M.C.Q (1 Marks)

Question 11 Mark

The degree of the differential equation $\left(\frac{d s}{d t}\right)^4+3 s \frac{d^2 s}{d t^2}=0$ is :

Answer

Corrrect option is (A)

Question 21 Mark

The degree of the differential equation $x y \frac{d^2 y}{d x^2}+x\left(\frac{d y}{d x}\right)^3-y \frac{d y}{d x}=0$ is :

Answer

(B)
The highest order derivative present in this differential equation is $\frac{d^2 y}{d x^2}$ and the highest exponent of $\frac{d^2 y}{d x^2}$ is 1 , therefore the degree of this differential equation is 1.

View full question & answer→

Question 31 Mark

The degree of differential equation $\sqrt{1+\left(\frac{d y}{d x}\right)^2}=$ $a\left(\frac{d^2 y}{d x^2}\right)^{\frac{1}{3}}:$

Answer

(B)
2

View full question & answer→

Question 41 Mark

The solution of the equation $\frac{d y}{d x}=\frac{e^x+e^{-x}}{e^x-e^{-x}}$ is :

Answer

(B)
$
\frac{d y}{d x}=\frac{e^x+e^{-x}}{e^x-e^{-x}}
$
Separating the variables
$
d y=\frac{e^x+e^{-x}}{e^x-e^{-x}} d x
$
Hence $\quad \int d y=\int \frac{e^x+e^{-x}}{e^x-e^{-x}} d x$ $
\Rightarrow \quad y=\log \left(e^x-e^{-x}\right)+c
$
Hence the correct choice is (B).

View full question & answer→

Question 51 Mark

The solution of the equation $\frac{d y}{d x}=\frac{1-\cos 2 y}{1+\cos 2 y}$ is :

Answer

(D)
$\frac{d y}{d x}=\frac{1-\cos 2 y}{1+\cos 2 y}=\frac{2 \sin ^2 y}{2 \cos ^2 y}=\tan ^2 y$
or $\quad \frac{d y}{\tan ^2 y}=d x \Rightarrow \cot ^2 y d y=d x$
or $\quad \int \cot ^2 y d y=\int d x$
or $\int\left(\operatorname{cosec}^2 y-1\right) d y=\int d x$
$-\cot y-y=x+c$
or $\quad x+\cot y+y=-c$ which is a constant
or $\quad x+\cot y+y=c$
Hence the correct choice is (D).

View full question & answer→

Question 61 Mark

The solution of the equation $\frac{d y}{d x}=\cos ^2 y$ is :

Answer

(B)
$
\begin{aligned}
\frac{d y}{d x} & =\cos ^2 y \\
\Rightarrow \quad \frac{1}{\cos ^2 y} d y & =d x \Rightarrow \sec ^2 y d y=d x
\end{aligned}
$
Hence
$
\begin{aligned}
\int \sec ^2 y d y & =\int d x \\
\tan y & =x+c
\end{aligned}
$
Therefore the correct choice is (B).

View full question & answer→

Question 71 Mark

The solution of the equation $\frac{d y}{d x}+2 y=4 x$ is :

Answer

(A)
$
\frac{d y}{d x}+2 y=4 x
$
Here $P =2$ and $Q =4 x$
$
\text { I.F. }=e^{\int P d x}=e^{\int 2 d x}=e^{2 x}
$
Required solution is
$
\begin{aligned}
& & y \times e^{2 x} & =\int e^{2 x} \cdot 4 x d x+c \\
\Rightarrow & & y \cdot e^{2 x} & =4 \int x e^{2 x} d x+c \\
\Rightarrow & & y \cdot e^{2 x} & =4\left(\frac{x \cdot e^{2 x}}{2}-\int \frac{1 \cdot e^{2 x}}{2} d x\right)+c \\
\Rightarrow & & y \cdot e^{2 x} & =2 x e^{2 x}-e^{2 x}+C \\
\Rightarrow & & y & =(2 x-1)+C e^{-2 x}
\end{aligned}
$
Hence the correct choice is (A).

View full question & answer→

Question 81 Mark

The solution of the equation $\frac{d y}{d x}=e^{x+y}+e^y x^2$ is :

Answer

(D)
$
\frac{d y}{d x}=e^{x+y}+e^y x^2
$
$
\begin{array}{rlrl}
\frac{d y}{d x} & =e^x \times e^y+e^y x^2 \\
\frac{d y}{d x} & =e^y\left(e^x+x^2\right) \\
\Rightarrow & \frac{d y}{e^y} & =\left(e^x+x^2\right) d x \\
\Rightarrow & e^{-y} d y & =\left(e^x+x^2\right) d x \\
\text { Hence } & \int e^{-y} d y & =\int\left(e^x+x^2\right) d x \\
& -e^{-y} & =e^x+\frac{x^3}{3}+c \\
\text { or } & -e^x+e^{-y}+\frac{x^3}{3} & =c
\end{array}
$
Therefore the correct choice is (D).

View full question & answer→

Question 91 Mark

The solution of the equation $\frac{d y}{d x}+\cos x \cdot \tan y=0$ is :

Answer

$\frac{d y}{d x}+\cos x \cdot \tan y=0$
$\Rightarrow \frac{d y}{d x}=-\cos x \cdot \tan y$
$\Rightarrow \frac{d y}{\tan y}=-\cos x\ d x$
or $\cot y\ d y=-\cos x\ d x$
Hence $\int \cot y\ d y=-\int \cos x\ d x$
$\log \sin y=-\sin x+c$
$\Rightarrow \log \sin y+\sin x=c$

View full question & answer→

Question 101 Mark

The solution of the equation $\frac{d y}{d x}=e^{x-y}$ is :

Answer

$\frac{d y}{d x}=e^{x-y}$
$=e^x \times e^{-y}$
$\Rightarrow \frac{d y}{e^{-y}}=e^x d x$
$\Rightarrow e^y d y=e^x d x$
So $\int e^y d y=\int e^x d x$
$\Rightarrow e^y=e^x+c$

View full question & answer→

Question 111 Mark

The solution of the equation $\frac{d y}{d x}+2 x=e^{3 x}$ is :

Answer

(A)
$
\begin{aligned}
& \\
\Rightarrow \quad \frac{d y}{d x}+2 x & =e^{3 x} \\
\frac{d y}{d x} & =e^{3 x}-2 x \\
d y & =\left(e^{3 x}-2 x\right) d x \\
\therefore \quad \int d y & =\int\left(e^{3 x}-2 x\right) d x \\
y & =\frac{1}{3} e^{3 x}-\frac{2 x^2}{2}+C \\
y & =\frac{1}{3} e^{3 x}-x^2+C \\
\Rightarrow \quad y+x^2 & =\frac{1}{3} e^{3 x}+C
\end{aligned}
$
Hence the correct choice is (A).

View full question & answer→

MATHS M.C.Q (1 Marks)

Take a timed test