Maths MCQ

Hint: $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$

... (1)

$\therefore \frac{1}{a^2} \times 2 x-\frac{1}{b^2} \times 2 y \frac{d y}{d x}=0$

$\therefore \frac{x}{a^2}-\frac{y}{b^2} \frac{d y}{d x}=0$

$\ldots$ (2)

and $\frac{1}{a^2} \times 1-\frac{1}{b^2}\left[y \frac{d^2 y}{d x^2}+\left(\frac{d y}{d x}\right)^2\right]=0$

Equations (1), (2) and (3) are consistent

$\therefore\left|\begin{array}{ccc}x^2 & -y^2 & 1 \\ x & -y \frac{d y}{d x} & 0 \\ 1 & -\left[y \frac{d^2 y}{d x^2}+\left(\frac{d y}{d x}\right)^2\right] & 0\end{array}\right|=0$

$\left.\therefore x y \frac{d^2 y}{d x^2}+x\left(\frac{d y}{d x}\right)^2-y \frac{d y}{d x}=0\right]$

Hint:

$\frac{d y}{d x}=\sec x - y \tan x$

$\therefore \frac{d y}{d x}+ y \tan x =\sec x$

I.F. $=e^{\int \tan x d x}=e^{\log \sec x}=\sec x$

∴ the solution is

y . sec x = ∫sec x . sec x dx + c

∴ y sec x = tan x + c

Hint:

I.F. $=e^{\int \frac{2}{x} d x}$

$=e^{2 \log x}$

$=x^2$

Hint:

$\frac{d y}{d x}+ y =\cos x -\sin x$

I.F. $=e^{\int 1 d x}=e^x$

$\therefore$ the solution is $y \cdot e ^{ x }=\int(\cos x -\sin x ) e ^{ x }+ c$

$\therefore y \cdot e ^{ x }= e ^{ x } \cos x + c$

$\operatorname{Hint}: \frac{1}{x} \frac{d y}{d x}=\tan ^{-1} x \quad \therefore d y=x \tan ^{-1} x d x$

$\therefore \int d y=\int\left(\tan ^{-1} x\right) \cdot x d x$

$\therefore y=\left(\tan ^{-1} x\right) \cdot \frac{x^2}{2}-\int \frac{d}{d x}\left(\tan ^{-1} x\right) \cdot \frac{x^2}{2} d x+c$

$=\frac{x^2 \tan ^{-1} x}{2}-\int \frac{1}{1+x^2} \times \frac{1}{2} x^2 d x+c$

$=\frac{x^2 \tan ^{-1} x}{2}-\frac{1}{2} \int \frac{\left(1+x^2\right)-1}{1+x^2} d x+c$

$=\frac{x^2 \tan ^{-1} x}{2}-\frac{1}{2} \int\left(1-\frac{1}{1+x^2}\right) d x+c$

$\left.=\frac{x^2 \tan ^{-1} x}{2}-\frac{1}{2}\left(x-\tan ^{-1} x\right)+c.\right]$

circles y $\frac{d y}{d x}+x=0$

$\therefore \int y d y+\int x d x=c$

$\therefore \frac{y^2}{2}+\frac{x^2}{2}=c$

$\therefore x ^2+ y ^2=2 c$ which is a circle.

Hint : Equation of the circle is

$(x-h)^2+(y-5)^2=5^2$.......(1)

$\therefore 2(x-h)+2(y-5) \frac{d y}{d x}=0$

$\therefore(x-h)^2=(y-5)^2\left(\frac{d y}{d x}\right)^2$

$\therefore 25-(y-5)^2=(y-5)^2\left(\frac{d y}{d x}\right)^2$

$[B y(1)]$

$\therefore(y-5)^2\left[1+\left(\frac{d y}{d x}\right)^2\right]=25$

Hint : $x^2+y^2=a^2 \quad \therefore 2 x+2 y \frac{d y}{d x}=0$

$\therefore \frac{d y}{d x}=-\frac{x}{y}$

$=x\left(-\frac{x}{y}\right)+a \sqrt{1+\frac{x^2}{y^2}}=-\frac{x^2}{y}+a \times \frac{a}{y}$

$=\frac{a^2-x^2}{y}=\frac{y^2}{y}=y$.

