RD Sharma Class 12 Exercise 21.8 Differential Equation solutions Maths

RD Sharma Class 12 Exercise 21.8 Differential Equation Solutions Maths - Download PDF Free Online

Updated on Jan 24, 2022 04:20 PM IST

The class 12 students need much focus and guidance in each subject while preparing for their exams. Jumping from one solution book to another for each topic or concept would be hectic. Especially when mathematics comes into play, the students feel hard to solve the sums in the Differential Equations concept. Authorized reference materials like the RD Sharma Class 12th Exercise 21.8 books are a boon for these students.

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RD Sharma Class 12 Solutions Chapter21 Differential Equation - Other Exercise
Differential Equations Excercise: 21.8
RD Sharma Chapter wise Solutions

RD Sharma Class 12 Solutions Chapter21 Differential Equation - Other Exercise

Differential Equations Excercise: 21.8

Differential equations exercise 21.8 question 1

Answer: $\tan^{- 1} (x + y + 1) = x + c$
Given: $\frac{d y}{d x} = {(x + y + 1)}^{2}$

To find :- solve the differential equation.

Hint :- differential equation of the form

$\frac{d y}{d x} = (a x + b y + c)$ can be reduced to variable separable form by the substitution $a x + b y + c = v$

Solution:-we have

$\frac{d y}{d x} = {(x + y + 1)}^{2}$ ........(i)

$L e t x + y + 1 = v$

Differentiating with respect to x, we get

$\begin{aligned} \Rightarrow 1 + \frac{d y}{d x} = \frac{d v}{d x} \\ \Rightarrow \frac{d y}{d x} = \frac{d v}{d x} - 1 \end{aligned}$ ....(ii)

Now, substituting equation (ii) in equation (i), we get,
$\begin{aligned} \frac{d v}{d x} - 1 = (x + y + 1)^{2} \\ \Rightarrow & \frac{d v}{d x} - 1 = v^{2} (∵ x + y + 1 = v) \\ \Rightarrow & \frac{d v}{d x} = v^{2} + 1 \\ \Rightarrow & \frac{d v}{v^{2} + 1} = d x \end{aligned}$ ( taking like variable on same side)
Integrating on both sides, we get,
$\int \frac{d v}{v^{2} + 1} = \int d x$
$\Rightarrow \tan^{- 1} v = x + c$ $(∵ \int \frac{d x}{a^{2 +} x^{2}} = \frac{1}{a} \tan^{- 1} (\frac{x}{a}))$
Putting the value of V, we get,
$\Rightarrow \tan^{- 1} (x + y + 1) = x + c$ , which is the required solution.

Differential equations exercise 21.8 question 2

Answer: $\cot \frac{x - y}{2} = y + C_{1}$
Gievn:- $\frac{d y}{d x} \cot (x - y) = 1$

To find :- solve the differential equation.

Hint :- We will reduce the differential equation to variable separable form by substitution .

Solution ;- We have,

$\frac{d y}{d x} \cot (x - y) = 1$ .....(i)

Let $x - y = v$

Differentiating with respect to x, we get,

$\frac{d}{d x} (x - y) = \frac{d}{d x} (v)$

$\Rightarrow 1 - \frac{d y}{d x} = \frac{d v}{d x}$

$\Rightarrow \frac{d y}{d x} = 1 - \frac{d v}{d x}$ (transposing)

Putting the value of $\frac{d y}{d x}$ in Equ. (i), we get

$\begin{aligned} (1 - \frac{d v}{d x}) \cos v = 1 (∵ x - y = v) \\ \Rightarrow 1 - \frac{d v}{d x} = \frac{1}{\cos v} \\ \Rightarrow 1 - \frac{d v}{d x} = \sec v \\ \Rightarrow \frac{d v}{d x} = 1 - \sec v \end{aligned}$

Now, taking like variable in same side, we get,

$\begin{aligned} \Rightarrow \frac{d v}{1 - \sec v} = d x \\ \Rightarrow dx = \frac{d v}{1 - \sec v} \\ \Rightarrow dx = (\frac{\cos v}{\cos v - 1}) d v \end{aligned}$

