Class 12 NCERT Solutions- Mathematics Part ii – Chapter 9– Differential Equations Exercise 9.3

For each of the differential equations in Exercises 1 to 10, find the general solution:

1. [Tex]\frac{dy}{dx}=\frac{1-cos\space x}{1+cos \space x}[/Tex]

Solution:

The given equation can be rewritten as:

[Tex]\frac{dy}{dx}=\frac{2sin^2\frac{x}2}{2cos^2\frac{x}2}\\⇒\frac{dy}{dx}=tan^2\frac{x}2\\⇒\frac{dy}{dx}=sec^2\frac{x}2-1\\⇒dy=[sec^2\frac{x}2-1]dx[/Tex]

On integrating both sides, we get,

[Tex]\int dy=\int[sec^2\frac{x}2-1]dx\\ ⇒y=\int sec^2\frac{x}2dx-\int 2dx\\ ⇒y=2tan\frac{x}2-x+C[/Tex]

2. [Tex]\frac{dy}{dx}=\sqrt{4-y^2}[/Tex]

Solution:

The given equation can be rewritten as:

[Tex]\frac{dy}{\sqrt{4-y^2}}=dx[/Tex]

On integrating both sides, we get,

[Tex]\int \frac{dy}{\sqrt{4-y^2}}=\int dx\\ ⇒sin^{-1}\frac{y}2=x+C[/Tex]

⇒ y/2 = sin(x +C)

⇒ y = 2sin(x + C)

3. [Tex]\frac{dy}{dx}+y=1[/Tex]

Solution:

The given equation can be rewritten as:

dy + ydx = dx

⇒ dy = (1 – y)dx

[Tex]⇒\frac{dy}{1-y}=dx[/Tex]

Integrating both sides, we get

[Tex]\int \frac{dy}{1-y}=\int dx[/Tex]

⇒ -log(y – 1) = x + log C

⇒ log C(y – 1) = -x

⇒ [Tex]y = 1 + \frac{1}Ce^{-x}[/Tex]

4. sec² x tan y dx + sec² y tan x dy = 0

Solution:

Dividing both sides by tan x tan y, we have:

[Tex]\frac{sec^2 x tan y dx + sec^2 y tan x dy}{tan x tany}= \frac{0}{tanxtany}\\ ⇒\frac{sec^2 x }{tan x }dx + \frac{sec^2 y} {tan y} dy= 0\\ ⇒\frac{sec^2 x }{tan x }dx =- \frac{sec^2 y} {tan y} dy[/Tex]

On integrating both sides, we get,

[Tex]\int \frac{sec^2 x }{tan x }dx =- \int\frac{sec^2 y} {tan y} dy\\ ⇒\frac{d}{dx}tanx=\frac{dt}{dx}\\ ⇒sec^2xdx=dt\\ \int \frac{sec^2 x }{tan x }dx =\int \frac{1}{t}dt\\=log\space t\\=log(tan\space x) \int\frac{sec^2 y} {tan y} dy=log(tan\space y)[/Tex]

From the above computations, we have:

tan x tan y = C

5. y = (e^x + e^-x)dy – (e^x – e^-x)dx = 0

Solution:

The given equation can be rewritten as:

(ex + e-x)dy = (ex – e-x)dx

[Tex]dy = [\frac{e^x-e^{-x}}{e^x+e^{-x}}]dx[/Tex]

On integrating both sides, we get,

[Tex]\int dy = \int[\frac{e^x-e^{-x}}{e^x+e^{-x}}]dx\\ \frac{d}{dx}(e^x-e^{-x})=\frac{dt}{dx}[/Tex]

From the above computations, we get:

y = log(e^x – e^-x) + C

6. [Tex]\frac{dy}{dx}=(1+x^2)(1+y^2)[/Tex]

Solution:

The given equation can be rewritten as:
[Tex]\frac{dy}{1+y^2}=(1+x^2)dx[/Tex]