(b) : Equation of family of circles touching $y$-axis at origin is :
$
\begin{aligned}
& (x-a)^2+y^2=a^2 \\
& \Rightarrow x^2+a^2-2 a x+y^2=a^2 \\
& \Rightarrow x^2+y^2-2 a x=0...(i)
\end{aligned}
$
Differentiate w.r.t. $x$, we get $2 x+2 y y^{\prime}-2 a=0 \Rightarrow x+y y^{\prime}=a$
Putting value of ' $a$ ' in (i), we get $x^2+y^2=2\left(x+y y^{\prime}\right) x$
$
\begin{aligned}
& \Rightarrow x^2+y^2=2 x^2+2 x y y^{\prime} \Rightarrow x^2+2 x y y^{\prime}-y^2=0 \\
& \Rightarrow\left(x^2-y^2\right)+2 x y \frac{d y}{d x}=0
\end{aligned}
$

(c) : We have, $\cos x(1+\cos y) d x-\sin y(1+\sin x) d y$ $=0$
$
\Rightarrow \frac{\cos x}{1+\sin x} d x=\frac{\sin y}{1+\cos y} d y
$
On integrating both sides, we get
$
\int \frac{\cos x}{1+\sin x} d x=\int \frac{\sin y}{1+\cos y} d y ...(i)
$
Let $1+\sin x=p$ and $1+\cos y=q$
$\Rightarrow \cos x d x=d p$ and $\sin y d y=-d q$
So, equation (i) becomes
$
\begin{aligned}
& \int \frac{1}{p} d p=\int \frac{-1}{q} d q \\
\Rightarrow & \log p=-\log q+C \\
\Rightarrow & \log (1+\sin x)+\log (1+\cos y)=C \\
\Rightarrow & \log [(1+\sin x)(1+\cos y)]=C
\end{aligned}
$
$\Rightarrow(1+\sin x)(1+\cos y)=e^C=c$, where $c$ is constant of integration.

(c) : We have given equation of parabola whose axes are parallel to $y$-axis.
$\therefore \quad$ Vertex is $(h, k)$
So, $(x-h)^2=4 a(y-k)$
On differentiating both sides, we get
$
2(x-h)=4 a \frac{d y}{d x}
$
Again on differentiating both sides, we get
$
2=4 a \frac{d^2 y}{d x^2}
$
Again, on differentiating both sides, we get $0=4 a \cdot \frac{d^3 y}{d x^3}$
$
\begin{aligned}
& \Rightarrow \frac{d^3 y}{d x^3}=0 \\
& \Rightarrow y_3=0
\end{aligned}
$

(c) : $\left(1+y^2\right) d x-x y d y=0$
$
\Rightarrow \frac{d y}{d x}=\frac{1+y^2}{x y} \Rightarrow \frac{1}{x} d x=\frac{y}{1+y^2} d y
$
On integrating both sides, we get
$
\int \frac{1}{x} d x=\int \frac{y}{1+y^2} d y
$
$
\Rightarrow \ln x+\ln c=\int \frac{y}{1+y^2} d y
$
Let $1+y^2=t \Rightarrow y d y=\frac{d t}{2} \Rightarrow \ln (x c)=\int \frac{1}{2 t} d t$
$
\begin{aligned}
& \Rightarrow \ln (x c)=\frac{1}{2} \ln \left(y^2+1\right) \\
& \Rightarrow \ln (x c)=\ln \left(y^2+1\right)^{1 / 2} \\
& \Rightarrow x c=\left(y^2+1\right)^{1 / 2} \Rightarrow x^2 c^2=\left(y^2+1\right)
\end{aligned}
$
Now, at $x=1, y=0$
$
\Rightarrow c^2=1 \Rightarrow y^2+1=x^2 \Rightarrow x^2-y^2=1
$
This represent a hyperbola.

