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

Solve the following differential equation.

x²y dx − (x³ + y³) dy = 0

Solve the following differential equation.

`dy /dx +(x-2 y)/ (2x- y)= 0`

Solve the following differential equation.

(x² − y² ) dx + 2xy dy = 0

Solve the following differential equation.

`xy  dy/dx = x^2 + 2y^2`

Find the equation of tangent and normal to the curve at the given points on it.

y = 3x² - x + 1 at (1, 3)

Find the equation of tangent and normal to the curve at the given points on it.

2x² + 3y² = 5 at (1, 1)

Find the equation of tangent and normal to the curve at the given points on it.

x² + y² + xy = 3 at (1, 1)

Find the equations of tangent and normal to the curve y = x² + 5 where the tangent is parallel to the line 4x − y + 1 = 0.

Find the equations of tangent and normal to the curve y = 3x² - 3x - 5 where the tangent is parallel to the line 3x − y + 1 = 0.

Solve the following differential equation.

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

Solve the following differential equation.

`dy/dx + y = e ^-x`

Solve the following differential equation.

`dy/dx + y` = 3

Solve the following differential equation.

`(x + y) dy/dx = 1`

Solve the following differential equation.

y dx + (x - y² ) dy = 0

Solve the following differential equation.

`dy/dx + 2xy = x`

Solve the following differential equation.

`(x + a) dy/dx = – y + a`

Solve the following differential equation.

dr + (2r)dθ= 8dθ

Choose the correct alternative.

The equation of tangent to the curve y = x² + 4x + 1 at (-1, -2) is 

Choose the correct alternative.

The equation of tangent to the curve x² + y² = 5 where the tangent is parallel to the line 2x – y + 1 = 0 are

Choose the correct alternative.

If elasticity of demand η = 1, then demand is