac{\mathrm{v}}{2}-2 \sin ^{2} \frac{\mathrm{v}}{2}}{2 \cos ^{2} \frac{\mathrm{v}}{2}}\right) \mathrm{dv}=\mathrm{dx}$
$\left(1-\tan ^{2} \frac{\mathrm{v}}{2}\right) \mathrm{dv}=2 \mathrm{dx}$
or, $\left\{1-\left(\sec ^{2} \frac{\mathrm{v}}{2}-1\right)\right\} \mathrm{dv}=2 \mathrm{dx}$
or, $\left(2-\sec ^{2} \frac{\mathrm{v}}{2}\right) \mathrm{dv}=2 \mathrm{dx}$
Integrating on both the sides
$2 \int \mathrm{dv}-\int \sec ^{2} \frac{\mathrm{v}}{2} \mathrm{dv}=2 \int \mathrm{d} \mathrm{x}+\mathrm{c}_{1}$
where $c_{1}$ is a constant.
$2 \mathrm{v}-2 \tan \frac{\mathrm{v}}{2}=2 \mathrm{x}+\mathrm{c}_{1}$
$2(x+y)-2 \tan \frac{x+y}{2}=2 x+c_{1}$
or, $\quad 2 \mathrm{x}+2 \mathrm{y}-2 \tan \frac{\mathrm{x}+\mathrm{y}}{2}=2 \mathrm{x}+\mathrm{c}_{1}$
$\mathrm{y}-\tan \left(\frac{\mathrm{x}+\mathrm{y}}{2}\right)=\frac{\mathrm{c}_{1}}{2}=\mathrm{c}$
[c is a constant]
so, $y-\tan \frac{x+y}{2}=c$

Topic: Mathematics, Differential Equations, NDA, NDA 2007 (Phase 1)
What is the solution of the differential equation
$\frac{\mathrm{dy}}{\mathrm{dx}}+\frac{\mathrm{y}}{\mathrm{x}}=0 ?$

Options:
A $x y=c$
B $x=\mathrm{cy}$
C $\mathrm{y}=\mathrm{cx}$
D None of the above

The correct answer is:
$x y=c$

Solution:
$\frac{\mathrm{dy}}{\mathrm{dx}}+\frac{\mathrm{y}}{\mathrm{x}}=0$
$\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{-\mathrm{y}}{\mathrm{x}} \Rightarrow \frac{\mathrm{dy}}{\mathrm{y}}=-\frac{\mathrm{dx}}{\mathrm{x}}$
$\Rightarrow \frac{\mathrm{dy}}{\mathrm{y}}+\frac{\mathrm{d} \mathrm{x}}{\mathrm{x}}=0 \Rightarrow \mathrm{O} \frac{\mathrm{dy}}{\mathrm{y}}+\mathrm{O} \frac{\mathrm{d} \mathrm{x}}{\mathrm{x}}=0$
$\Rightarrow \log \mathrm{y}+\log \mathrm{x}=\log \mathrm{c}$
$\Rightarrow \log x y=\log c$
$\Rightarrow x y=c$

Topic: Mathematics, Differential Equations, NDA, NDA 2012 (Phase 1)
What is the solution of the differential equation
$\frac{\mathrm{dy}}{\mathrm{dx}}=\mathrm{xy}+\mathrm{x}+\mathrm{y}+1 ?$

Options:
A $\mathrm{y}=\frac{\mathrm{x}^{2}}{2}+\mathrm{x}+\mathrm{c}$
B $\log (\mathrm{y}+1)=\frac{\mathrm{x}^{2}}{2}+\mathrm{x}+\mathrm{c}$
C $y=x^{2}+x+c$
D $\log (y+1)=x^{2}+x+c$

The correct answer is:
$\log (\mathrm{y}+1)=\frac{\mathrm{x}^{2}}{2}+\mathrm{x}+\mathrm{c}$

Solution:
The given differential equation is:
$\frac{d y}{d x}=x y+x+y+1 \Rightarrow \frac{d y}{d x}=(x+1)(y+1)$
Separating variables,
$\Rightarrow \frac{1}{(1+y)} d y=(x+1) d x \Rightarrow \log (1+y)=\frac{x^{2}}{2}+x+c$

