1
MHT CET 2023 9th May Morning Shift
MCQ (Single Correct Answer)
+2
-0

The particular solution of the differential equation $$\left(1+y^2\right) \mathrm{d} x-x y \mathrm{~d} y=0$$ at $$x=1, y=0$$, represents

A
circle
B
pair of straight lines
C
hyperbola
D
ellipse
2
MHT CET 2022 11th August Evening Shift
MCQ (Single Correct Answer)
+2
-0

The general solution of the differential equation $$\frac{\mathrm{d} y}{\mathrm{~d} x}=\frac{3 x+y}{x-y}$$ is (where $$C$$ is a constant of integration.)

A
$$\tan ^{-1}\left(\frac{y}{x}\right)+\log \left(\frac{y^2+3 x^2}{x^2}\right)=\log (x)+C$$
B
$$\frac{1}{\sqrt{3}} \tan ^{-1}\left(\frac{y}{x \sqrt{3}}\right)+\log \left(\frac{y^2+3 x^2}{x^2}\right)^{\frac{1}{2}}=\log (x)+C$$
C
$$\frac{1}{\sqrt{3}} \tan ^{-1}\left(\frac{y}{x \sqrt{3}}\right)-\log \left(\frac{y^2+3 x^2}{x^2}\right)^{\frac{1}{2}}=\log (x)+C$$
D
$$\tan ^{-1}\left(\frac{x}{y}\right)+\log \left(\frac{y^2+3 x^2}{x^2}\right)=\log (x)+C$$
3
MHT CET 2022 11th August Evening Shift
MCQ (Single Correct Answer)
+2
-0

The differential equation $$\frac{\mathrm{d} y}{\mathrm{~d} x}=\frac{\sqrt{1-y^2}}{y}$$ determines a family of circles with

A
fixed radius of 1 unit and variable centres along the $$X$$-axis
B
fixed radius of 1 unit and variable centres along the $$Y$$-axis
C
variable radii and a fixed centre at $$(0,1)$$
D
variable radii and a fixed centre at $$(0,-1)$$
4
MHT CET 2021 24th September Evening Shift
MCQ (Single Correct Answer)
+2
-0

The particular solution of the differential equation $$\left(1+e^{2 x}\right) d y+e^x\left(1+y^2\right) d x=0$$ at $$x=0$$ and y = 1 is

A
$$\tan ^{-1} e^x-\tan ^{-1} y=0$$
B
$$\tan ^{-1} e^x+\tan ^{-1} y=\frac{\pi}{2}$$
C
$$2 \tan ^{-1} e^x+\tan ^{-1} y=\frac{3 \pi}{4}$$
D
$$\tan ^{-1} e^x-\tan ^{-1} y=\frac{3 \pi}{4}$$
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