Integrating in both side, we get,

$\begin{aligned} \int d x = \int - (\frac{\cos v}{1 - \cos v}) d v \\ \Rightarrow x = - \int \frac{\cos v}{1 - Cos v} \times \frac{1 + \cos v}{1 + \cos v} d v \\ \Rightarrow x = - \int \frac{\cos v + \cos^{2} v}{1 - \cos^{2} v} d v \\ \Rightarrow x = - \int (\frac{\cos v + \cos^{2} v}{\sin^{2} v}) d v \\ \Rightarrow x = - \int (\frac{\cos v}{\sin^{2} v} + \frac{\cos^{2} v}{\sin^{2} v}) d v \\ \Rightarrow x = - \int (\cot v Cosec v + \cot^{2}) d v \end{aligned}$

$\begin{aligned} \Rightarrow x = - \int (\cot v cosec v + {cosec}^{2} v - 1) d v \\ \Rightarrow x = - (- Cosec v - \cot v - v) + C \\ \Rightarrow x = Cosec v + \cot v + v + C_{1} \\ \Rightarrow x = \frac{1}{\sin v} + \frac{cosv}{\sin v} + v + C_{1} \\ \Rightarrow x - v - C_{1} = \frac{1 + cosv}{\sin v} \end{aligned}$

$\begin{aligned} \Rightarrow x - v - C_{1} = \frac{2 {los}^{2} \frac{v}{2}}{2 \cdot 8 s_{2}^{T} \sin \frac{q}{2}} \\ \Rightarrow x - v - C_{1} = \cot \frac{v}{2} \end{aligned}$ $(put C = C_{1} and using 1 + Cos 2 x = 2 {Cos}^{2} x; \sin x = 2 \sin \frac{x}{2} \cos \frac{x}{2})$

Putting $v = x - y$

$\Rightarrow x - / x + / y + C_{1} = \cot (\frac{x - y}{2})$

$\Rightarrow \cot (\frac{x - y}{2}) = y + C_{1}$

(This is the required solution)

Differential equations exercise 21.8 question 3

Answer: $2 (x - y) + l o g (x - y + 2) = x + C$
Given: $\frac{d y}{d x} = \frac{(x - y) + 3}{2 (x - y) + 5}$
Hint : - first, we will separate the variables and then solve .
Solution:

\frac{d y}{d x} = \frac{(x - y) + 3}{2 (x - y) + 5}

Let

x - y = v

Differentiating with respect to x , we get,

\begin{aligned} \frac{d}{d x} (x - y) = \frac{d v}{d x} \\ \Rightarrow 1 - \frac{d y}{d x} = \frac{d v}{d x} \\ \Rightarrow \frac{d y}{d x} = 1 - \frac{d v}{d x} \end{aligned}

Putting $\frac{d y}{d x} = 1 - \frac{d v}{d x}$ and

x - y = v

in equation (i), we get,

\begin{aligned} 1 - \frac{d v}{d x} = \frac{v + 3}{2 v + 5} \\ \Rightarrow \frac{d v}{d x} = 1 - (\frac{v + 3}{2 v + 5}) \\ \Rightarrow \frac{d v}{d x} = \frac{2 v + 5 - v - 3}{2 v + 5} \\ \Rightarrow \frac{d v}{d x} = \frac{v + 2}{2 v + 5} \end{aligned}

Taking like variables in the same side,

\Rightarrow \frac{2 v + 5}{v + 2} d v = d x

Now, integrating in both sides, we get,

\Rightarrow \int \frac{2 v + 5}{v + 2} d v = d x

\begin{aligned} \Rightarrow \int (\frac{2 v + 4 + 1}{v + 2}) d v = \int d x \\ \Rightarrow \int (\frac{2 (v + 2)}{v + 2} + \frac{1}{v + 2}) d v = \int d x \\ \Rightarrow \int (2 + \frac{1}{v + 2}) d v = \int d x \\ \Rightarrow 2 v + \log | v + 2 | = x + c \end{aligned}

Putting

v = x - y

\Rightarrow 2 (x - y) + l o g | x - y + 2 | = x + c

( this is the required solution).