(d) : Given family of curves is $y^2=2 c(x+\sqrt{c})$ ...(i)
On differentiating (i) w.r.t. to $x$, both sides, we get
$
2 y \frac{d y}{d x}=2 c \Rightarrow c=y \frac{d y}{d x}
$
From (i), we have
$
\begin{aligned}
& y^2=2 y \frac{d y}{d x}\left\{x+\left(y \frac{d y}{d x}\right)^{1 / 2}\right\} \\
\Rightarrow & y^2-2 x y \frac{d y}{d x}=2\left(y \frac{d y}{d x}\right)^{3 / 2}
\end{aligned}
$
On squaring both sides, we get
$
\left(y^2-2 x y \frac{d y}{d x}\right)^2=4\left(y \frac{d y}{d x}\right)^3
$
$\therefore$ Order of differentiatial eqation $=1$
Degree of differentiatial eqation $=3$

(d) : $y=e^x(a \cos x+b \sin x)$
Differentiating both sides w.r.t. ' $x$ ', we get
$
\begin{aligned}
& \frac{d y}{d x}=e^x(a \cos x+b \sin x)+e^x(b \cos x-a \sin x) \\
\Rightarrow & \frac{d y}{d x}=y+e^x(b \cos x-a \sin x) ...(i)
\end{aligned}
$
Again differentiating both sides w.r.t. ' $x$ ', we get
$
\begin{aligned}
& \frac{d^2 y}{d x^2}=\frac{d y}{d x}+e^x(b \cos x-a \sin x)+e^x(-b \sin x-a \cos x) \\
& =\frac{d y}{d x}+\frac{d y}{d x}-y-e^x(a \cos x+b \sin x) \quad \text { [Using (i)] } \\
& =\frac{2 d y}{d x}-2 y \Rightarrow \frac{d^2 y}{d x^2}-2 \frac{d y}{d x}+2 y=0
\end{aligned}
$

$\begin{aligned} & \text {(b) : } \frac{d y}{d x}=e^{x+y}+x^2 e^{x^3+y} \\ & \Rightarrow \frac{d y}{d x}=e^x \cdot e^y+x^2 e^{x^3} \cdot e^y\end{aligned}$
$
\Rightarrow e^{-y} d y=\left(e^x+x^2 e^{x^3}\right) d x
$
By integrating on both sides, we get
$
\begin{aligned}
& \int e^{-y} d y=\int e^x d x+\int x^2 e^{x^3} d x \\
& \Rightarrow-e^{-y}=e^x+\int x^2 e^{x^3} d x ...(i)
\end{aligned}
$
Let $I=\int x^2 e^{x^3} d x$
Let $x^3=t \Rightarrow x^2 d x=\frac{d t}{3}$
$
\therefore \quad I=\frac{1}{3} \int e^t d t=\frac{1}{3} e^t+C=\frac{1}{3} e^{x^3}+C
$
Substituting the value of $I$ in equation (i), we get
$
\begin{aligned}
& -e^{-y}=e^x+\frac{1}{3} e^{x^3}+C \\
& \Rightarrow e^x+e^{-y}+\frac{1}{3} e^{x^3}=C_1 \quad\left[C_1=-C\right]
\end{aligned}
$

(a) : The given equation can be written as
$
\begin{aligned}
& 2 y \sin x \frac{d y}{d x}+y^2 \cos x=\sin 2 x \Rightarrow \frac{d}{d x}\left(y^2 \sin x\right)=\sin 2 x \\
& \Rightarrow y^2 \sin x=(-1 / 2) \cos 2 x+C ...(i) \\
& \text { So }(y(\pi / 2))^2 \sin (\pi / 2)=(-1 / 2) \cos (2 \pi / 2)+C \\
& \Rightarrow C=1 / 2...(ii)
\end{aligned}
$
From (i) and (ii), we have
$
y^2 \sin x=(1 / 2)(1-\cos 2 x)=\sin ^2 x \Rightarrow y^2=\sin x
$

