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

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

Edited By Lovekush kumar saini | 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.

## Differential Equations Excercise: 21.8

Differential equations exercise 21.8 question 1

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

To find :- solve the differential equation.

Hint :- differential equation of the form

$\frac{dy}{dx}=\left ( ax+by+c \right )$can be reduced to variable separable form by the substitution $ax+by+c=v$

Solution:-we have

$\frac{dy}{dx}=\left ( x+y+1 \right )^{2}$ ........(i)

$Let\: x+y+1=v$

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} \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \quad(\because 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,
$\int \frac{dv}{v^{2}+1}=\int dx$
$\Rightarrow \tan ^{-1}v=x+c$ $\left ( \because \int\frac{dx}{a^{2+}x^{2}}=\frac{1}{a}\tan ^{-1}\left ( \frac{x}{a} \right )\right )$
Putting the value of V, we get,
$\Rightarrow \tan ^{-1}\left ( x+y+1 \right )=x+c$ , which is the required solution.

Differential equations exercise 21.8 question 2

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

To find :- solve the differential equation.

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

Solution ;- We have,

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

Let $x-y=v$

Differentiating with respect to x, we get,

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

$\Rightarrow 1-\frac{dy}{dx}=\frac{dv}{dx}$

$\Rightarrow \frac{dy}{dx}=1-\frac{dv}{dx}$ (transposing)

Putting the value of $\frac{dy}{dx}$ in Equ. (i), we get

\begin{aligned} &\left(1-\frac{d v}{d x}\right) \cos v=1 \; \; \; \; \; \; \; \; \; \; \; \; \; \quad(\because \mathrm{x}-\mathrm{y}=\mathrm{v}) \\ &\Rightarrow \quad 1-\frac{d v}{d x}=\frac{1}{\cos v} \\ &\Rightarrow \quad 1-\frac{d v}{d x}=\operatorname{sec} \mathrm{v} \\ &\Rightarrow \frac{d v}{d x}=1-\operatorname{sec} \mathrm{v} \end{aligned}

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

\begin{aligned} &\Rightarrow \frac{d v}{1-\sec v}=d x \\ &\Rightarrow \mathrm{dx}=\frac{d v}{1-\sec v} \\ &\Rightarrow \mathrm{dx}=\left(\frac{\cos v}{\cos v-1}\right) \mathrm{d} \mathrm{v} \end{aligned}

Integrating in both side, we get,

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

\begin{aligned} &\Rightarrow \mathrm{x}=-\int\left(\cot v \operatorname{cosec} v+\operatorname{cosec}^{2} v-1\right) d v \\ &\Rightarrow \mathrm{x}=-(-\operatorname{Cosec} v-\cot v-v)+C \\ &\Rightarrow \mathrm{x}=\operatorname{Cosec} \mathrm{v}+\cot \mathrm{v}+\mathrm{v}+\mathrm{C}_{1} \\ &\Rightarrow \mathrm{x}=\frac{1}{\sin \mathrm{v}}+\frac{\operatorname{cosv}}{\sin \mathrm{v}}+\mathrm{v}+\mathrm{C}_{1} \\ &\Rightarrow \mathrm{x}-\mathrm{v}-\mathrm{C}_{1}=\frac{1+\operatorname{cosv}}{\sin \mathrm{v}} \end{aligned}

\begin{aligned} &\Rightarrow \mathrm{x}-\mathrm{v}-\mathrm{C}_{1}=\frac{2 \operatorname{los}^{2} \frac{\mathrm{v}}{2}}{2 \cdot 8 \mathrm{~s}_{2}^{\mathrm{T}} \sin \frac{\mathrm{q}}{2}} \\ &\Rightarrow \mathrm{x}-\mathrm{v}-\mathrm{C}_{1}=\operatorname{cot} \frac{\mathrm{v}}{2} \end{aligned}$\text { (put } \left.\mathrm{C}=\mathrm{C}_{1} \text { and using } 1+\operatorname{Cos} 2 \mathrm{x}=2 \operatorname{Cos}^{2} \mathrm{x} ; \sin \mathrm{x}=2 \sin \frac{\mathrm{x}}{2} \cos \frac{\mathrm{x}}{2}\right)$