Differential equations exercise 21.8 question 4

Answer:- $x + y = \tan (x + c)$
Given: $\frac{d y}{d x} = {(x + y)}^{2}$
Hint : - Differential equation of the form $\frac{d y}{d x} = \int (a x + b y + c)$ can be reduced to variable separable form by substitution

a x + b y + c = v

Solution : - We have,

$\frac{d y}{d x} = {(x + y)}^{2}$

Let $x + y = v$

Differentiating with respect to x, we get,

$\begin{aligned} \frac{d}{d x} (x + y) = \frac{d}{d x} (v) \\ \Rightarrow 1 + \frac{d y}{d x} = \frac{d v}{d x} \\ \Rightarrow \frac{d y}{d x} = \frac{d v}{d x} - 1 \end{aligned}$

Putting the value of $\frac{d y}{d x}$ in equation (i), we get

$\frac{d v}{d x} - 1 = (x + y)^{2}$

$\Rightarrow \frac{d v}{d x} - 1 = v^{2}$ $(∵ x + y = v)$

$\Rightarrow \frac{d v}{d x} = v^{2} + 1$

$\Rightarrow \frac{d v}{v^{2} + 1} = d x$
Integrating on both the sides, we get,

$\int \frac{d v}{v^{2} + 1} = \int d x$

$\Rightarrow \tan^{- 1} v = x + c$ $(∵ \frac{d x^{2}}{x^{2} + 1} = \tan^{- 1} x)$

Putting $v = x + y,$ we get

$\Rightarrow \tan^{- 1} (x + y) = x + c$

$\Rightarrow x + y = \tan (x + c)$

(This is the required solution).

Differential equations exercise 21.8 question 5

Answer: $y - \tan^{- 1} (x + y) = c$
Given: ${(x + y)}^{2} \frac{d y}{d x} = 1$

Hint : - Use variable separable form by substitution.

Solution : - We have,

${(x + y)}^{2} \frac{d y}{d x} = 1$ ......(i)

Let $x + y = v$

Differentiating with respect to x, we get,

$\frac{d}{d x} (x + y) = \frac{d v}{d x}$

$\Rightarrow 1 + \frac{d y}{d x} = \frac{d v}{d x}$

$\Rightarrow \frac{d y}{d x} = \frac{d v}{d x} - 1$ .....(ii)

Now, substituting equation (ii) in equation (i), we get

$(x + y)^{2} (\frac{d v}{d x} - 1) = 1$

$\Rightarrow v^{2} (\frac{d v}{d x} - 1) = 1$

$\Rightarrow \frac{d v}{d x} - 1 = \frac{1}{v^{2}}$

$\Rightarrow \frac{d v}{d x} = \frac{1}{v^{2}} + 1$

$\Rightarrow \frac{d v}{d x} = \frac{1 + v^{2}}{v^{2}}$

Taking like variables on same side, we get,

$\Rightarrow \frac{v^{2}}{1 + v^{2}} d v = d x$

Integrating on both the sides, we get,

$\int (\frac{v^{2}}{1 + v^{2}}) d v = \int d x$

$\Rightarrow \int (\frac{v^{2} + 1 - 1}{1 + v^{2}}) d v = \int d x$

$\Rightarrow \int (\frac{v^{2} + 1}{v^{2} + 1} - \frac{1}{v^{2} + 1}) d v = \int d x$

$\Rightarrow \int (1 - \frac{1}{v^{2} + 1}) d v = \int d x$

$\Rightarrow \int d v - \int \frac{1}{v^{2} + 1} d v = \int d x$

$\Rightarrow v - \tan^{- 1} v = x + c (∵ \int (\frac{1}{x^{2} + 1}) d x = \tan^{- 1} x)$

Putting $v = x + y,$ we have

$\Rightarrow (x + y) - \tan^{- 1} (x + y) = x + c$

$\Rightarrow y - \tan^{- 1} (x + y) = c$

(This is the required solution).

