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1
COMEDK 2023 Morning Shift
MCQ (Single Correct Answer)
+1
-0

The solution of the differential equation $$\left(\frac{d y}{d x}\right) \tan y=\sin (x+y)+\sin (x-y)$$ is

A
$$\sec x=-2 \sec y+C$$
B
$$\sec y=2 \cos y+C$$
C
$$\sec y=-2 \cos x+C$$
D
$$\sec x=-2 \cos y+C$$
2
COMEDK 2023 Evening Shift
MCQ (Single Correct Answer)
+1
-0

The particular solution of $$e^{\frac{d y}{d x}}=2 x+1$$ given that $$y=1$$ when $$x=0$$ is

A
$$ y=\left(x+\frac{1}{2}\right) \log |2 x+1|-x+1 $$
B
$$ y=(x+1) \log |2 x+1|-x+1 $$
C
$$ y=\left(x+\frac{1}{2}\right) \log |2 x+1|-\frac{1}{2} x+1 $$
D
$$ y=\left(x-\frac{1}{2}\right) \log |2 x+1|-x-1 $$
3
COMEDK 2023 Evening Shift
MCQ (Single Correct Answer)
+1
-0

The general solution of the differential equation $$\left(1+y^2\right) d x=\left(\tan ^{-1} y-x\right) d y$$

A
$$ x=\tan ^{-1} y-1+c e^{\tan ^{-1} y} $$
B
$$ x=\tan ^{-1} y-1+c e^{-\tan ^{-1} y} $$
C
$$ x=\tan ^{-1} y+c e^{\tan ^{-1} y} $$
D
$$ x=c \tan ^{-1} y+e^{-\tan ^{-1} y} $$
4
COMEDK 2023 Evening Shift
MCQ (Single Correct Answer)
+1
-0

The solution of the differential equation $$\frac{d y}{d x}+y \cos x=\frac{1}{2} \sin 2 x$$

A
$$ y e^{\sin x}=e^{\sin x}(\sin x+1)+c $$
B
$$ y e^{\sin x}=e^{\sin x}(\sin x-1)+c $$
C
$$ y e^{\sin 2 x}=e^{\sin 2 x}(\sin x-1)+c $$
D
$$ y e^{\cos x}=e^{\sin x}(\cos x-1)+c $$
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