Putting $v=x-y$

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

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

(This is the required solution)

Differential equations exercise 21.8 question 3

Answer: $2\left ( x-y \right )+log\left ( x-y+2 \right )=x+C$
Given:$\frac{dy}{dx}=\frac{\left ( x-y \right )+3}{2\left ( x-y \right )+5}$
Hint : - first, we will separate the variables and then solve .
Solution:$\frac{dy}{dx}=\frac{\left ( x-y \right )+3}{2\left ( x-y \right )+5}$
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 \quad 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 $\frac{dy}{dx}=1-\frac{dv}{dx}$ 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-\left(\frac{v+3}{2 v+5}\right) \\ &\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,
$\Rightarrow \frac{2v+5}{v+2}dv=dx$
Now, integrating in both sides, we get,
$\Rightarrow\int \frac{2v+5}{v+2}dv=dx$
\begin{aligned} &\Rightarrow \int\left(\frac{2 v+4+1}{v+2}\right) \mathrm{d} \mathrm{v}=\int \mathrm{d} \mathrm{x} \\ &\Rightarrow \int\left(\frac{2(v+2)}{v+2}+\frac{1}{\mathrm{v}+2}\right) \mathrm{d} \mathrm{v}=\int \mathrm{d} \mathrm{x} \\ &\Rightarrow \quad \int\left(2+\frac{1}{v+2}\right) \mathrm{d} \mathrm{v}=\int \mathrm{d} \mathrm{x} \\ &\Rightarrow 2 \mathrm{v}+\log |\mathrm{v}+2|=\mathrm{x}+\mathrm{c} \end{aligned}
Putting $v=x-y$
$\Rightarrow 2\left ( x-y \right )+log |x-y+2|=x+c$
( this is the required solution).

Differential equations exercise 21.8 question 4

Answer:- $x+y=\tan \left ( x+c \right )$
Given:$\frac{dy}{dx}=\left ( x+y \right )^{2}$
Hint : - Differential equation of the form$\frac{dy}{dx}=\int \left ( ax+by+c \right )$can be reduced to variable separable form by substitution $ax+by+c=v$

Solution : - We have,

$\frac{dy}{dx}=\left ( x+y \right )^{2}$

Let $x+y=v$

Differentiating with respect to x, we get,

\begin{aligned} \; \; \; \; \; \; &\quad \frac{d }{d x}(\mathrm{x}+\mathrm{y})=\frac{d}{d x}(v) \\ &\Rightarrow \quad 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 $\frac{dy}{dx}$ in equation (i), we get

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

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

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

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

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

$\Rightarrow \tan ^{-1}v=x+c$ $\left ( \because \frac{dx^{2}}{x^{2}+1}=\tan ^{-1}x \right )$

Putting $v=x+y,$ we get

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

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

(This is the required solution).

Differential equations exercise 21.8 question 5

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

Hint : - Use variable separable form by substitution.

Solution : - We have,

$\left ( x+y \right )^{2}\frac{dy}{dx}=1$ ......(i)

Let $x+y=v$

Differentiating with respect to x, we get,

$\frac{d}{dx}\left ( x+y \right )=\frac{dv}{dx}$

$\Rightarrow 1+\frac{dy}{dx}=\frac{dv}{dx}$

$\Rightarrow \frac{dy}{dx}=\frac{dv}{dx}-1$ .....(ii)

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

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

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

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

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

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

Taking like variables on same side, we get,

$\Rightarrow \frac{v^{2}}{1+v^{2}}dv=dx$

Integrating on both the sides, we get,

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

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

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

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

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

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

Putting $v=x+y,$ we have

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

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

(This is the required solution).