Differential equations exercise 21.8 question 6

Answer: $x = \tan (x - 2 y) + c$
Given: $\cos^{2} (x - 2 y) = 1 - 2 \frac{d y}{d x}$
Hint : - first, we will separate variables and then solve.
Solution : - We have,

$\cos^{2} (x - 2 y) = 1 - 2 \frac{d y}{d x}$ .....(i)

Let $x - 2 y = v$
Differentiating with respect to x, we get,

$\frac{d}{d x} (x - 2 y) = \frac{d v}{d x}$

$\Rightarrow 1 - 2 \frac{d y}{d x} = \frac{d v}{d x}$

$\Rightarrow 2 \frac{d y}{d x} = 1 - \frac{d v}{d x}$
Now, substituting equation (ii) in equation (i), we get

$\begin{aligned} \cos^{2} (x - 2 y) = 1 - (1 - \frac{d v}{d x}) \\ \Rightarrow & \cos^{2} v = 1 - 1 + \frac{d v}{d x} (∵ v = x - 2 y) \\ \Rightarrow & \cos^{2} v = \frac{d v}{d x} \end{aligned}$
Taking like variables on same side, we get,

$\Rightarrow d x = \frac{d v}{\cos^{2} v}$

$\Rightarrow d x = \sec^{2} v d v$
Integrating on both the sides, we get,

$\int d x = \int \sec^{2} v d v$

$\Rightarrow x = \tan (x - 2 y) + c$
(This is the required solution).

Differential equations exercise 21.8 question 7

Answer: $y = \tan (\frac{x + y}{2}) + c$
Given: $\frac{d y}{d x} = \sec (x + y)$

Hint : - first, we will separate variables and then solve.

Solution : - We have,

$\frac{d y}{d x} = \sec (x + y)$

Let $x + y = v$

Differentiating with respect to x, we get,

$\begin{aligned} \frac{d}{d x} (x + y) = \frac{d v}{d x} \\ \Rightarrow 1 + \frac{d y}{d x} = \frac{d v}{d x} \\ \Rightarrow \frac{d y}{d x} = \frac{d v}{d x} - 1 \end{aligned}$

Putting x + y = v and dydx = dvdx -1 in the given differential equation, we get,

$∴ \frac{d v}{d x} - 1 = \sec v$

$\Rightarrow \frac{d v}{d x} - 1 = \frac{1}{\cos v}$

$\Rightarrow \frac{d v}{d x} = \frac{\cos v + 1}{\cos v}$

Taking like variables on same side, we get,

$\Rightarrow \frac{\cos v}{\cos v + 1} d v = d x$

$\Rightarrow \frac{\cos v (1 - \cos v)}{(1 + \cos v) (1 - \cos v)} d v = d x$

$\Rightarrow \frac{\cos v (1 - \cos v)}{1 - \cos^{2} v} d v = d x$

$\Rightarrow \frac{\cos v (1 - \cos v)}{\sin^{2} v} d v = d x$

Integrating on both the sides, we get,

$\int \frac{\cos v (1 - \cos v)}{\sin^{2} v} d v = \int d x$

$\Rightarrow \int (\frac{\cos v}{\sin^{2} v} - \frac{\cos^{2} v}{\sin^{2} v}) d v = \int d x$

$\Rightarrow \int (cotv cosec v - \cot^{2} v) d v = \int d x$

$\Rightarrow \int (cotv cosec v d v - ({cosec}^{2} v - 1) d v = \int d x (∵ u sing \cot^{2} v = {cosec}^{2} v - 1)$

$\Rightarrow - cosec v + \cot v + v) = x + c$

Putting $v = x + y$ ,we have

$\Rightarrow - cosec (x + y) + \cot (x + y) + x + y = x + c$

$\Rightarrow - cosec (x + y) + \cot (x + y) + y = c$

$\Rightarrow - \frac{1}{\sin (x + y)} + \frac{\cos (x + y)}{\sin (x + y)} + y = c$

$\Rightarrow - (\frac{1 - \cos (x + y)}{\sin (x + y)}) + y = c$

$\Rightarrow - \frac{2 \sin^{2} \frac{x + y}{2}}{2 \cos \frac{x + y}{2} \sin \frac{x + y}{2}} + y = c$

$\Rightarrow - \tan \frac{(x + y)}{2} + y = c$

$\Rightarrow y = \tan \frac{(x + y)}{2} + c$

(This is the required solution)

Differential equations exercise 21.8 question 8

Answer: $y - x + l o g | \sin (x + y) + \cos (x + y) | = c$

Given: $\frac{d y}{d x} = \tan (x + y)$

Hint : - first, we will separate variables and then solve.

