RD Sharma Class 12 Exercise 21.4 Differential Equation Solutions Maths

Lovekush kumar sainiUpdated on 24 Jan 2022, 04:20 PM IST

Class 12 students spend most of their time preparing for the public exams. The right solution book is required to clarify the doubts of the students in every subject, especially mathematics. Chapters like the Differential Equations are very challenging for most of the students. The RD Sharma Class 12th Exercise 21.4 reference book is very helpful for the students to complete their homework and assignments. It also lends them a helping hand in preparing for their daily tests and public exams.

RD Sharma Class 12 Solutions Chapter21 Differential Equation - Other Exercise

Differential Equations Excercise: 1.3

Differential equations excercise 21.4 question 1

Answer: $y=\log x$ is the solution of given function.
Hint:
Just differentiate the function & then put value
Given:
$y=\log x$ is the function.
Solution:
Differentiate with respect to x
$\begin{aligned} &\Rightarrow \frac{d y}{d x}=\frac{d}{d x}(\log x) \\ &\Rightarrow \frac{d y}{d x}=\frac{1}{x} \\ &\Rightarrow x \frac{d y}{d x}=1 \end{aligned}$
Hence the function satisfies the equation.
Also, when
$\begin{aligned} &x=1 \\ &y=\log 1=0 \end{aligned}$
Thus $y(1)=0$ satisfies the initial value problem.

Differential equation exercise 21.4 question 2

Answer: $y=e^{x}$ is the solution of given function
Hint: Differentiate the function with respect to x
Given:
$y=e^{x}$ is the function.
Solution:
Differentiate with respect to x
$\begin{aligned} &\Rightarrow \frac{d y}{d x}=\frac{d e^{x}}{d x} \\ &\Rightarrow \frac{d y}{d x}=e^{x} \\ &\Rightarrow \frac{d y}{d x}=y\left[\because y=e^{x}\right] \end{aligned}$
Thus, $y\left(e^{x}\right)$ satisfies the equation.
Now , When
$\begin{aligned} &x=0 \\ &y=e^{0}=1 \end{aligned}$
Thus, $\lambda(0)=1$ also satisfies the given equation.

Differential equation excercise 21.4 question 3

Answer: $y=\sin x$ is the solution of given function.
Hint:
Take derivative of the function and check that the solution is satisfying or not.
Given:
$y=\sin x$ is the function.
Solution:
Differentiate with respect to x
$\begin{aligned} &\Rightarrow \frac{d y}{d x}=\frac{d}{d x}(\sin x)\\ &\Rightarrow \frac{d y}{d x}=\cos x \cdots(i)\\ &\text { Differentiating eq(i) }\\ &\Rightarrow \frac{d^{2} y}{d x^{2}}=-\sin x\\ &\left.\Rightarrow \frac{d^{2} y}{d x^{2}}=-y \quad \because \because y=\sin x \text { is given }\right]\\ &\Rightarrow \frac{d^{2} y}{d x^{2}}+y=0 \end{aligned}$
Thus , $y=\sin x$ satisfies the initial value problem
Now,
$\begin{aligned} y(0) &=0 \\ x &=0 \\ \text { then } y &=\sin 0=0 \end{aligned}$
Thus $y(0)=0$ also satisfies initial value problem
$\begin{aligned} &\text { Now, }\\ &\begin{aligned} y^{\prime}(0) &=1 \\ i e, y^{\prime} &=\cos x \\ & \therefore x=0, y^{\prime}=\cos 0=1 \end{aligned} \end{aligned}$

Thus $y^{\prime}(0)=1$also satisfies initial value problem.

Differential equations excercise 21.4 question 4

Answer: $y=e^{x}+1$ is the solution of given function
Hint:
Take the differentiation of the function $y=e^{x}+1$
Given:
$y=e^{x}+1$ is the function.
Solution:
Differentiating with respect to x
$\Rightarrow \frac{d y}{d x}=e^{x} \cdots(i)$
Again differentiating eq(i)
$\begin{aligned} &\Rightarrow \frac{d^{2} y}{d x^{2}}=e^{x} \\ &\Rightarrow \frac{d^{2} y}{d x^{2}}=\frac{d y}{d x} \quad\left[\because \frac{d y}{d x}=e^{x}\right] \\ &\Rightarrow \frac{d^{2} y}{d x^{2}}-\frac{d y}{d x}=0 \end{aligned}$
Thus, $y=e^{x}+1$ satisfies the function.
Now,
When $x=0$
$\begin{aligned} y &=e^{0}+1 \\ &=1+1 \\ &=2 \end{aligned}$
Now,
When $x=0$
$y^{\prime}=e^{x}=e^{0}=1$
Thus both $y(0)=2 \text { and } y^{\prime}(0)=1$ satisfies the initial value problem.