(a): We have, $\frac{d y}{d x}-f(x) \cdot y=0$
$
\Rightarrow \frac{d y}{y}=f(x) d x \Rightarrow \ln y=\int f(x) d x \Rightarrow y_1(x)=e^{\int f(x) d x}
$
For required equation, I.F. $=e^{\int f(x) d x}$
Hence solution is $y \cdot y_1(x)=\int r(x) \cdot y_1(x) d x$
i.e., $y=\frac{1}{y_1(x)} \int r(x) \cdot y_1(x) d x$

(a): We have, $y_1 y_3=3 y_2{ }^2$
$
\begin{aligned}
& \Rightarrow \frac{y_3}{y_2}=3 \frac{y_2}{y_1} \Rightarrow \ln y_2=3 \ln y_1+\ln c \\
& \Rightarrow y_2=c y_1{ }^3 \Rightarrow \frac{y_2}{y_1^2}=c y_1 \quad \therefore-\frac{1}{y_1}=c y+c_2 \\
& \Rightarrow \frac{d x}{d y}=-c y-c_2 \Rightarrow x=-\frac{c y^2}{2}-c_2 y+c_3 \\
& \therefore x=A_1 y^2+A_2 y+A_3
\end{aligned}
$

(b) : Equation of a circle having radius $=4$ is
$
(x-a)^2+(y-b)^2=4^2 ...(i)
$
where $(a, b)$ is the centre of the circle.
$
\Rightarrow \quad(x-a)^2+(y-b)^2=16
$
On differentiating w.r.t. ' $x$ ', we get
$
2(x-a)+2(y-b) y^{\prime}=0 \Rightarrow(x-a)+(y-b) y^{\prime}=0 ...(ii)
$
On differentiating again w.r.t. ' $x$ ', we get
$
\begin{array}{r}
1+\left(y^{\prime}\right)^2+(y-b) y^{\prime \prime}=0 \\
\Rightarrow y-b=\frac{-\left(1+\left(y^{\prime}\right)^2\right)}{y^{\prime \prime}}
\end{array}
$
Substituting value of $(y-b)$ in (ii), we get
$
(x-a)-\frac{\left(1+\left(y^{\prime}\right)^2\right)}{y^{\prime \prime}} y^{\prime}=0 \Rightarrow(x-a)=\frac{\left(1+\left(y^{\prime}\right)^2\right) y^{\prime}}{y^{\prime \prime}}
$
Now, substituting value of $(x-a)$ and $(y-b)$ in (i), we get
$
\begin{aligned}
& {\left[\frac{\left(1+\left(y^{\prime}\right)^2\right) y^{\prime}}{y^{\prime \prime}}\right]^2+\left[\frac{-\left(1+\left(y^{\prime}\right)^2\right)}{y^{\prime \prime}}\right]^2=4^2} \\
& \Rightarrow\left(\left(1+\left(y^{\prime}\right)^2\right) y^{\prime}\right)^2+\left(1+\left(y^{\prime}\right)^2\right)^2=16\left(y^{\prime \prime}\right)^2 \\
& \therefore \text { Order is 2 }
\end{aligned}
$

(a) : We have, $\frac{d \theta}{d t}=-k\left(\theta-\theta_0\right)$
$
\begin{aligned}
& \Rightarrow \int \frac{d \theta}{\theta-\theta_0}=-k \int d t \\
& \Rightarrow \log \left(\theta-\theta_0\right)=-k t+c \Rightarrow \theta-\theta_0=e^{-k t+c}=e^{-k t} \cdot e^c \\
& \Rightarrow \theta=\theta_0+a e^{-k t} ; \text { where } a=e^c
\end{aligned}
$

