# NCERT solution class 12 chapter 3 Differential Equations exercise 3.6 mathematics part 2

## EXERCISE 3.6

#### Question 1:

The given differential equation is

This is in the form of

The solution of the given differential equation is given by the relation,

Therefore, equation (1) becomes:

This is the required general solution of the given differential equation.

#### Question 2:

The given differential equation is

The solution of the given differential equation is given by the relation,

This is the required general solution of the given differential equation.

#### Question 3:

The given differential equation is:

The solution of the given differential equation is given by the relation,

This is the required general solution of the given differential equation.

#### Question 4:

The given differential equation is:

The general solution of the given differential equation is given by the relation,

#### Question 5:

The given differential equation is:

This equation is in the form of:

The general solution of the given differential equation is given by the relation,

Therefore, equation (1) becomes:

#### Question 6:

The given differential equation is:

This equation is in the form of a linear differential equation as:

The general solution of the given differential equation is given by the relation,

#### Question 7:

The given differential equation is:

This equation is the form of a linear differential equation as:

The general solution of the given differential equation is given by the relation,

Substituting the value of in equation (1), we get:

This is the required general solution of the given differential equation.

#### Question 8:

This equation is a linear differential equation of the form:

The general solution of the given differential equation is given by the relation,

#### Question 9:

This equation is a linear differential equation of the form:

The general solution of the given differential equation is given by the relation,

#### Question 10:

This is a linear differential equation of the form:

The general solution of the given differential equation is given by the relation,

#### Question 11:

This is a linear differential equation of the form:

The general solution of the given differential equation is given by the relation,

#### Question 12:

This is a linear differential equation of the form:

The general solution of the given differential equation is given by the relation,

#### Question 13:

The given differential equation is

This is a linear equation of the form:

The general solution of the given differential equation is given by the relation,

Now,

Therefore,

Substituting C = –2 in equation (1), we get:

Hence, the required solution of the given differential equation is

#### Question 14:

This is a linear differential equation of the form:

The general solution of the given differential equation is given by the relation,

Now, y = 0 at x = 1.

Therefore,

Substituting  in equation (1), we get:

This is the required general solution of the given differential equation.

#### Question 15:

The given differential equation is

This is a linear differential equation of the form:

The general solution of the given differential equation is given by the relation,

Now,

Therefore, we get:

Substituting C = 4 in equation (1), we get:

This is the required particular solution of the given differential equation.

#### Question 16:

Find the equation of a curve passing through the origin given that the slope of the tangent to the curve at any point (xy) is equal to the sum of the coordinates of the point.

Let F (xy) be the curve passing through the origin.

At point (xy), the slope of the curve will be

According to the given information:

This is a linear differential equation of the form:

The general solution of the given differential equation is given by the relation,

Substituting in equation (1), we get:

The curve passes through the origin.

Therefore, equation (2) becomes:

1 = C

⇒ C = 1

Substituting C = 1 in equation (2), we get:

Hence, the required equation of curve passing through the origin is

#### Question 17:

Find the equation of a curve passing through the point (0, 2) given that the sum of the coordinates of any point on the curve exceeds the magnitude of the slope of the tangent to the curve at that point by 5.

Let F (xy) be the curve and let (xy) be a point on the curve. The slope of the tangent to the curve at (xy) is

According to the given information:

This is a linear differential equation of the form:

The general equation of the curve is given by the relation,

Therefore, equation (1) becomes:

The curve passes through point (0, 2).

Therefore, equation (2) becomes:

0 + 2 – 4 = Ce0

⇒ – 2 = C

⇒ C = – 2

Substituting C = –2 in equation (2), we get:

This is the required equation of the curve.

#### Question 18:

The integrating factor of the differential equation is

A. ex

B. ey

C.

D. x

The given differential equation is:

This is a linear differential equation of the form:

The integrating factor (I.F) is given by the relation,

Hence, the correct answer is C.

#### Question 19:

The integrating factor of the differential equation.

is

A.

B.

C.

D.