Topic: Mathematics, Differential Equations, NDA, NDA 2008 (Phase 1)
What is the solution of the differential equation
$\frac{\mathrm{dx}}{\mathrm{dy}}=\frac{\mathrm{x}+\mathrm{y}+1}{\mathrm{x}+\mathrm{y}-1} ?$

Options:
A $y-x+4 \ln (x+y)=c$
B $y+x+c \ln (x+y)=c$
C $\mathrm{y}-\mathrm{x}+\ln (\mathrm{x}+\mathrm{y})=\mathrm{c}$
D $\mathrm{y}+\mathrm{x}+2 \ln (\mathrm{x}+\mathrm{y})=\mathrm{c}$

The correct answer is:
$\mathrm{y}-\mathrm{x}+\ln (\mathrm{x}+\mathrm{y})=\mathrm{c}$

Solution:
Let $t=x+y \Rightarrow \frac{d t}{d y}=\frac{d x}{d y}+1$
So, $\frac{d t}{d y}-1=\frac{t+1}{t-1}$
$\Rightarrow \quad \frac{d t}{d y}=\frac{t+1+t-1}{t-1}=\frac{2 t}{t-1}$
$\Rightarrow \quad \int \frac{t-1}{t} d t=2 \int d y$
$\Rightarrow \quad t-\log t=2 y+\mathrm{C}_{1}$
$\Rightarrow \quad x+y-\log (x+y)-2 y=\mathrm{C}_{1}$
$\Rightarrow \quad y-x+\log (x+y)=-\mathrm{C}_{1}=\mathrm{C}$

Topic: Mathematics, Differential Equations, NDA, NDA 2018 (Phase 2)
What is the solution of the differential equation
$\ln \left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)-\mathrm{a}=0 ?$

Options:
A $\mathrm{y}=\mathrm{xe}^{\mathrm{a}}+\mathrm{c}$
B $x=y e^{a}+c$
C $\mathrm{y}=\ln \mathrm{x}+\mathrm{c}$
D $x=\ln y+c$

The correct answer is:
$\mathrm{y}=\mathrm{xe}^{\mathrm{a}}+\mathrm{c}$

Solution:
$\ln \left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)-\mathrm{a}=0$
$\Rightarrow \ln \left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)=\mathrm{a} \quad \quad \Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=\mathrm{e}^{\mathrm{a}}$
$\Rightarrow \int \mathrm{dy}=\int \mathrm{e}^{\mathrm{a}} \cdot \mathrm{dx} \quad \Rightarrow \mathrm{y}=\mathrm{e}^{\mathrm{a}} \cdot \mathrm{x}+\mathrm{c}$

Topic: Mathematics, Differential Equations, NDA, NDA 2017 (Phase 1)
What is the solution of the differential equation $\ln$ $\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)=\mathrm{ax}+\mathrm{by} ?$

Options:
A $a e^{a x}+b e^{b y}=c$
B $\frac{1}{\mathrm{a}} \mathrm{e}^{\mathrm{ax}}+\frac{1}{\mathrm{~b}} \mathrm{e}^{\mathrm{by}}=\mathrm{c}$
C $a \mathrm{e}^{\mathrm{ax}}+\mathrm{b} \mathrm{e}^{-\mathrm{by}}=\mathrm{c}$
D $\frac{1}{\mathrm{a}} \mathrm{e}^{\mathrm{ax}}+\frac{1}{\mathrm{~b}} \mathrm{e}^{-\mathrm{by}}=\mathrm{c}$

The correct answer is:
$\frac{1}{\mathrm{a}} \mathrm{e}^{\mathrm{ax}}+\frac{1}{\mathrm{~b}} \mathrm{e}^{-\mathrm{by}}=\mathrm{c}$

Solution:
$\frac{d y}{d x}=e^{a x+b y}=e^{a x} \cdot e^{b y}$
Integrating both sides,
$\frac{1}{a} e^{a x}+\frac{1}{b} e^{-b y}=c$

Topic: Mathematics, Differential Equations, NDA, NDA 2018 (Phase 2)
What is the solution of the differential equation $\sin \left(\frac{d y}{d x}\right)-a=0 ?$

Options:
A $y=x \sin ^{-1} a+c$
B $x=y \sin ^{-1} a+c$
C $y=x+x \sin ^{-1} a+c$
D $y=\sin ^{-1} a+c$
where $c$ is an arbitrary constant.

