NCERT Solutions
Class 12 - Mathematics Part-2 - Chapter 9: DIFFERENTIAL EQUATIONS
Exercise 9.3

Question. 1

For the differential equation, find the general solution:

\(\frac{dy}{dx} = \frac{1-\cos x}{1+\cos x}\)

Answer:

\(y = 2\tan\frac{x}{2} - x + C\)

Question. 2

For the differential equation, find the general solution:

\(\frac{dy}{dx} = \sqrt{4-y^2}\) \((-2<y<2)\)

Answer:

\(y = 2\sin(x + C)\)

Question. 3

For the differential equation, find the general solution:

\(\frac{dy}{dx} + y = 1\) \((y \ne 1)\)

Answer:

\(y = 1 + Ae^{-x}\)

Question. 4

For the differential equation, find the general solution:

\(\sec^2 x\,\tan y\,dx + \sec^2 y\,\tan x\,dy = 0\)

Answer:

\(\tan x\,\tan y = C\)

Question. 5

For the differential equation, find the general solution:

\((e^x + e^{-x})\,dy - (e^x - e^{-x})\,dx = 0\)

Answer:

\(y = \log(e^x + e^{-x}) + C\)

Question. 6

For the differential equation, find the general solution:

\(\frac{dy}{dx} = (1+x^2)(1+y^2)\)

Answer:

\(\tan^{-1}y = x + \frac{x^3}{3} + C\)

Question. 7

For the differential equation, find the general solution:

\(y\log y\,dx - x\,dy = 0\)

Answer:

\(y = e^{cx}\)

Question. 8

For the differential equation, find the general solution:

\(x^5\frac{dy}{dx} = -y^5\)

Answer:

\(x^{-4} + y^{-4} = C\)

Question. 9

For the differential equation, find the general solution:

\(\frac{dy}{dx} = \sin^{-1} x\)

Answer:

\(y = x\sin^{-1}x + \sqrt{1-x^2} + C\)

Question. 10

For the differential equation, find the general solution:

\(e^x\tan y\,dx + (1-e^x)\sec^2 y\,dy = 0\)

Answer:

\(\tan y = C(1 - e^x)\)

Question. 11

Find a particular solution satisfying the given condition:

\((x^3 + x^2 + x + 1)\frac{dy}{dx} = 2x^2 + x\), \(y = 1\) when \(x = 0\)

Answer:

\(y = \frac{1}{4}\log\left[(x+1)^2(x^2+1)^3\right] - \frac{1}{2}\tan^{-1}x + 1\)

Question. 12

Find a particular solution satisfying the given condition:

\(x(x^2 - 1)\frac{dy}{dx} = 1\), \(y = 0\) when \(x = 2\)

Answer:

\(y = \frac{1}{2}\log\left(\frac{x^2-1}{x^2}\right) - \frac{1}{2}\log\left(\frac{3}{4}\right)\)

Question. 13

Find a particular solution satisfying the given condition:

\(\cos\left(\frac{dy}{dx}\right) = a\) \((a \in \mathbb{R})\), \(y = 1\) when \(x = 0\)

Answer:

\(\cos\left(\frac{y-2}{x}\right) = a\)

Question. 14

Find a particular solution satisfying the given condition:

\(\frac{dy}{dx} = y\tan x\), \(y = 1\) when \(x = 0\)

Answer:

\(y = \sec x\)

Question. 15

Find the equation of a curve passing through the point \((0,0)\) and whose differential equation is \(y' = e^x\sin x\).

Answer:

\(2y - 1 = e^x(\sin x - \cos x)\)

Question. 16

For the differential equation \(xy\frac{dy}{dx} = (x+2)(y+2)\), find the solution curve passing through the point \((1,-1)\).

Answer:

\(y - x + 2 = \log\left(x^2(y+2)^2\right)\)

Question. 17

Find the equation of a curve passing through the point \((0,-2)\) given that at any point \((x,y)\) 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.

Answer:

\(y^2 - x^2 = 4\)

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 \((-2,1)\).

Answer:

\((x+4)^2 = y + 3\)

Question. 19

The volume of a spherical balloon being inflated changes at a constant rate. If initially its radius is 3 units and after 3 seconds it is 6 units, find the radius of the balloon after \(t\) seconds.

Answer:

\((63t + 27)^{\frac{1}{3}}\)

Question. 20

In a bank, principal increases continuously at the rate of \(r\%\) per year. Find the value of \(r\) if Rs 100 doubles itself in 10 years \((\log_e 2 = 0.6931)\).

Answer:

\(6.93\%\)

Question. 21

In a bank, principal increases continuously at the rate of 5% per year. An amount of Rs 1000 is deposited with this bank, how much will it be worth after 10 years \((e^{0.5} = 1.648)\).

Answer:

Rs 1648

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?

Answer:

\(\frac{2\log 2}{\log\left(\frac{11}{10}\right)}\)

Question.  23

The general solution of the differential equation \(\frac{dy}{dx} = e^{x+y}\) is

(A)

\(e^x + e^{-y} = C\)

(B)

\(e^x + e^y = C\)

(C)

\(e^{-x} + e^y = C\)

(D)

\(e^{-x} + e^{-y} = C\)

NCERT Solutions Class 12 – Mathematics Part-2 – Chapter 9: DIFFERENTIAL EQUATIONS – Exercise 9.3 | Detailed Answers