MHT CET 2025 26th April Evening Shift | Differential Equations Question 1 | Mathematics

The equation of the curve passing through the point $(0,2)$ given that the sum of the ordinate and abscissa of any point exceeds the slope of the tangent to the curve at that point by 5 is

$y=x-4-2 \mathrm{e}^x$

$y=4-x-2 \mathrm{e}^x$

$y=4+x-2 \mathrm{e}^x$

$y=4-x+2 \mathrm{e}^x$

MHT CET 2025 26th April Evening Shift

MCQ (Single Correct Answer)

-0

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

$y(1+x)=x+\mathrm{ce}^x$, where c is the constant of integration

$\quad y(1+x)=\mathrm{ce}^x$, where c is the constant of integration

$y(1-x)=x-\mathrm{ce}^x$, where c is the constant of integration

$y(1+x)=x$, where c is the constant of integration

MHT CET 2025 26th April Evening Shift

MCQ (Single Correct Answer)

-0

The differential equation representing the family of parabolas having vertex at the origin and axis along the positive Y -axis is

$x \frac{\mathrm{~d} y}{\mathrm{~d} x}-2 y=0$

$\frac{\mathrm{d} y}{\mathrm{~d} x}+x y=0$

$\quad x \frac{\mathrm{~d} y}{\mathrm{~d} x}+y=0$

$x^2 \frac{\mathrm{~d} y}{\mathrm{~d} x}+y=0$

MHT CET 2025 26th April Evening Shift

MCQ (Single Correct Answer)

-0

The population of towns A and B increase at the rate proportional to their population present at that time. At the end of the year 1984, the population of both the towns was 20,000 . At the end of the year 1989, the population of town A was 25,000 and that of town B was 28,000 . The difference of populations of towns A and B at the end of 1994 was

5950

8000

7950

6950

MHT CET Subjects