The correct answer is:
$y=x \sin ^{-1} a+c$

Solution:
$\sin \left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)-\mathrm{a}=0$
$\sin \left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)=\mathrm{a}$
$\Rightarrow \frac{d y}{d x}=\sin ^{-1} a$
$\mathrm{dy}=\sin ^{-1} \mathrm{a} \mathrm{dx}$
Now integrating both sides, $\int \mathrm{dy}=\int \sin ^{-1} \mathrm{a} \mathrm{dx}$
$\mathrm{y}=\mathrm{x} \sin ^{-1} \mathrm{a}+\mathrm{c}$
$\therefore$ Option (a) is correct.

Topic: Mathematics, Differential Equations, NDA, NDA 2015 (Phase 1)
What is the solution of the differential equation $x d y-y d x=0 ?$

Options:
A $x y=c$
B $\mathrm{y}=\mathrm{cx}$
C $x+y=c$
D $x-y=c$

The correct answer is:
$\mathrm{y}=\mathrm{cx}$

Solution:
$x d y-y d x=0$
$\Rightarrow x d y=y d x$
$\quad \Rightarrow \frac{d y}{y}=\frac{d x}{x}$
$\Rightarrow \log y=\log x+\log c$
$\Rightarrow \log y-\log x=\log c$
$\Rightarrow \log \left(\frac{y}{x}\right)=\log c$
$\Rightarrow \frac{y}{x}=c \Rightarrow y=c x$

Topic: Mathematics, Differential Equations, NDA, NDA 2018 (Phase 1)
What is the solution of the differential equation $x d y-y d x=x y^{2} d x ? \quad$

Options:
A $y+x^{-2}=c$
B $y^{2}+2 x^{-1}=c$
C $y+x^{-1}=c$
D $x^{2}+2 x y^{-1}=c$

The correct answer is:
$x^{2}+2 x y^{-1}=c$

Solution:
Given differential equation is $x d y-y d x=x y^{2} d x$
Which can be rewritten as $\frac{x d y-y d x}{y^{2}}=x d x$
$\Rightarrow-d\left(\frac{x}{y}\right)=x d x$
On integrating both sides, we get
$\int-d\left(\frac{x}{y}\right)=\int x d x$
$-\frac{x}{y}=\frac{-c}{2}+\left(\frac{-c}{2}\right) \quad\left(\because \frac{-c}{2}\right.$ is constant $)$
$\Rightarrow \frac{c}{2}=\frac{x^{2}}{2}+\frac{x}{y} \Rightarrow \frac{c}{2}=\frac{x^{2} y+2 x}{2 y}$
$\Rightarrow c=\frac{x^{2} y}{y}+\frac{2 x}{y} \Rightarrow c=x^{2}+\frac{2 x}{y}$
$\Rightarrow x^{2}+2 x y^{-1}=c$

Topic: Mathematics, Differential Equations, NDA, NDA 2009 (Phase 2)
What is the solution of the differential equation $x d y-y d x=x y^{2} d x ? \quad$

Options:
A $y x^{2}+2 x=2 c y$
B $y^{2} x+2 y=2 c x$
C $y^{2} x^{2}+2 x=2 c y$
D None of these

The correct answer is:
$y x^{2}+2 x=2 c y$

Solution:
Given, $x d y-y d x=x y^{2} d x$
$\Rightarrow \frac{x d y-y d x}{y^{2}}=x d x \Rightarrow \frac{y d x-x d y}{y^{2}}=-x d x$
$\Rightarrow \int d\left(\frac{x}{y}\right)=-\int x d x \Rightarrow \frac{x}{y}=\frac{-x^{2}}{2}+c=\frac{-x^{2}+2 c}{2}$
$\Rightarrow \quad y x^{2}+2 x=2 c y$

Topic: Mathematics, Differential Equations, NDA, NDA 2008 (Phase 2)
What is the solution of the differential equation $(\mathrm{x}+\mathrm{y})(\mathrm{dx}-\mathrm{dy})=\mathrm{dx}+\mathrm{dy} ?$

Options:
A $2 \log (\mathrm{x}+\mathrm{y})=\mathrm{c}(\mathrm{y}-\mathrm{x})$
B $(y-x)+\log (x+y)=c$
C $\left(\frac{y}{x}\right)+\left[\log \left(\frac{y}{x}\right)\right]=c$
D None of these

