1
MHT CET 2023 14th May Morning Shift
MCQ (Single Correct Answer)
+2
-0

General solution of the differential equation $$\cos x(1+\cos y) \mathrm{d} x-\sin y(1+\sin x) \mathrm{d} y=0$$ is

A
$$(1+\cos x)(1+\sin y)=\mathrm{c}$$, where $$\mathrm{c}$$ is a constant of integration.
B
$$1+\sin x+\cos y=\mathrm{c}$$, where $$\mathrm{c}$$ is a constant of integration.
C
$$(1+\sin x)(1+\cos y)=\mathrm{c}$$, where $$\mathrm{c}_{\text {is }}$$ constant of integration.
D
$$1+\sin x \cos y=\mathrm{c}$$, where $$\mathrm{c}$$ is a constant of integration.
2
MHT CET 2023 14th May Morning Shift
MCQ (Single Correct Answer)
+2
-0

The differential equation of $$y=\mathrm{e}^x(\mathrm{a} \cos x+\mathrm{b} \sin x)$$ is

A
$$\frac{\mathrm{d}^2 y}{\mathrm{~d} x^2}-2 \frac{\mathrm{d} y}{\mathrm{~d} x}-y=0$$
B
$$\frac{\mathrm{d}^2 y}{\mathrm{~d} x^2}+2 \frac{\mathrm{d} y}{\mathrm{~d} x}+2 y=0$$
C
$$\frac{\mathrm{d}^2 y}{\mathrm{~d} x^2}-2 \frac{\mathrm{d} y}{\mathrm{~d} x}+y=0$$
D
$$\frac{\mathrm{d}^2 y}{\mathrm{~d} x^2}-2 \frac{\mathrm{d} y}{\mathrm{~d} x}+2 y=0$$
3
MHT CET 2023 13th May Evening Shift
MCQ (Single Correct Answer)
+2
-0

If $$x d y=y(d x+y d y), y(1)=1, y(x)>0$$, then $$y(-3)$$ is

A
1
B
2
C
3
D
4
4
MHT CET 2023 13th May Evening Shift
MCQ (Single Correct Answer)
+2
-0

The solution of $$(1+x y) y d x+(1-x y) x d y=0$$ is

A
$$\log \left(\frac{x}{y}\right)+\frac{1}{x y}=k$$, where $$k$$ is constant of integration
B
$$\log \left(\frac{x}{y}\right)=\frac{1}{x y}+k$$, where $$k$$ is constant of integration
C
$$\log \left(\frac{x}{y}\right)+{x y}=k$$, where $$k$$ is constant of integration
D
$$\log \left(\frac{x}{y}\right)={x y}+k$$, where $$k$$ is constant of integration
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