On integrating both sides, we get,

[Tex]\int\frac{dy}{1+y^2}=\int(1+x^2)dx\\ ⇒tan^{-1}y=\int dx+\int x^2dx\\ ⇒tan^{-1}y=x+\frac{x^3}x+C[/Tex]

7. y log y dx – x dy = 0

Solution:

The given equation can be rewritten as:

[Tex]\frac{dy}{y\space log\space y}=\frac{dx}{x}[/Tex]

On integrating both sides, we get,

[Tex]\int \frac{dy}{y\space log\space y}=\int\frac{dx}{x}\\ ⇒\frac{1}{y}(dy)=dt[/Tex]

Substituting these values in the original equation, we have:

log t = log x + log C

⇒ log(log y) = log Cx

⇒ log y = Cx

⇒ y = Ce^x

8. [Tex]x^5\frac{dy}{dx}=-y^5[/Tex]

Solution:

The given equation can be rewritten as:

[Tex]\frac{dx}{x^5}+\frac{dy}{y^5}=0[/Tex]

On integrating both sides, we get,

[Tex]\int\frac{dx}{x^5}+\int\frac{dy}{y^5}=k\\ ⇒\int x^{-5}dx+\int y^{-5}dy=k\\ ⇒\frac{x^{-4}}{-4}+\frac{y^{-4}}{-4}=k[/Tex]

⇒ x^-4 + y^-4 = C

9. [Tex]\frac{dy}{dx}=sin^{-1}x[/Tex]

Solution:

The given equation can be rewritten as:

dy = sin^-1x dx

On integrating both sides, we get,

[Tex]\int dy = \int sin^{-1}x dx\\ ⇒y=\int(sin^{-1}x.1)dx\\ ⇒y=sin^{-1}x.\int(1)dx-\int[(\frac{d}{dx}(sin^{-1}x)\int(1)dx)]dx\\ ⇒y=xsin^{-1}x+\int\frac{x}{\sqrt{1-x^2}}dx[/Tex]

Assuming 1 – x² = t, we have:

[Tex]\frac{d}{dx}(1-x^2)=\frac{dt}{dx}\\ ⇒-2x=\frac{dt}{dx}\\ ⇒x\space dx=-\frac{1}2dt[/Tex]

Putting these values back in the original equation, we have:

[Tex]y=xsin^{-1}x+\int\frac{1}{2\sqrt{t}}dt\\ ⇒y=xsin^{-1}x+\sqrt{t}+C[/Tex]

⇒ y = xsin^-1x + (1 – x²)^1/2 + C

10. e^x tan y dx + (1 – e^x)sec²y dy = 0

Solution:

The given equation can be rewritten as:

[Tex]\frac{sec^2\space y}{tan\space y}dy=\frac{-e^x}{1-e^x}dx[/Tex]

On integrating both sides, we get,

[Tex]\int\frac{sec^2\space y}{tan\space y}dy=\int\frac{-e^x}{1-e^x}dx[/Tex]

Let tan y = u.

[Tex]⇒\frac{d}{dy}(tan\space y)=\frac{du}{dy}\\ ⇒sec^2\space y=\frac{du}{dy}\\ ⇒\int\frac{-e^x}{1-e^x}dx=\int\frac{dt}{t}[/Tex]

⇒ tan y = C(1 – e^x)

For each of the differential equations in Exercises 11 to 14, find a particular solution satisfying the given condition:

11. (x³ + x² + x + 1)dy/dx = 2x² + x; y = 1 when x = 0

Solution:

The given equation can be rearranged as:

[Tex]\frac{dy}{dx}=\frac{2x^2+x}{x^3+x^2+x+1}\\ ⇒dy=\frac{2x^2+x}{x^3+x^2+x+1}dx[/Tex]

On integrating both sides, we get,

[Tex]\int dy=\int\frac{2x^2+x}{x^3+x^2+x+1}dx\\ Let\space \frac{2x^2+x}{x^3+x^2+x+1}dx=\frac{A}{x+1}+\frac{Bx+C}{x^2+1}\\ ⇒2x^2+x=(A+B)x^2+(B+C)x+(A+C)[/Tex]