The correct answer is:
$(y-x)+\log (x+y)=c$

Solution:
Given differential equation is:
$(x+y)(d x-d y)=d x+d y$
$\Rightarrow(\mathrm{x}+\mathrm{y}) \mathrm{dx}-(\mathrm{x}+\mathrm{y}) \mathrm{dy}=\mathrm{dx}+\mathrm{dy}$
$\Rightarrow(x+y-1) d x=(x+y+1) d y$
$\Rightarrow \frac{d y}{d x}=\frac{x+y-1}{x+y+1}$
Let $x+y=v$ and $\frac{d y}{d x}=\frac{d v}{d x}-1$
$\therefore \quad \frac{d v}{d x}-1=\frac{v-1}{v+1}$
$\Rightarrow \frac{d v}{d x}=\frac{v-1}{v+1}+1$
$\Rightarrow \frac{d v}{d x}=\frac{v-1+v+1}{v+1}$
$\Rightarrow \frac{v+1}{2 v} d v=d x$
$\Rightarrow \frac{1}{2} \int 1 d v+\frac{1}{2} \int \frac{1}{v} d v=\int 1 d x$
$\Rightarrow \frac{1}{2} v+\frac{1}{2} \log v=x+c_{1}$
$\Rightarrow x+y+\log (x+y)=2 x+c \quad\left(\because 2 c_{1}=c=\right.$ costant $)$
$\Rightarrow(y-x)+\log (x+y)=c$

Topic: Mathematics, Differential Equations, NDA, NDA 2007 (Phase 2)
What is the solution of the differential equation $(\mathrm{x}+\mathrm{y})$ $(\mathrm{dx}-\mathrm{dy})=\mathrm{dx}+\mathrm{dy} ? \quad$

Options:
A $x+y+\ln (x+y)=c$
B $x-y+\ln (x+y)=c$
C $\mathrm{y}-\mathrm{x}+\ln (\mathrm{x}+\mathrm{y})=\mathrm{c}$
D $y-x-\ln (x-y)=c$

The correct answer is:
$\mathrm{y}-\mathrm{x}+\ln (\mathrm{x}+\mathrm{y})=\mathrm{c}$

Solution:
Differential equation is $(x+y)(d x-d y)=d x+d y$
dividing by $\mathrm{dx}$ on both the sides
$(x+y)\left(1-\frac{d y}{d x}\right)=1+\frac{d y}{d x}$
Putting $\mathrm{x}+\mathrm{y}=\mathrm{v}$
$1+\frac{d y}{d x}=\frac{d v}{d x}$ and $\frac{d y}{d x}=\frac{d v}{d x}-1$
The equation changes to
$\mathrm{v}\left\{1-\left(\frac{\mathrm{dv}}{\mathrm{dx}}-1\right)\right\}=\frac{\mathrm{dv}}{\mathrm{dx}}$
$v\left(2-\frac{d v}{d x}\right)=\frac{d v}{d x}$
$2 \mathrm{v}-\mathrm{v} \frac{\mathrm{dv}}{\mathrm{dx}}=\frac{\mathrm{dv}}{\mathrm{dx}}$
$2 \mathrm{v}=(1+\mathrm{v}) \frac{\mathrm{d} \mathrm{v}}{\mathrm{d} \mathrm{x}}$
$\left(\frac{1+\mathrm{v}}{\mathrm{v}}\right) \mathrm{d} \mathrm{v}=2 \mathrm{dx}$
or,$\left(\frac{1}{\mathrm{v}}+1\right) \mathrm{dv}=2 \mathrm{dx}$
Integrating on both the sides,
$\int \frac{\mathrm{dv}}{\mathrm{v}}+\int \mathrm{dv}=2 \int \mathrm{d} \mathrm{x}+\mathrm{c}$
$\log _{v}+v=2 x+c$
Putting $\mathrm{v}=\mathrm{x}+\mathrm{y}$
$\log (x+y)+x+y=2 x+c$
or, $\log (x+y)+y-x=c$
or, $y-x+\log (x+y)=c$

Topic: Mathematics, Differential Equations, NDA, NDA 2007 (Phase 1)
What is the solution of the differential equation
$\frac{y d x-x d y}{y^{2}}=0 ?$

Options:
A $\quad x y=c$
B $y=c x$
C $x+y=c$
D $x-y=c$ where $c$ is an arbitrary constant.

