MHT CET 2026 19th April Morning Shift | Differential Equations Question 10 | Mathematics

The population $p$ of the city at time $t$ is given by $\frac{\mathrm{dp}}{\mathrm{dt}}=\frac{\mathrm{p}}{2}-100$. If initial population is 100 then $\mathrm{p}=$

$200+100 \mathrm{e}^{\frac{\mathrm{t}}{2}}$

$200-100 \mathrm{e}^{\frac{1}{2}}$

$300-100 \mathrm{e}^{\frac{1}{2}}$

$300+100 \mathrm{e}^{\frac{1}{2}}$

MHT CET 2025 5th May Evening Shift

MCQ (Single Correct Answer)

-0

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

$x=\log (x+y+2)+\mathrm{c}$, where c is the constant of integration

$x=\log (x+y-2)+\mathrm{c}$, where c is the constant of integration

$y=\log (x+y+2)+c$, where $c$ is the constant of integration

$y=\log (x+y-2)+\mathrm{c}$, where c is the constant of integration

MHT CET 2025 5th May Evening Shift

MCQ (Single Correct Answer)

-0

The solution of $\log \left(\frac{\mathrm{d} y}{\mathrm{~d} x}\right)=2 x-5 y, y(0)=0$ is

$\quad 2 \mathrm{e}^{2 x}+5 \mathrm{e}^{5 y}=6$

$\quad 5 \mathrm{e}^{2 x}-2 \mathrm{e}^{5 y}=3$

$\quad 2 \mathrm{e}^{2 x}-5 \mathrm{e}^{5 y}=6$

$5 \mathrm{e}^{2 x}+2 \mathrm{e}^{5 y}=3$

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