Solution : - We have,

$\frac{d y}{d x} = \tan (x + y)$

$L e t x + y = v$

Differentiating with respect to x, we get,

$\frac{d}{d x} (x + y) = \frac{d v}{d x}$

$\Rightarrow 1 + \frac{d y}{d x} = \frac{d v}{d x}$

$\Rightarrow \frac{d y}{d x} = \frac{d v}{d x} - 1$

Putting $x + y = v$ and $\frac{d y}{d x} = \frac{d v}{d x} - 1$ in the given differential equation, we get,

$∴ \frac{d y}{d x} = \frac{d v}{d x} - 1 = \tan v$

$\Rightarrow \frac{d v}{d x} = 1 + \tan v$

Taking like variables on same side, we get,

$\begin{aligned} \Rightarrow \frac{d v}{1 + \tan v} = d x \\ \Rightarrow (\frac{1}{1 + \frac{\sin v}{\cos v}}) d v = d x \\ \Rightarrow (\frac{\cos v}{\cos v + \sin v}) d v = d x \end{aligned}$

Multiplying 2 on both the sides, we get,

$\begin{aligned} (\frac{2 \cos v}{\cos v + \sin v}) d v = 2 d x \\ \Rightarrow (1 + \frac{\cos v - \sin v}{\cos v + \sin v}) d v = 2 d x \end{aligned}$

Integrating on both sides, we get,

$\begin{aligned} \Rightarrow \int (1 + \frac{\cos v - \sin v}{\cos v + \sin v}) d v = 2 d x \\ \Rightarrow \int d v + \int \frac{\cos v - \sin v}{\cos v + \sin v} d v = 2 d x \\ \Rightarrow v + \log | \cos v + \sin v | = 2 x + c \end{aligned}$

Putting v = x + y, we get,

$\begin{aligned} \Rightarrow (x + y) + \log ∣ \cos (x + y) + \sin (x + y) = 2 x + c \\ \Rightarrow y - x + \log | \cos (x + y) + \sin (x + y) | = c \end{aligned}$

(This is the required solution).

Differential equations exercise 21.8 question 9

Answer: $\frac{1}{2} (y - x) + \frac{1}{2} l o g | x + y | = c$
Given: $(x + y) (d x - d y) = d x + d y$

Hint : - first, we will separate variables and then solve.

Solution : - We have,

$(x + y) (d x - d y) = d x + d y$

$\Rightarrow x d x - x d y + y d x - y d y = d x + d y$

$\Rightarrow x d x + y d x - d x = d y + x d y + y d y$

$\Rightarrow (x + y - 1) d x = (1 + x + y) d y$

$\Rightarrow \frac{dy}{dx} = \frac{x + y - 1}{x + y + 1}$

Let $x + y = v$

Differentiating with respect to x, we get,

$\frac{d}{d x} (x + y) = \frac{d v}{d x}$

$\Rightarrow 1 + \frac{d y}{d x} = \frac{d v}{d x}$

$\Rightarrow \frac{d y}{d x} = \frac{d v}{d x} - 1$

Substituting (ii) in equation (i), we get,

$∴ \frac{d v}{d x} - 1 = \frac{x + y - 1}{x + y + 1}$

$\Rightarrow \frac{d v}{d x} - 1 = \frac{v - 1}{v + 1} [∵ x + y = v]$

$\Rightarrow \frac{d v}{d x} = \frac{v - 1}{v + 1} + 1$

$\Rightarrow \frac{d v}{d x} = \frac{v - 1 + v + 1}{v + 1}$

$\Rightarrow \frac{d v}{d x} = \frac{2 v}{v + 1}$

Taking like variables on the same side, we get,

$\Rightarrow \frac{v + 1}{2 v} d v = d x$

Integrating on both sides, we get,

$\begin{aligned} \Rightarrow \int \frac{v + 1}{2 v} d v = \int d x \\ \Rightarrow \frac{1}{2} \int (1 + \frac{1}{v}) d v = \int d x \\ \Rightarrow \frac{1}{2} \int (v + \log | v |) = x + c \end{aligned}$
Putting v = x + y, we get,
$\begin{aligned} \Rightarrow \frac{1}{2} (x + y + \log | x + y |) = x + c \\ \Rightarrow (x + y + \log | x + y |) = 2 (x + c) \\ \Rightarrow (x + y + \log | x + y |) = 2 x + 2 c \\ \Rightarrow y - x + \log | x + y | = 2 c \\ \Rightarrow \frac{1}{2} (y - x) + \frac{1}{2} \log | x + y | = c \end{aligned}$
(This is the required solution).