The correct answer is:
$y=c x$

Solution:
$\frac{y \mathrm{dx}-\mathrm{x} \mathrm{dy}}{\mathrm{y}^{2}}=0$
$\therefore \quad y d x-x d y=0$
$\therefore \quad \frac{d y}{y}=\frac{d x}{x}$
Now integrating both sides,
$\int \frac{\mathrm{dy}}{\mathrm{y}}=\int \frac{\mathrm{dx}}{\mathrm{x}}$
$\Rightarrow \log y=\log x+\log c$
$\Rightarrow \log \mathrm{y}=\log \mathrm{c} \mathrm{x}$
$\mathrm{y}=\mathrm{cx}$
$\therefore$ Option (b) is correct.

Topic: Mathematics, Differential Equations, NDA, NDA 2015 (Phase 1)
What is the solution of the equation $\quad$
$\ln \left(\frac{d y}{d x}\right)+x=0 ?$

Options:
A $y+e^{x}=c$
B $y-e^{-x}=c$
C $y+e^{-x}=c$
D $y-e^{x}=c$

The correct answer is:
$y+e^{-x}=c$

Solution:
Consider the given differential equation
$\ln \left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)+\mathrm{x}=0$
$\Rightarrow \ln \left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)=-\mathrm{x}$
$\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=\mathrm{e}^{-\mathrm{x}}$
on separating the variables, we get $\mathrm{dy}=\mathrm{e}^{-\mathrm{x}} \mathrm{dx}$
on integrating both sides, we get
$\int \mathrm{dy}=\int \mathrm{e}^{-\mathrm{x}} \mathrm{dx}$
$\Rightarrow \mathrm{y}=\frac{\mathrm{e}^{-\mathrm{x}}}{-1}+\mathrm{C}=-\mathrm{e}^{-\mathrm{x}}+\mathrm{C}$

Topic: Mathematics, Differential Equations, NDA, NDA 2014 (Phase 2)
What is the solution of $y^{\prime}=1+x+y^{2}+x y^{2}, y(0)=0$?

Options:
A $y=\tan ^{2}\left(\frac{x^{2}}{2}+x\right)$
B $\mathrm{y}=\tan ^{2}\left(\mathrm{x}^{2}+\mathrm{x}\right)$
C $y=\tan \left(x^{2}+x\right)$
D $\quad y=\tan \left(\frac{x^{2}}{2}+x\right)$

The correct answer is:
$\quad y=\tan \left(\frac{x^{2}}{2}+x\right)$

Solution:
Given differential equation is
$\frac{d y}{d x}=1+x+y^{2}+x y^{2}$
$\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=(1+\mathrm{x})\left(1+\mathrm{y}^{2}\right)$
$\Rightarrow \frac{d y}{1+y^{2}}=(1+x) d x$
$\Rightarrow \tan ^{-1} y=\frac{x^{2}}{2}+x+c$
Given that when $\mathrm{x}=0, \mathrm{y}(0)=0 .$ Hence, $\mathrm{c}=0$
$\Rightarrow \quad y=\tan \left(\frac{x^{2}}{2}+x\right)$

Topic: Mathematics, Application of Derivatives, NDA, NDA 2006 (Phase 2)
What is the solution set for the equation $x^{4}-26 x^{2}+25=0$

Options:
A $\{-5,-1,1,5\}$
B $\{-5,-1\}$
C $\{1,5\}$
D $\{-5,0,1,5\}$

The correct answer is:
$\{-5,-1,1,5\}$

Solution:
Consider, $x^{4}-26 x^{2}+25=0$ $\Rightarrow \quad x^{4}-25 x^{2}-x^{2}+25=0$
$\quad=x^{2}\left(x^{2}-25\right)-\left(x^{2}-25\right)=0$
$\Rightarrow \quad\left(x^{2}-25\right)\left(x^{2}-1\right)=0$
$\Rightarrow \quad(x-5)(x+5)(x-1)(x+1)=0$
$\Rightarrow \quad x=5,-5,1,-1$
$\therefore$ Solution set for given equation is $\{5,-5,-1,1\}$

Topic: Mathematics, Quadratic Equation, NDA, NDA 2011 (Phase 1)
What is the speed of a transverse wave on the string stretched between two rigid supports, given that it vibrates in its fundamental mode with a frequency of 45 Hz, a mass of 3.5 × ${{10}^{-2}}$kg, and a linear mass density of 4.0 × ${{10}^{-2}}$kg${{m}^{-1}}$?