Comparing the coefficients of x² and x, we have:

A + B = 2

B + C = 2

A + C = 0

Thus, A = 1/2, B = 3/2 and C = -1/2

Substituting these values in the original equation, we have:

[Tex]y=\frac{1}2log(x+1)+\frac{3}2\int\frac{x}{x^2+1}dx-\frac{1}2\int\frac{x}{x^2+1}dx\\ ⇒y=\frac{1}4[log(x+1)^2(x^2+1)^3]-\frac{1}2tan^{-1}x+C[/Tex]

Now, y = 1 when x = 0

[Tex]⇒1=\frac{1}4log(1)-\frac{1}2tan^{-1}0+C\\ ⇒1=\frac{1}4\times 0-\frac{1}2\times0+C[/Tex]

⇒ C = 1

Thus, [Tex]y=\frac{1}4[log(x+1)^2(x^2+1)^3]-\frac{1}2tan^{-1}x+1[/Tex]

12. [Tex]x(x^2-1)\frac{dy}{dx}=1[/Tex]; y=0 when x = 2

Solution:

The given equation can be rearranged as:

[Tex]dy=\frac{dx}{x(x^2-1)}\\ ⇒dy=\frac{1}{x(x-1)(x+1)}dx[/Tex]

On integrating both sides, we get,

[Tex]\int dy=\int\frac{1}{x(x-1)(x+1)}dx\\ Let\space \frac{1}{x(x-1)(x+1)}=\frac{A}x+\frac{B}{x-1}+\frac{C}{x+1}\\ ⇒\frac{1}{x(x-1)(x+1)}=\frac{(A+B+C)x^2+(B-C)x-A}{x(x-1)(x+1)}[/Tex]

Comparing the coefficients of x² and x, we have:

A = -1

B – C = 0

A + B + C = 0

Thus, A = -1, B = 1/2 and C = 1/2

Substituting these values in the original equation, we have:

[Tex]\frac{1}{x(x-1)(x+1)}=\frac{-1}x+\frac{1}{2(x-1)}+\frac{1}{2(x+1)}[/Tex]

Hence, [Tex]y=\frac{1}2log[\frac{k^2(x-1)(x+1)}{x^2}][/Tex]

Now, y = 0 when x = 2.

[Tex]⇒0=\frac{1}2log[\frac{k^2(2-1)(2+1)}{2^2}]\\ ⇒log[\frac{3k^2}4]=0\\ ⇒k^2=\frac{4}3\\ Thus, \space y=\frac{1}2log[\frac{x^2-1}{x^2}]-\frac{1}2log\frac{3}4[/Tex]

13. [Tex]cos(\frac{dy}{dx})=a(a∈R);y=1[/Tex] when x = 0

Solution:

The given equation can be rearranged as:

[Tex]\frac{dy}{dx})=cos^{-1}a\\ ⇒dy=cos^{-1}adx[/Tex]

On integrating both sides, we get,

[Tex]\int dy=\int cos^{-1}adx\\ ⇒y=xcos^{-1}a+C[/Tex]

Now, y = 1 when x = 0

⇒ 1 = 0.cos^-1 a + C

⇒ C = 1

Thus, y = xcos^-1a + 1

⇒ [Tex]cos[\frac{y-1}x]=a[/Tex]

14. [Tex]\frac{dy}{dx}=y\space tan\space x; y= 1[/Tex] when x = 0

Solution:

The given equation can be rearranged as:

[Tex]\frac{dy}y=tan\space xdx[/Tex]

On integrating both sides, we get,

[Tex]\int\frac{dy}y=\int tan\space xdx[/Tex]

⇒ log y = log(sec x) + log C

⇒ log y = log(sec x.C)