Differential equations exercise 21.4 question 5

Answer: $y=e^{-x}+2$ is the solution of given function
Hint:
Differentiate the function and then obtain the value and satisfy initial value of problem.
Given:
$y=e^{-x}+2$ is the function.
Solution:
Differentiating on both sides with respect to x
$\Rightarrow \frac{d y}{d x}=-e^{-x}$
$\begin{aligned} &\text { For } e^{-x}=y-2 \quad\left[\therefore y=e^{-x}+2\right] \\ &\Rightarrow \frac{d y}{d x}=-(y-2) \\ &\Rightarrow \frac{d y}{d x}=-y+2 \\ &\Rightarrow \frac{d y}{d x}+y=2 \end{aligned}$
Thus,
$y=e^{-x}+2$ satisfies the equation
$\begin{aligned} &\text { Now, When }\\ &x=0\\ &y=e^{-(0)}+2\\ &y=1+2\\ &y=3 \end{aligned}$
Thus, $y(0)=3$ solves initial value problem.

Differential equations excercise 21.4 question 6

Answer: $y=\sin x+\cos x$ is the solution of given function
Hint:
Differentiate the function and then place value of x as per given.
Given:
$y=\sin x+\cos x$ is the function.
Solution:
Differentiating on both sides with respect to x
$\Rightarrow \frac{d y}{d x}=\cos x-\sin x \cdots(i)$
$\begin{aligned} &\text { Differentiating eq(i) w.r.t } x\\ &\Rightarrow \frac{d^{2} y}{d x^{2}}=-\sin x-\cos x\\ &\Rightarrow \frac{d^{2} y}{d x^{2}}=-(\sin x+\cos x)\\ &\Rightarrow \frac{d^{2} y}{d x^{2}}=-y\\ &\Rightarrow \frac{d^{2} y}{d x^{2}}+y=0 \end{aligned}$
Thus, $y=\sin x+\cos x$ satistfies the equation
$\begin{aligned} &\text { When, } x=0\\ &\begin{aligned} y &=\sin 0+\cos 0 \\ &=0+1 \\ &=1 \end{aligned} \end{aligned}$
$when, x=0.$
$\begin{aligned} y^{\prime} &=\sin 0-\cos 0 \\ &=-1 \end{aligned}$
$\therefore \text { For both } y(0)=1 \text { and } y^{\prime}(0)=-1 \text { the functions satisfies the initial value problem }$.

Differential equations exercise 21.4 question 7

Answer: $y=e^{x}+e^{-x}$ is the solution of given function
Hint:
Differentiate the function and then substitute
Given:
$y=e^{x}+e^{-x}$ is the function.
Solution:
Differentiating $y=e^{x}+e^{-x}$ with respect to x
$\begin{aligned} &\Rightarrow \frac{d y}{d x}=e^{x}-e^{-x} \cdots(i)\\ &\text { Again differentiating eq(i) }\\ &\Rightarrow \frac{d^{2} y}{d x^{2}}=e^{x}+e^{-x}\\ &\Rightarrow \frac{d^{2} y}{d x^{2}}=y\\ &\Rightarrow \frac{d^{2} y}{d x^{2}}-y=0 \end{aligned}$
Thus $y=e^{x}+e^{-x}$ satisfies the equation
Now, when $x=0$
$\begin{aligned} \mathrm{y} &=\boldsymbol{e}^{0}+e^{-(0)} \\ &=1+1=2 \end{aligned}$
Now, when $x=0$
$\begin{aligned} \mathrm{y}^{\prime} &=e^{0}-e^{-(0)} \\ &=1-1=0 \end{aligned}$
Thus both $y(0)=2 \text { and } y^{\prime}(0)=0$ satisfies the equation.

