1
MHT CET 2021 23th September Morning Shift
+2
-0

The general solution of the differential equation $$\frac{d x}{d t}=\frac{x \log x}{t}$$ is

A
$$\log x-x=c$$
B
$$e^{c t}+x=0$$
C
$$\log t=x+c$$
D
$$e^{c t}=x$$
2
MHT CET 2021 23th September Morning Shift
+2
-0

The particular solution of differential equation $$(x+y) d y+(x-y) d x=0$$ at $$x=y=1$$ is

A
$$\log \left|\frac{x^2+y^2}{2}\right|=\frac{\pi}{2}-2 \tan ^{-1}\left(\frac{y}{x}\right)$$
B
$$\log \left|x^2+y^2\right|=\frac{\pi}{2}-2 \tan ^{-1}\left(\frac{y}{x}\right)$$
C
$$\log \left|x^2+y^2\right|=\frac{\pi}{2}-2 \tan ^{-1}\left(\frac{y}{x}\right)$$
D
$$\log \left|\frac{x^2+y^2}{2}\right|=\frac{\pi}{4}-2 \tan ^{-1}\left(\frac{y}{x}\right)$$
3
MHT CET 2021 23th September Morning Shift
+2
-0

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

A
$$2^x-2^y=c$$
B
$$\frac{1}{2^x}-\frac{1}{2^y}=c$$
C
$$\frac{1}{2^x}+\frac{1}{2^y}=c$$
D
$$2^x+2^y=c$$
4
MHT CET 2021 23th September Morning Shift
+2
-0

If the surrounding air is kept at $$25^{\circ} \mathrm{C}$$ and a body cools from $$80^{\circ} \mathrm{C}$$ to $$50^{\circ} \mathrm{C}$$ in 30 minutes, then temperature of the body after one hour will be

A
$$31.72^{\circ} \mathrm{C}$$ approximately
B
$$34.74^{\circ} \mathrm{C}$$ approximately
C
$$32.36^{\circ} \mathrm{C}$$ approximately
D
$$36.36^{\circ} \mathrm{C}$$ approximately
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