(b): We have, $y d x-x d y=x y d x$
$
\begin{aligned}
& \Rightarrow y d x-x y d x=x d y \Rightarrow y(1-x) d x=x d y \\
& \Rightarrow\left(\frac{1-x}{x}\right) d x=\frac{d y}{y} \Rightarrow \int\left(\frac{1-x}{x}\right) d x=\int \frac{d y}{y} \\
& \Rightarrow \int \frac{1}{x} d x-\int 1 d x=\log y+c_1 \Rightarrow \log x-x=\log y+c_1 \\
& \Rightarrow \log x-\log y=x+c_1 \Rightarrow \log \left(\frac{x}{y}\right)=x+c_1 \\
& \Rightarrow \frac{x}{y}=e^{x+c_1}=e^x \cdot e^{c_1} \Rightarrow x=c y e^x\left[\text { where } c=e^{c_1}\right]
\end{aligned}
$

(d) : We have, $\log \left(\frac{d y}{d x}\right)=x$
$
\begin{aligned}
& \Rightarrow \frac{d y}{d x}=e^x \Rightarrow d y=e^x d x \\
& \Rightarrow \int d y=\int e^x d x \Rightarrow y=e^x+c
\end{aligned}
$
Now, at $x=0, y=1$
$
\therefore \quad 1=1+c \Rightarrow c=0
$
$\therefore \quad$ Particular solution is $y=e^x$.

(c) : We are given that $x \frac{d y}{d x}=y-x \tan \left(\frac{y}{x}\right)$
$
\Rightarrow \frac{d y}{d x}=\frac{y}{x}-\tan \left(\frac{y}{x}\right)
$
Clearly, the given differential equation is homogeneous.
$
\begin{aligned}
& \text { Put } y=v x \Rightarrow \frac{d y}{d x}=v+x \frac{d v}{d x} \\
& \therefore v+x \frac{d v}{d x}=v-\tan v \Rightarrow x \frac{d v}{d x}=-\tan v \\
& \Rightarrow \int \cot v d v=-\int \frac{d x}{x} \Rightarrow \log \sin v=-\log x+\log c \\
& \Rightarrow \sin v=\frac{c}{x} \Rightarrow \sin \left(\frac{y}{x}\right)=\frac{c}{x} \Rightarrow x \sin \left(\frac{y}{x}\right)=c
\end{aligned}
$

(d) : Equation of the family of such parabolas is
$
(y-k)^2=4 a(x-h) ...(i)
$
where $h$ and $k$ are arbitrary constants.
On differentiating (i) w.r.t. $x$, we get
$
(y-k) \frac{d y}{d x}=2 a ...(ii)
$
On differentiating (ii) w.r.t. $x$, we get
$
(y-k) \frac{d^2 y}{d x^2}+\left(\frac{d y}{d x}\right)^2=0 ...(ii)
$
On putting value of $(y-k)$ from (ii) in (iii), we get
$
2 a \frac{d^2 y}{d x^2}+\left(\frac{d y}{d x}\right)^3=0
$
Hence the order of required differential equation is 2.

(a) : We have $d x / d y=\cos (x+y)$...(i)
Let $x+y=v$
On differentiating both sides w.r.t. $y$, we get
$
\frac{d x}{d y}+1=\frac{d v}{d y} \Rightarrow \frac{d x}{d y}=\frac{d v}{d y}-1
$
Put in (i), we get
$
\therefore \quad \frac{d v}{d y}-1=\cos v \Rightarrow \frac{d v}{d y}=1+\cos v \Rightarrow \frac{d v}{d y}=2 \cos ^2 \frac{v}{2}
$
On integrating both sides, we get
$
\begin{aligned}
& \int \frac{d v}{\cos ^2 \frac{v}{2}}=2 \int d y \Rightarrow \int \sec ^2 \frac{v}{2} d v=2 y+K \\
& \Rightarrow \quad 2 \tan \left(\frac{v}{2}\right)=2 y+K \Rightarrow 2 \tan \left(\frac{x+y}{2}\right)=2 y+K \\
& \Rightarrow \quad \tan \left(\frac{x+y}{2}\right)=y+c \quad \quad(\text { where } c=K / 2)
\end{aligned}
$
(where $c=K / 2$ )

