NCERT Solutions for Class 12 Maths Chapter 9 Differential Equations Ex 9.4 are part of NCERT Solutions for Class 12 Maths. Here we have given NCERT Solutions for Class 12 Maths Chapter 9 Differential Equations Ex 9.4.

- Differential Equations Class 12 Ex 9.1
- Differential Equations Class 12 Ex 9.2
- Differential Equations Class 12 Ex 9.3
- Differential Equations Class 12 Ex 9.5
- Differential Equations Class 12 Ex 9.6

Board |
CBSE |

Textbook |
NCERT |

Class |
Class 12 |

Subject |
Maths |

Chapter |
Chapter 9 |

Chapter Name |
Differential Equations |

Exercise |
Ex 9.4 |

Number of Questions Solved |
23 |

Category |
NCERT Solutions |

## NCERT Solutions for Class 12 Maths Chapter 9 Differential Equations Ex 9.4

**For each of the following D.E in Q. 1 to 10 find the general solution:**

**Question 1.**

**Solution:**

integrating both sides, we get

**Question 2.**

**Solution:**

**Question 3.**

**Solution:**

which is required solution

**Question 4.**

sec² x tany dx+sec² y tanx dy = 0

**Solution:**

we have

sec² x tany dx+sec² y tanx dy = 0

**Question 5.**

**Solution:**

we have

Integrating on both sides

**Question 6.**

**Solution:**

integrating on both side we get

which is required solution

**Question 7.**

y logy dx – x dy = 0

**Solution:**

integrating we get

**Question 8.**

**Solution:**

**Question 9.**

solve the following

**Solution:**

integrating both sides we get

**Question 10.**

**Solution:**

we can write in another form

**Find a particular solution satisfying the given condition for the following differential equation in Q.11 to 14.**

**Question 11.**

**Solution:**

here

integrating we get

**Question 12.**

**Solution:**

**Question 13.**

**Solution:**

**Question 14.**

**Solution:**

=> logy = logsecx + C

When x = 0, y = 1

=> log1 = log sec0 + C => 0 = log1 + C

=> C = 0

∴ logy = log sec x

=> y = sec x.

**Question 15.**

Find the equation of the curve passing through the point (0,0) and whose differential equation

**Solution:**

**Question 16.**

For the differential equation find the solution curve passing through the point (1,-1)

**Solution:**

The differential equation is

or xydy=(x + 2)(y+2)dx

**Question 17.**

Find the equation of a curve passing through the point (0, -2) given that at any point (pc, y) on the curve the product of the slope of its tangent and y-coordinate of the point is equal to the x-coordinate of the point

**Solution:**

According to the question

0, – 2) lies on it.c = 2

∴ Equation of the curve is : x² – y² + 4 = 0.

**Question 18.**

At any point (x, y) of a curve the slope of the tangent is twice the slope of the line segment joining the point of contact to the point (-4,-3) find the equation of the curve given that it passes through (- 2,1).

**Solution:**

Slope of the tangent to the curve =

slope of the line joining (x, y) and (- 4, – 3)

**Question 19.**

The volume of a spherical balloon being inflated changes at a constant rate. If initially its radius is 3 units and offer 3 seconds it is 6 units. Find the radius of balloon after t seconds.

**Solution:**

Let v be volume of the balloon.

**Question 20.**

In a bank principal increases at the rate of r% per year. Find the value of r if Rs 100 double itself in 10 years

**Solution:**

Let P be the principal at any time t.

According to the problem

**Question 21.**

In a bank principal increases at the rate of 5% per year. An amount of Rs 1000 is deposited with this bank, how much will it worth after 10 years

**Solution:**

Let p be the principal Rate of interest is 5%

**Question 22.**

In a culture the bacteria count is 1,00,000. The number is increased by 10% in 2 hours. In how many hours will the count reach 2,00,000 if the rate of growth of bacteria is proportional to the number present

**Solution:**

Let y denote the number of bacteria at any instant t • then according to the question

**Question 23.**

The general solution of a differential equation is

(a)

(b)

(c)

(d)

**Solution:**

(a)

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