Differential equations exercise 21.8 question 10

Answer: $x = c e^{y} y - 2$
Given: $(x + y + 1) \frac{d y}{d x} = 1$
Hint : - Differential equation of the form

\frac{d y}{d x} = \int (a x + b y + c)

can be reduced to variable separable form by substitution

a x + b y + c = v

Solution : - We have,
$(x + y + 1) \frac{d y}{d x} = 1$ ........(i)
Let $x + y = v$
Differentiating with respect to x, we get,

\frac{d}{d x} (x + y) \frac{d v}{d x}

\Rightarrow 1 + \frac{d}{d x} = \frac{d v}{d x}

\Rightarrow \frac{d}{d x} = \frac{d v}{d x} - 1

Putting dydx = dvdx -1 and x + y = v in equation (i), we get,

$\begin{aligned} (v + 1) (\frac{d v}{d x} - 1) = 1 \\ \Rightarrow (v + 1) \frac{d v}{d x} - (v + 1) = 1 \\ \Rightarrow (v + 1) \frac{d v}{d x} = 1 + 1 + v \\ \Rightarrow (v + 1) \frac{d v}{d x} = 2 + v \end{aligned}$

Taking like variables on the same side, we get,

$\Rightarrow \frac{v + 1}{v + 2} d v = d x$

Integrating on both sides, we get,

$\Rightarrow \int \frac{v + 1}{v + 2} d v = \int d x$

$\Rightarrow \int (1 - \frac{1}{v + 2}) d v = \int d x$

$\Rightarrow (v - \log | v + 2 |) = x + \log c_{1}$

Putting v = x + y, we get,

$\Rightarrow x + y - \log | x + y + 2 | = x + \log c_{1}$

$\Rightarrow x + y - x = \log | x + y + 2 | + \log c_{1}$

$\Rightarrow y = \log c_{1} | x + y + 2 |$

$\Rightarrow e^{y} = c_{1} (x + y + 2) (∵ c = \frac{1}{c_{1}})$

$\Rightarrow x = c e^{y} - y - 2$
(This is the required solution).

Differential equations exercise 21.8 question 11

Answer: $- e^{- (x + y)} = x + c$
Given: $\frac{d y}{d x} + 1 = e^{x + y}$
Hint: Differential equation of the form

\frac{d y}{d x} = \int (a x + b y + c)

can be reduced to variable separable form by substitution

a x + b y + c = v

Solution: We have,
$\frac{d y}{d x} + 1 = e^{x + y}$ .......(i)
Let $x + y = v$
Differentiating with respect to x, we get,

\frac{d}{dx} (x + y) = \frac{d v}{d x}

\Rightarrow 1 + \frac{d y}{d x} = \frac{d v}{d x} \dots (ii)

Substituting (ii) in equation (i), we get,

\frac{d v}{d x} = e^{x + y}

\Rightarrow \frac{d v}{d x} = e^{v} [∵ x + y = v]

Taking like variables on the same side, we get,

\Rightarrow \frac{d v}{e^{v}} = d x

\Rightarrow e^{- v} d v = d x

Integrating on both sides, we get,

\Rightarrow \int e^{- v} d v = \int d x

\Rightarrow - e^{- v} = x + c

Putting

v = x + y

, we get,

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

(This is the required solution).

The class 12 mathematics portion consists of Differential Equations as its 21st chapter. There are around eleven exercises in the chapter, mostly revolving around the same concept in different methods. The eighth exercise, ex 21.8, has eleven questions present in the textbook. Again, the concept of this exercise is to solve the given differential equations. The sums are given under Level 1, and there is no Level 2 part. The students can use the RD Sharma Class 12 Chapter 21 Exercise 21.8 reference material for answer verification and clarify the other doubts they possess.

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