(b) : We have, $x d y+2 y d x=0$
$
\Rightarrow \frac{d y}{y}+2 \frac{d x}{x}=0
$
Integrating both sides, we get
$\log y+2 \log x=\log C \Rightarrow x^2 y=C$...(i)
Put $x=2 ; y=1$ in (i), we get $C=4$
$\therefore \quad$ Particular solution is $x^2 y=4$

(b) : We have, $\frac{d y}{d x}=\tan \left(\frac{y}{x}\right)+\left(\frac{y}{x}\right)$ Put $\frac{y}{x}=v \Rightarrow y=v x \Rightarrow \frac{d y}{d x}=v+x \frac{d v}{d x}$
$\therefore \quad$ The given equation becomes
$
v+x \frac{d v}{d x}=\tan v+v \Rightarrow \frac{1}{\tan v} d v=\frac{1}{x} d x
$
Integrating both sides, we get, $\int \cot v d v=\int \frac{1}{x} d x$
$
\begin{aligned}
& \Rightarrow \log |\sin v|=\log x+\log c=\log (x c) \\
& \Rightarrow \sin v=x c \Rightarrow \sin \left(\frac{y}{x}\right)=x c
\end{aligned}
$

(d) : $\frac{d y}{d x}+P y=Q$
Since, I.F. $=\sin x$
$
\Rightarrow e^{\int P d x}=\sin x=e^{\log \sin x} \Rightarrow \int P d x=\log \sin x
$
Differentiate both sides w.r.t. $x$, we get
$P=\frac{1}{\sin x} \cdot \cos x=\cot x
$

$\begin{aligned} & \text { (a) : } y(1+\log x) \frac{d x}{d y}-x \log x=0 \\
& \Rightarrow \quad y \frac{d x}{d y}=\frac{x \log x}{1+\log x} \Rightarrow \int \frac{d y}{y}=\int\left(\frac{1+\log x}{x \log x}\right) d x \\
& \Rightarrow \log y=\log (x \log x)+\log c \Rightarrow y=c x \log x \\
& \text { When } y=e^2, x=e \Rightarrow e^2=c e \log e \Rightarrow c e=e^2 \Rightarrow c=e \\
& \therefore \quad y=e x \log x\end{aligned}$

(b) : Substitute $x+y=z \Rightarrow \frac{d y}{d x}=\frac{d z}{d x}-1$
So, the given equation becomes
$
\begin{aligned}
& \frac{d z}{d x}=\sec ^2 z \Rightarrow d x=\cos ^2 z d z \\
\Rightarrow & d x=\left(\frac{1+\cos 2 z}{2}\right) d z \\
\Rightarrow & x=\frac{1}{2}\left(z+\frac{\sin 2 z}{2}\right)+c_1 (Integrating) \\
\Rightarrow & 2 x=x+y+\frac{\sin 2(x+y)}{2}+c_2 \\
\Rightarrow & 2(x-y)=\sin 2(x+y)-c \\
\Rightarrow & \sin 2(x+y)=2(x-y)+c
\end{aligned}
$

(d) : $\frac{d y}{d x}=\frac{y}{x}\left(1+\ln \frac{y}{x}\right)$
Substitute $y=v x \Rightarrow \frac{d y}{d x}=\frac{x d v}{d x}+v$
So, given equation becomes $\frac{x d v}{d x}+v=v(1+\ln v)$
$
\begin{aligned}
& \Rightarrow \int \frac{d v}{v \ln v}=\int \frac{d x}{x} \Rightarrow \ln \left(\ln \frac{y}{x}\right)=\ln x+\ln c \\
& \Rightarrow \ln \frac{y}{x}=c x .
\end{aligned}
$