Options:
A 79 m/s
B 81 m/s
C 66m/s
D 60 m/s

The correct answer is:
79 m/s

Solution:
Correct Option is : (A)
79 m/s

Topic: Chemistry, Waves and Sound, NEET
What is the speed of sound in air at standard temperature and pressure, where the mass of 1 mole of air is 29.0 ×${{10}^{-3}}$kg?

Options:
A 280 m/s
B 274m/s
C 331m/s
D 319m/s

The correct answer is:
280 m/s

Solution:
Correct Option is : (A)
280 m/s

Topic: Chemistry, Waves and Sound, NEET
What is the speed of sound in steel if a steel rod 100 cm long, clamped at its middle, has a fundamental frequency of longitudinal vibrations of 2.53 kHz?

Options:
A 5.06 km/s
B 1.01 km/s
C 2.52 km/s
D 5.91 km/s

The correct answer is:
5.06 km/s

Solution:
Correct Option is : (A)
5.06 km/s

Topic: Chemistry, Waves and Sound, NEET
What is the square root of $\frac{1}{2}-\mathrm{i} \frac{\sqrt{3}}{2}$ ?

Options:
A $\pm\left(\frac{\sqrt{3}}{2}+\frac{\mathrm{i}}{2}\right)$
B $\pm\left(\frac{\sqrt{3}}{2}-\frac{i}{2}\right)$
C $\pm\left(\frac{1}{2}+\mathrm{i} \frac{\sqrt{3}}{2}\right)$
D $\pm\left(\frac{1}{2}-i \frac{\sqrt{3}}{2}\right)$

The correct answer is:
$\pm\left(\frac{\sqrt{3}}{2}+\frac{\mathrm{i}}{2}\right)$

Solution:
Let $\sqrt{\frac{1}{2}-\frac{\mathrm{i} \sqrt{3}}{2}}=\mathrm{x}-\mathrm{iy}$
Squaring both the sides,
$\Rightarrow \frac{1}{2}-\frac{i \sqrt{3}}{2}=(x-i y)^{2} \Rightarrow \frac{1}{2}-\frac{i \sqrt{3}}{2}=x^{2}-y^{2}-2 i x y$
Comparing real and imaginary parts, we get
$\mathrm{x}^{2}-\mathrm{y}^{2}=\frac{1}{2}$
and $2 x y=\frac{\sqrt{3}}{2}$
and $\left(x^{2}+y^{2}\right)^{2}=\left(x^{2}-y^{2}\right)^{2}+4 x^{2} y^{2}=\frac{1}{4}+\frac{3}{4}=1$
$\Rightarrow \mathrm{x}^{2}+\mathrm{y}^{2}=1$
On solving eqs. (1) and (3) we get
$\mathrm{x}^{2}=\frac{3}{4}$ and $\mathrm{y}^{2}=\frac{1}{4}$
$\Rightarrow x=\pm \frac{\sqrt{3}}{2}, y=\pm \frac{1}{2}$
$\therefore \sqrt{\frac{1}{2}-\frac{\mathrm{i} \sqrt{3}}{2}}=\pm\left(\frac{\sqrt{3}}{2}+\frac{\mathrm{i}}{2}\right)$

Topic: Mathematics, Complex Number, NDA, NDA 2008 (Phase 1)
What is the square root of $i$, where $i=\sqrt{-1}$ ?

Options:
A $\frac{1+i}{2}$
B $\frac{1-i}{2}$
C $\frac{1+i}{\sqrt{2}}$
D None of these

