1
MHT CET 2023 11th May Morning Shift
+2
-0

The solution of the differential equation $$\frac{\mathrm{d} y}{\mathrm{~d} x}=\frac{1+y^2}{1+x^2}$$ is

A
$$x+y=\mathrm{c}$$, where $$\mathrm{c}$$ is a constant of integration.
B
$$x-y=\mathrm{c}(x y)$$, where $$\mathrm{c}$$ is a constant of integration.
C
$$x+y=\mathrm{c}(1+x y)$$, where $$\mathrm{c}$$ is a constant of integration.
D
$$y-x=\mathrm{c}(1+x y)$$, where $$\mathrm{c}$$ is a constant of integration.
2
MHT CET 2023 11th May Morning Shift
+2
-0

A curve passes through the point $$\left(1, \frac{\pi}{6}\right)$$. Let the slope of the curve at each point $$(x, y)$$ be $$\frac{y}{x}+\sec \left(\frac{y}{x}\right), x>0$$, then, the equation of the curve is

A
$$\sin \left(\frac{y}{x}\right)=\log (x)+\frac{1}{2}$$
B
$$\operatorname{cosec}\left(\frac{y}{x}\right)=\log (x)+2$$
C
$$\sec \left(\frac{2 y}{x}\right)=\log (x)+2$$
D
$$\cos \left(\frac{2 y}{x}\right)=\log (x)+\frac{1}{2}$$
3
MHT CET 2023 10th May Evening Shift
+2
-0

The general solution of the differential equation $$\frac{\mathrm{d} y}{\mathrm{~d} x}+\left(\frac{3 x^2}{1+x^3}\right) y=\frac{1}{x^3+1}$$ is

A
$$y\left(1+x^3\right)=x^3+\mathrm{c}$$, where $$\mathrm{c}$$ is a constant of integration.
B
$$y\left(1+x^3\right)=x+\mathrm{c}$$, where $$\mathrm{c}$$ is a constant of integration.
C
$$y\left(1+x^3\right)=x^2+\mathrm{c}$$, where $$\mathrm{c}$$ is a constant of integration.
D
$$y\left(1+x^3\right)=2 x+\mathrm{c}$$, where $$\mathrm{c}$$ is a constant of integration.
4
MHT CET 2023 10th May Evening Shift
+2
-0

The differential equation of all circles, passing through the origin and having their centres on the $$\mathrm{X}$$-axis, is

A
$$y^2=x^2+x y \frac{\mathrm{d} y}{\mathrm{~d} x}$$
B
$$x^2=y^2+2 x y \frac{\mathrm{d} y}{\mathrm{~d} x}$$
C
$$y^2=x^2+2 x y \frac{\mathrm{d} y}{\mathrm{~d} x}$$
D
$$x^2=y^2-x y \frac{\mathrm{d} y}{\mathrm{~d} x}$$
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