⇒ y = C sec x

Now, y = 1 when x = 0

⇒ 1 = C x sec 0

⇒ 1 = C x 1

⇒ C = 1

Thus, y = sec x

15. Find the equation of a curve passing through the point (0, 0) and whose differential equation is y′ = e^xsin x.

Solution:

The given equation can be rearranged as:

dy = e^x sin x dx

On integrating both sides, we get,

[Tex]\int dy=\int e^xsinxdx\\ ⇒y=sinx\int e^xdx-\int[\frac{d}{dx}(sinx).\int e^xdx]dx\\ ⇒y=sinx.e^x-\int cosx.e^xdx\\ ⇒y=\frac{e^x(sinx-cosx)}2+C[/Tex]

Since the curve passes through the origin, we have:

[Tex]⇒0=\frac{e^0(sin0-cos0)}2+C[/Tex]

⇒ C = 1/2

Thus, the required equation is 2y – 1 = e^x(sin x – cos x).

16. For the differential equation [Tex]xy\frac{dy}{dx}=(x+2)(y+2)[/Tex], find the solution curve passing through the point (1, –1).

Solution:

The given equation can be rearranged as:

[Tex][\frac{y}{y+2}]dy=[\frac{x+2}{x}]dx[/Tex]

On integrating both sides, we get,

[Tex]\int[\frac{y}{y+2}]dy=\int[\frac{x+2}{x}]dx\\ ⇒\int dy-2\int\frac{1}{y+2}dy=\int dx+2\int\frac{1}xdx [/Tex]

⇒ y – 2log(y + 2) = x + 2 log x + C

⇒ y – x – C = log x² + log(y + 2)²

⇒ y – x – C = log[x²(y + 2)²]

Since the curve passes through (1, –1), we have:

1 – 1 – C = log[1²(-1 + 2)²]

⇒ C = -2

Thus, y – x + 2 = log[x²(y + 2)²] is the required equation.

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.

Solution:

Since, [Tex]y.\frac{dy}{dx}=x\\ ⇒ydy=xdx[/Tex]

On integrating both sides, we get,

[Tex]⇒\int ydy=\int xdx\\ ⇒\frac{y^2}2=\frac{x^2}2+C[/Tex]

⇒ y² – x² = 2C

Since the curve passes through (0, –2), we have:

(-2)² – 0² = 2C

C = 2

Thus, y² – x² = 4 is the required equation.

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 m₂ is tangent to dy/dx.

Since, m₂ = 2m₁, we have:

[Tex]\frac{dy}{dx}=\frac{2(y+3)}{x+4}\\ ⇒\frac{dy}{y+3}=\frac{2(dx)}{x+4}[/Tex]

On integrating both sides, we get,

[Tex]\int\frac{dy}{y+3}=2\int\frac{dx}{x+4}[/Tex]

⇒ log(y + 3) = 2 log(x + 4) + log C

⇒ log(y + 3) = log C(x + 4)²

⇒ y + 3 = C(x + 4)²

Since the curve passes through (–2, 1), we have:

1 + 3 = C(-2 + 4)²

⇒ C = 1

Thus, y + 3 = (x + 4)² is the required equation.

19. The volume of 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 balloon after t seconds.

Solution:

Suppose the rate of change of volume of balloon be k.

[Tex]⇒\frac{dV}{dt}=k\\ ⇒\frac{d}{dt}[\frac{4}3\pi r^3]=k\\ ⇒\frac{4}3\pi .3r^2.\frac{dr}{dt}=k\\ ⇒4\pi r^2dr=kdt[/Tex]

On integrating both sides, we get,

[Tex]⇒\int4\pi r^2dr=\int kdt\\ ⇒4\pi\frac{r^3}3=kt +C[/Tex]

⇒ 4πr³ = 3(kt + C)

At t = 0, r = 3, we have:

⇒ 4π x 27 = 3(k x 0 + C)

⇒ C = 36π

At t = 3 and r = 6, we have:

⇒ 4π x 216 = 3(k x 3 + C)

⇒ C = 84π

Hence, 4πr³ = 3(84πt + 36π)

4πr³ = 4π(63t + 27)

r = (63t + 27)^1/3

20. In a bank, principal increases continuously at the rate of r% per year. Find the value of r if Rs 100 double itself in 10 years (log_e2 = 0.6931).