Differential equations excercise 21.4 question 8

Answer:

$y=e^{x}+e^{2 x}$ is the solution of given function
Hint:
Differentiate the function
Given:
$y=e^{x}+e^{2 x}$ is the given function
Solution:
Differentiating with respect to x
$\Rightarrow \frac{d y}{d x}=e^{x}+e^{2 x} \cdots(i)$
Differentiating with respect to x
$\begin{aligned} &\Rightarrow \frac{d^{2} y}{d x^{2}}=\frac{d}{d x}\left(e^{x}+2 e^{2 x}\right) \\ &\Rightarrow \frac{d^{2} y}{d x^{2}}=e^{x}+2\left(2 e^{2 x}\right) \\ &\Rightarrow \frac{d^{2} y}{d x^{2}}=e^{x}+4 e^{2 x} \cdots(i i) \end{aligned}$
Put value (i) and (ii) in different equations
$\begin{aligned} \frac{d^{2} y}{d x^{2}} &-3 \frac{d y}{d x}+2 y=0 \\ L H S &=\frac{d^{2} y}{d x^{2}}-3 \frac{d y}{d x}+2 y \\ &=\left(e^{x}+4 e^{2 x}\right)-3\left(e^{x}+2 e^{2 x}\right)+2\left(e^{x}+e^{2 x}\right) \\ &=e^{x}+4 e^{2 x}-3 e^{x}-6 e^{2 x}+2 e^{x}+2 e^{2 x} \\ &=0 \\ &=R H S \end{aligned}$

Thus,
$y=e^{x}+e^{2 x}$ satisfies the diffrential equation
Now,when $x=0$
$\begin{aligned} \mathrm{y} &=e^{0}+e^{2(0)} \\ &=1+e^{0} \\ &=1+1 \\ &=2 \end{aligned}$
Now,when $x=0$
$\begin{aligned} y^{\prime} &=e^{0}+2 e^{2(0)} \\ &=1+2(1) \\ y^{\prime} &=3 \end{aligned}$
Thus, $y(0)=2 \text { and } y^{\prime}(0)=3$ satisfies the initial value problem.

Differential equation exercise 21.4 question 9

Answer: $y=x e^{x}+e^{x}$ is the solution of given function
Hint:
Differentiate the function
Given:
$y=x e^{x}+e^{x}$ is the function.
Solution:
Differentiate with respect to x
$\begin{aligned} &\Rightarrow \frac{d y}{d x}=\frac{d}{d x}\left(x e^{x}+e^{x}\right) \\ &\Rightarrow \frac{d y}{d x}=\frac{d}{d x}\left(x e^{x}\right)+\frac{d}{d x}\left(e^{x}\right) \\ &\Rightarrow \frac{d y}{d x}=x e^{x}+e^{x}+e^{x} \\ &\Rightarrow \frac{d y}{d x}=x e^{x}+2 e^{x} \cdots(i) \end{aligned}$
Differentiate (i) with respect to x
$\begin{aligned} &\Rightarrow \frac{d^{2} y}{d x^{2}}=\frac{d}{d x}\left(x e^{x}+2 e^{x}\right) \\ &\Rightarrow \frac{d^{2} y}{d x^{2}}=x e^{x}+e^{x}+2 e^{2 x} \\ &\Rightarrow \frac{d^{2} y}{d x^{2}}=x e^{x}+3 e^{2 x} \cdots(i i) \end{aligned}$
$\begin{aligned} &\text { Put value (i) and (ii) in given equation }\\ &\begin{aligned} \frac{d^{2} y}{d x^{2}}-2 \frac{d y}{d x}+y &=0 \\ L H S &=\frac{d^{2} y}{d x^{2}}-2 \frac{d y}{d x}+y \\ &=\left(x e^{x}+3 e^{2 x}\right)-2\left(x e^{x}+2 e^{x}\right)+\left(x e^{x}+e^{x}\right) \\ &=x e^{x}+3 e^{x}-2 x e^{x}-4 e^{x}+x e^{x}+e^{x} \\ &=0 \\ &=R H S \end{aligned} \end{aligned}$
Thus,$y=x e^{x}+e^{x}$ satisfies the equation
Now, when $x = 0$
$\begin{aligned} \mathrm{y} &=0 e^{0}+e^{0} \\ &=0+1 \\ &=1 \end{aligned}$
Now, when $x = 0$
$\begin{aligned} \mathrm{y}^{\prime} &=0 e^{0}+2 e^{0} \\ &=0+2 \\ \mathrm{y}^{\prime} &=2 \end{aligned}$
Thus $y(0)=1 \text { and } y^{\prime}(0)=2$ satisfies the equation and initial problem

The topics in class 12 mathematics, chapter 21, Differential Equations, are complex when compared to the other concepts in the book. There are eleven exercises in this chapter, ex 21.1 to ex 21.11. The fourth exercise, ex 21.4, consists of concepts with various methods in which the Differential Equations can be solved. The RD Sharma Class 12 Chapter 21 Exercise 21.4 book contains solved sums for the nine questions given in this exercise.

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