Differential equations exercise 21.8 question 6

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

$\cos ^{2}\left ( x-2y \right )=1-2\frac{dy}{dx}$ .....(i)

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

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

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

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

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

$\Rightarrow dx=\frac{dv}{\cos ^{2}v}$

$\Rightarrow dx=\sec ^{2}vdv$
Integrating on both the sides, we get,

$\int dx=\int \sec ^{2}v\; dv$

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

Differential equations exercise 21.8 question 7

Answer: $y=\tan \left ( \frac{x+y}{2} \right )+c$
Given:$\frac{dy}{dx}=\sec \left ( x+y \right )$

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

Solution : - We have,

$\frac{dy}{dx}=\sec \left ( x+y \right )$

Let $x+y=v$

Differentiating with respect to x, we get,

\begin{aligned} &\quad \frac{d}{d x}(\mathrm{x}+\mathrm{y})=\frac{d v}{d x} \\ &\Rightarrow \quad 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,

$\therefore \frac{d v}{d x}-1=\sec \mathrm{v}$

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

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

Taking like variables on same side, we get,

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

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

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

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

Integrating on both the sides, we get,

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

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

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

$\Rightarrow \int\left(\operatorname{cotv} \operatorname{cosec} v d v-\left(\operatorname{cosec}^{2} v-1\right) d v=\int d x \quad\left(\because u \operatorname{sing} \cot ^{2} v=\operatorname{cosec}^{2} v-1\right)\right. \\$

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

Putting $v=x+y$,we have

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

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

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

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

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

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

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

(This is the required solution)

Differential equations exercise 21.8 question 8

Answer: $y-x+log|\sin \left ( x+y \right )+\cos \left ( x+y \right )|=c$

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

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

Solution : - We have,

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

$Let\: x+y=v$

Differentiating with respect to x, we get,

$\frac{d}{dx}\left ( x+y \right )=\frac{dv}{dx}$

$\Rightarrow 1+\frac{dy}{dx}=\frac{dv}{dx}$

$\Rightarrow \frac{dy}{dx}=\frac{dv}{dx}-1$

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

$\therefore \frac{dy}{dx}=\frac{dv}{dx}-1=\tan v$

$\Rightarrow \frac{dv}{dx}=1+\tan v$

Taking like variables on same side, we get,

\begin{aligned} &\Rightarrow \frac{d v}{1+\tan v}=\mathrm{d} \mathrm{x} \\ &\Rightarrow\left(\frac{1}{1+\frac{\sin v}{\cos v}}\right) d v=d x \\ &\Rightarrow\left(\frac{\cos v}{\cos v+\sin v}\right) d v=d x \end{aligned}

Multiplying 2 on both the sides, we get,

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

Integrating on both sides, we get,

\begin{aligned} &\Rightarrow \int\left(1+\frac{\cos v-\sin v}{\cos v+\sin v}\right) \mathrm{d} \mathrm{v}=2 \mathrm{~d} \mathrm{x} \\ &\Rightarrow \int \mathrm{d} \mathrm{v}+\int \frac{\cos v-\sin v}{\cos v+\sin v} \mathrm{~d} \mathrm{v}=2 \mathrm{~d} \mathrm{x} \\ &\Rightarrow \mathrm{v}+\log |\operatorname{cos} \mathrm{v}+\sin \mathrm{v}|=2 \mathrm{x}+\mathrm{c} \end{aligned}

Putting v = x + y, we get,

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

(This is the required solution).

