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 :
A
3
✓
1
C
$0$
D
2
Answer
Correct option: B.
1
(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.
The solution of the equation $\frac{d y}{d x}=\frac{1-\cos 2 y}{1+\cos 2 y}$ is :
A
$x+\tan y+y=c$
B
$x+\cot y-y=c$
C
$x+\tan y-y=c$
✓
$x+\cot y+y=c$
Answer
Correct option: D.
$x+\cot y+y=c$
(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).
The solution of the equation $\frac{d y}{d x}=\cos ^2 y$ is :
A
$x+\tan y=c$
✓
$\tan y=x+c$
C
$\sin y+x=c$
D
$\sin y-x=c$
Answer
Correct option: B.
$\tan y=x+c$
(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).
The solution of the equation $\frac{d y}{d x}=e^{x-y}$ is :
A
$e^x=e^{-y}+c$
B
$e^y=e^{-x}+c$
✓
$e^y=e^x+c$
D
$e^{-x}=e^{-y}+c$
Answer
Correct option: C.
$e^y=e^x+c$
(C) $ \begin{array}{l} \frac{d y}{d x}=e^{x-y}=e^x \times e^{-y} \\ \Rightarrow \quad \frac{d y}{e^{-y}}=e^x d x \\ \Rightarrow \quad e^y d y=e^x d x \\ \text { So } \quad \int e^y d y=\int e^x d x \\ \Rightarrow \quad e^y=e^x+c \end{array} $ Hence the correct choice is (C).
The solution of the equation $\frac{d y}{d x}+\cos x \cdot \tan y=0$ is :
✓
$\log \sin y+\sin x=c$
B
$\log \sin x \sin y=c$
C
$\log y+\log \sin x=c$
D
$\sin x \cdot \sin y=c$
Answer
Correct option: A.
$\log \sin y+\sin x=c$
(A) $ \begin{array}{l} \frac{d y}{d x}+\cos x \cdot \tan y=0 \\ \Rightarrow \quad \frac{d y}{d x}=-\cos x \cdot \tan y \\ \Rightarrow \quad \frac{d y}{\tan y}=-\cos x d x \\ \text { or } \quad \cot y d y=-\cos x d x \\ \text { Hence } \quad \int \cot y d y=-\int \cos x d x \\ \log \sin y=-\sin x+c \\ \Rightarrow \quad \log \sin y+\sin x=c \end{array} $ Therefore the correct choice is (A).
The solution of the equation $\frac{d y}{d x}=\frac{e^x+e^{-x}}{e^x-e^{-x}}$ is :
A
$y=\log \left(e^x+e^{-x}\right)+c$
✓
$y=\log \left(e^x-e^{-x}\right)+c$
C
$y=\log \left(e^x+1\right)+c$
D
$y=\log \left(1-e^{-x}\right)+c$
Answer
Correct option: B.
$y=\log \left(e^x-e^{-x}\right)+c$
(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).