Solution:

Given: [Tex]\frac{dp}{dt}=[\frac{r}{100}]p\\ ⇒\frac{dp}{p}=[\frac{r}{100}]dt[/Tex]

On integrating both sides, we get,

[Tex]\int\frac{dp}{p}=[\frac{r}{100}]\int dt\\ ⇒log p=\frac{rt}{100}+k\\ ⇒p=e^{\frac{rt}{100}+k}[/Tex]

Now, p = 100 when t = 0.

⇒ 100 = e^k

If t = 10 then p = 2 x 100 = 200.

[Tex]⇒200=e^{\frac{r}{10}+k}\\ ⇒200=e^\frac{r}{10}.e^k\\ ⇒200=e^\frac{r}{10}.100\\ ⇒e^\frac{r}{10}=2\\ ⇒\frac{r}{10}=log_e2\\ ⇒\frac{r}{10}=0.6931[/Tex]

⇒ r = 6.931%

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 worth after 10 years(e^0.5 = 1.648).

Solution:

Given: [Tex]\frac{dp}{dt}=\frac{5}{100}p\\ ⇒\frac{dp}{p}=\frac{dt}{20}[/Tex]

On integrating both sides, we get,

[Tex]\int\frac{dp}{p}=\frac{1}{20}\int dt\\ ⇒logp=\frac{t}{20}+C\\ ⇒p=e^{\frac{t}{20}+C}[/Tex]

Now, p = 1000 when t = 0.

Thus, 1000 = e^C

Now, when t = 10 and e^C = 1000.

[Tex]⇒p=e^{\frac{10}{20}+C}[/Tex]

⇒ p = e^0.5 x e^C

⇒ p = 1.648 x 1000

⇒ p = 1648

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 the number of bacteria in the culture at instant t be y.

Given:

[Tex]\frac{dy}{dt}=ky\\ ⇒\frac{dy}y=kdt[/Tex]

On integrating both sides, we get,

[Tex]\int\frac{dy}y=k\int dt[/Tex]

⇒log y = kt + C

At t = 0,

[Tex]log\space y_0=C\\ ⇒log[\frac{y}{y_0}]=kt[/Tex]

Given that the number is increased by 10% in 2 hours.

[Tex]y=\frac{110}{100}y_0\\ ⇒\frac{y}{y_0}=\frac{11}{10}\\ ⇒log[\frac{y}{y_0}]=log[\frac{11}{10}]\\ ⇒kt=log[\frac{11}{10}]\\ ⇒k=\frac{1}2log[\frac{11}{10}]\\ ⇒t=\frac{2log[\frac{y}{y_0}]}{log[\frac{11}{10}]}\\ ⇒t_1=\frac{2log[\frac{y}{y_0}]}{log[\frac{11}{10}]}\\ ⇒t_1=\frac{2log[\frac{200000}{100000}]}{log[\frac{11}{10}]}\\ ⇒t_1=\frac{2log2}{log[\frac{11}{10}]}[/Tex]

23. The general solution of the differential equation 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

Solution:

[Tex]\frac{dy}{dx} = e^x \cdot e^y[/Tex]

Now, we can separate variables:

[Tex]\frac{dy}{e^y} = e^x dx[/Tex]

Integrating both sides:

[Tex]\int \frac{1}{e^y} dy = \int e^x dx[/Tex]

[Tex]-e^{-y} = e^x + C[/Tex]

e^x + e^-y = -C

So, the correct option is (A)

Related Articles-

• Differential Equations
• Chapter 9 Differential Equations-Exercise -9.1
• Chapter 9 Differential Equations-Exercise -9.2