Differential equations exercise 21.8 question 9

Answer: $\frac{1}{2}\left ( y-x \right )+\frac{1}{2}log|x+y|=c$
Given:$\left ( x+y \right )\left ( dx-dy \right )=dx+dy$

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

Solution : - We have,

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

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

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

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

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

Let $x+y=v$

Differentiating with respect to x, we get,

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

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

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

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

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

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

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

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

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

Taking like variables on the same side, we get,

$\Rightarrow \frac{v+1}{2v}dv=dx$

Integrating on both sides, we get,

\begin{aligned} &\Rightarrow \int \frac{v+1}{2 v} d v=\int d \mathrm{x} \\ &\Rightarrow \frac{1}{2} \int\left(1+\frac{1}{v}\right) d v=\int \mathrm{d} \mathrm{x} \\ &\Rightarrow \frac{1}{2} \int(\mathrm{v}+\log |v|)=\mathrm{x}+\mathrm{c} \end{aligned}
Putting v = x + y, we get,
\begin{aligned} &\Rightarrow \frac{1}{2}(x+y+\log |x+y|)=x+c \\ &\Rightarrow \quad(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:$\left ( x+y+1 \right )\frac{dy}{dx}=1$
Hint : - Differential equation of the form $\frac{dy}{dx}=\int \left ( ax+by+c \right )$can be reduced to variable separable form by substitution $ax+by+c=v$
Solution : - We have,
$\left ( x+y+1 \right )\frac{dy}{dx}=1$ ........(i)
Let $x+y=v$
Differentiating with respect to x, we get,
$\frac{d}{dx}\left ( x+y \right )\frac{dv}{dx}$
$\Rightarrow 1+\frac{d}{dx}=\frac{dv}{dx}$
$\Rightarrow \frac{d}{dx}=\frac{dv}{dx}-1$

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

\begin{aligned} &(\mathrm{v}+1)\left(\frac{\mathrm{d} \mathrm{v}}{\mathrm{d} \mathrm{x}}-1\right)=1 \\ &\Rightarrow \quad(\mathrm{v}+1) \frac{\mathrm{d} \mathrm{v}}{\mathrm{d} \mathrm{x}}-(\mathrm{v}+1)=1 \\ &\Rightarrow(\mathrm{v}+1) \frac{\mathrm{d} \mathrm{v}}{\mathrm{d} \mathrm{x}}=1+1+\mathrm{v} \\ &\Rightarrow \quad(\mathrm{v}+1) \frac{\mathrm{d} \mathrm{v}}{\mathrm{d} \mathrm{x}}=2+\mathrm{v} \end{aligned}

Taking like variables on the same side, we get,

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

Integrating on both sides, we get,

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

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

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

Putting v = x + y, we get,

$\Rightarrow \mathrm{x}+\mathrm{y}-\log |\mathrm{x}+\mathrm{y}+2|=\mathrm{x}+\log \mathrm{c}_{1} \\$

$\Rightarrow \mathrm{x}+\mathrm{y}-\mathrm{x}=\log |\mathrm{x}+\mathrm{y}+2|+\log \mathrm{c}_{1} \\$

$\Rightarrow \mathrm{y}=\log \mathrm{c}_{1}|\mathrm{x}+\mathrm{y}+2| \\$

$\Rightarrow \mathrm{e}^{\mathrm{y}}=\mathrm{c}_{1}(\mathrm{x}+\mathrm{y}+2) \quad\left(\because \mathrm{c}=\frac{1}{\mathrm{c}_{1}}\right) \\$

