HSC Commerce (English Medium) १२ वीं कक्षा - Maharashtra State Board Question Bank Solutions for Mathematics and Statistics

Find the equation of tangent and normal to the following curve.

x = `1/"t",  "y" = "t" - 1/"t"`,  at t = 2

The differential equation of `y = k_1e^x+ k_2 e^-x` is ______.

Find the equation of tangent and normal to the following curve.

y = x³ - x² - 1 at the point whose abscissa is -2.

Find the equation of normal to the curve y = `sqrt(x - 3)` which is perpendicular to the line 6x + 3y – 4 = 0.

The solution of `dy/ dx` = 1 is ______

The solution of `dy/dx + x^2/y^2 = 0` is ______

Choose the correct alternative.

The solution of `x dy/dx = y` log y is

Choose the correct alternative.

Bacteria increases at the rate proportional to the number present. If the original number M doubles in 3 hours, then the number of bacteria will be 4M in

Choose the correct alternative.

The integrating factor of `dy/dx -  y = e^x `is e^x, then its solution is

A solution of a differential equation which can be obtained from the general solution by giving particular values to the arbitrary constants is called ___________ solution.

The integrating factor of the differential equation `dy/dx - y = x` is e^−x.

State whether the following is True or False:

The degree of a differential equation is the power of the highest ordered derivative when all the derivatives are made free from negative and/or fractional indices if any.

Solve the differential equation:

`e^(dy/dx) = x`

Solve the differential equation:

dr = a r dθ − θ dr

Solve:

(x + y) dy = a² dx

Solve

`dy/dx + 2/ x y = x^2`

y2 dx + (xy + x²)dy = 0

x²y dx – (x³ + y³) dy = 0

`xy dy/dx  = x^2 + 2y^2`

 `dy/dx = log x`