1
AP EAPCET 2025 - 21st May Evening Shift
MCQ (Single Correct Answer)
+1
-0

The general solution of the differential equation $\frac{d y}{d x}+x y=4 x-2 y+8$ is

A

$y=4-c e^{-\frac{(x+2)^2}{2}}$

B

$y=8+c e^{-\frac{x^2}{2}-2 x}$

C

$y=c e^{-(x+2)^2}+x$

D

$y+2 x=c e^{-\frac{x}{2}-2 x}$

2
AP EAPCET 2025 - 21st May Evening Shift
MCQ (Single Correct Answer)
+1
-0

The general solution of the differential equation $\left(x+2 y^3\right) \frac{d y}{d x}-y=0, y>0$ is

A

$y=x^3+c y$

B

$x=y^3+c y$

C

$y(1-x y)=c x$

D

$x(1-x y)=c y$

3
AP EAPCET 2025 - 21st May Evening Shift
MCQ (Single Correct Answer)
+1
-0

The general solution of the differential equation $\frac{d y}{d x}+\frac{x+y+1}{x-3 y+5}=0$ is

A

$3(y-1)^2-2(x+2)(y-1)-(x+2)^2=C$

B

$x^2-3 y^2-4 x y-2 x-10 y=C$

C

$3(y+1)^2+2(x-2)(y+1)-(x-2)^2=C$

D

$x^2+3 y^2+4 x y+2 x+10 y=C$

4
AP EAPCET 2025 - 21st May Evening Shift
MCQ (Single Correct Answer)
+1
-0

If the maximum and minimum temperatures at a place on a day are measured as $44^{\circ} \mathrm{C} \pm 0.5^{\circ} \mathrm{C}$ and $22^{\circ} \mathrm{C} \pm 0.5^{\circ} \mathrm{C}$ respectively, then the temperature difference is

A

$22^{\circ} \mathrm{C} \pm 1^{\circ} \mathrm{C}$

B

$22^{\circ} \mathrm{C} \pm 0.5^{\circ} \mathrm{C}$

C

$22^{\circ} \mathrm{C} \pm 0.25^{\circ} \mathrm{C}$

D

$22^{\circ} \mathrm{C} \pm 1.5^{\circ} \mathrm{C}$