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

Differential equations exercise 21.8 question 11

Answer: $-e^{-\left ( x+y \right )}=x+c$
Given: $\frac{dy}{dx}+1=e^{x+y}$
Hint: Differential equation of the form $\frac{dy}{dx}=\int \left ( ax+by+c \right )$ can be reduced to variable separable form by substitution $ax+by+c=v$
Solution: We have,
$\frac{dy}{dx}+1=e^{x+y}$ .......(i)
Let $x+y=v$
Differentiating with respect to x, we get,
$\frac{\mathrm{d}}{\mathrm{dx}}(\mathrm{x}+\mathrm{y})=\frac{\mathrm{d} \mathrm{v}}{\mathrm{d} \mathrm{x}} \\$
$\Rightarrow 1+\frac{d y}{d x}=\frac{\mathrm{d} \mathrm{v}}{\mathrm{d} \mathrm{x}} \quad \cdots \text { (ii) } \\$
Substituting (ii) in equation (i), we get,
$\frac{\mathrm{d} \mathrm{v}}{\mathrm{d} \mathrm{x}}=\mathrm{e}^{\mathrm{x}+\mathrm{y}} \\$
$\Rightarrow \frac{\mathrm{d} \mathrm{v}}{\mathrm{d} x}=\mathrm{e}^{\mathrm{v}} \quad[\because \mathrm{x}+\mathrm{y}=\mathrm{v}]$
Taking like variables on the same side, we get,
$\Rightarrow \frac{\mathrm{d} v}{\mathrm{e}^{\mathrm{v}}}=\mathrm{d} \mathrm{x} \\$
$\Rightarrow \mathrm{e}^{-\mathrm{v}} \mathrm{d} \mathrm{v}=\mathrm{d} \mathrm{x} \\$
Integrating on both sides, we get,
$\Rightarrow \int \mathrm{e}^{-\mathrm{v}} \mathrm{d} \mathrm{v}=\int \mathrm{d} \mathrm{x} \\$
$\Rightarrow-\mathrm{e}^{-\mathrm{v}}=\mathrm{x}+\mathrm{c} \\$
Putting $v=x+y$, we get,
$\Rightarrow-\mathrm{e}^{-(x+y)}=\mathrm{x}+\mathrm{c}$

(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.

Even though the chapter relies on the same concept for the sums, there are differences in how each equation is being solved. Therefore, students need to concentrate on identifying the proper method to be solved for every equation. If a doubt strikes them, they can very well use the RD Sharma Class 12th Exercise 21.8 for reference. The CBSE students find RD Sharma books more optimal as the solutions follow the NCERT pattern.

Experts and teachers from the mathematics field have contributed their time and knowledge to this set of best solution books. The Class 12 RD Sharma Chapter 21 Exercise 21.8 Solution contains numerous sets of practice questions. When students practice more differential equation sums, they tend to score more marks than the others. To learn this concept effortlessly is a must for the students to have the RD Sharma Class 12 Solutions Differential Equation Ex 21.8 by their side.

This book provides an in-depth knowledge of the concept and makes the students skilled enough to compete in the public exams. And the benefit that excites most of the students is that the RD Sharma Class 12th Exercise 21.8 book can be downloaded without any payment from the Career 360 website. As the best solutions are available for free, why would anyone wish to prefer other sources? Therefore, download the RD Sharma solution materials that you require from the Career 360 website.

Another significant advantage is that the practice questions in the RD Sharma Class 12 Solutions Chapter 21 Ex 21.8 can be asked in the public exams. Hence, preparing with this set of books from day one will help the students score marks effortlessly.

## RD Sharma Chapter wise Solutions

1. Why are RD Sharma solution books so popular?

Many schools recommend the RD Sharma solutions books due to the clarity and accuracy of answer keys.

2. Which books can the class 12 CBSE students use to clarify their doubts in mathematics, chapter 21?

The CBSE students can use the RD Sharma Class 12th Exercise 21.8 to clarify their doubts in this chapter.

3. Is there a chance to access the RD Sharma solution books for free?

4. Where can I find the practice questions for exercise 21.8 in mathematics?

More practice questions are given in the RD Sharma Class 12th Exercise 21.8 book to work.

5. What are the benefits of using RD Sharma solution books?
• Students can gain in-depth knowledge.

• They can easily cross their benchmark score.

## Upcoming School Exams

#### National Means Cum-Merit Scholarship

Application Date:05 August,2024 - 25 September,2024

#### National Rural Talent Scholarship Examination

Application Date:05 September,2024 - 20 September,2024

#### National Institute of Open Schooling 12th Examination

Admit Card Date:13 September,2024 - 07 October,2024

#### National Institute of Open Schooling 10th examination

Admit Card Date:13 September,2024 - 07 October,2024

#### Nagaland Board High School Leaving Certificate Examination

Application Date:17 September,2024 - 30 September,2024

Get answers from students and experts