The correct answer is:
$\frac{1+i}{\sqrt{2}}$

Solution:
Let $\sqrt{\mathrm{i}}=\mathrm{x}+\mathrm{iy}$
$\mathrm{i}=(\mathrm{x}+\mathrm{iy})^{2}$
$x^{2}-y^{2}+2 x y i=0+i$
$x^{2}-y^{2}=0 ; 2 x y=1$
Now, $\left(x^{2}+y^{2}\right)^{2}=\left(x^{2}-y^{2}\right)^{2}+4 x^{2} y^{2}$
$\left(x^{2}+y^{2}\right)^{2}=0+1$
$x^{2}+y^{2}=1$
$x^{2}-y^{2}=0$
$x^{2}+y^{2}=1$
$2 \mathrm{x}^{2}=1$
$\mathrm{x}^{2}=\frac{1}{2} \Rightarrow \mathrm{x}=\pm \frac{1}{\sqrt{2}}$
$\mathrm{y}^{2}=\frac{1}{2} \Rightarrow \mathrm{y}=\pm \frac{1}{\sqrt{2}}$
$\therefore \sqrt{\mathrm{i}}=\pm\left(\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}} \mathrm{i}\right)$
$=\frac{1}{\sqrt{2}}(1+\mathrm{i})$ or $\frac{-1}{\sqrt{2}}(1+\mathrm{i}) .$

Topic: Mathematics, Complex Number, NDA, NDA 2015 (Phase 1)
What is the square root of the complex number $-5+12 \mathrm{i}$ ?

Options:
A $2-3 \mathrm{i}$
B $2+3 \mathrm{i}$
C $-2+3 \mathrm{i}$
D $\sqrt{-5}+\sqrt{12 \mathrm{i}}$

The correct answer is:
$2+3 \mathrm{i}$

Solution:
Let $\sqrt{-5+12 \mathrm{i}}=\mathrm{x}+\mathrm{iy}$
$\Rightarrow(\mathrm{x}+\mathrm{iy})^{2}=-5+12 \mathrm{i}$
$\Rightarrow \mathrm{x}^{2}-\mathrm{y}^{2}+\mathrm{i} 2 \mathrm{xy}=-5+12 \mathrm{i}$
$\Rightarrow \mathrm{x}^{2}-\mathrm{y}^{2}=-5 \mathrm{and} 2 \mathrm{xy}=12 \Rightarrow \mathrm{xy}=6$
$\quad\left(\mathrm{x}^{2}-\mathrm{y}^{2}\right)^{2}+4 \mathrm{x}^{2} \mathrm{y}^{2}=\left(\mathrm{x}^{2}+\mathrm{y}^{2}\right)^{2}$
$\Rightarrow(-5)^{2}+4 \times(6)^{2}=\left(\mathrm{x}^{2}+\mathrm{y}^{2}\right)^{2}$
$\Rightarrow\left(\mathrm{x}^{2}+\mathrm{y}^{2}\right)^{2}=25+144=169$
$\Rightarrow \mathrm{x}^{2}+\mathrm{y}^{2}=\pm 13$ and $\mathrm{x}^{2}-\mathrm{y}^{2}=-5$
Adding both $2 \mathrm{x}^{2}=\pm 13-5=8$ or $-18$
$\Rightarrow \mathrm{x}^{2}=4($ -ve discarded $)$
$\Rightarrow \mathrm{x}=\pm 2$ and $\mathrm{y}^{2}=\pm 13-4=9$ or $-13$
Discard $\mathrm{y}^{2}=-13, \mathrm{y}^{2}=9$
$\Rightarrow \mathrm{y}=\pm 3$
$\Rightarrow \mathrm{x}+\mathrm{iy}=\pm(2+3 \mathrm{i})$
$\Rightarrow \sqrt{-5+12 \mathrm{i}}=(2+3 \mathrm{i})$

Topic: Mathematics, Complex Number, NDA, NDA 2007 (Phase 2)
What is the standard cell Potential E Cell o for following cell reaction?
2Fe(s) +  O 2 (g) +  2H 2 O(l) ⇌  2Fe 2+ (aq) +  4OH - (aq)

Given
E o Fe 2+ (aq) | Fe = − 0.44 V
E o O 2 (g) |  H 2 O | OH = 0.4 V

Options:
A E cell o = −0.48 V
B E cell o = −0.04 V
C E cell o = + 0.84 V
D Ecell0=Â +Â 1.28V

The correct answer is:
E cell o = + 0.84 V

Solution:
2Fe(s) +  O 2 (g) +  2H 2 O(l)⇌ 2Fe 2+ (aq) +  4OH − (aq)
For the given cell reaction
E cell o  =  E cathode o − E anode o
= 0.40− −0.44  = 0.84 V

Topic: Chemistry, Electrochemistry, NEET