Exercise 9.2 (Revised) - Chapter 9 - Differential Equations - Ncert Solutions class 12 - Maths

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NCERT Class 12 Maths Solutions: Chapter 9 - Differential Equations

In each of the Questions 1 to 6 verify that the given functions (explicit) is a solution of the corresponding differential equation:
Ex 9.2 Question 1.

$y=e^x+1: y^{\prime \prime}-y^{\prime}=0$

Answer.

Given: $y=e^x+1$
To prove: $\mathrm{y}$ is a solution of the differential equation $y^{\prime \prime}-y^{\prime}=0$ .(ii)

Proof: From eq. (i), $y^{\prime}=e^x+0=e^x$ and $y^{\prime \prime}=e^x$
$\therefore$ L.H.S. of eq. (ii), $y^{\prime \prime}-y^{\prime}=e^x-e^x=0=$ R.H.S.
Hence, $y$ given by eq. (i) is a solution of $y^{\prime \prime}-y^{\prime}=0$.

Ex 9.2 Question 2.

$y=x^2+2 x+\mathrm{C}: y^{\prime}-2 x-2=0$

Answer.

Given: $y=x^2+2 x+\mathrm{C}$
To prove: $\mathrm{y}$ is a solution of the differential equation $y^{\prime}-2 x-2=0$ $\qquad$
Proof:From, eq. (i),
$
y^{\prime}=2 x+2
$
L.H.S. of eq. (ii),
$
\begin{aligned}
& =y^{\prime}-2 x-2 \\
& =(2 x+2)-2 x-2 \\
& \quad=2 x+2-2 x-2=0=\text { R.H.S. }
\end{aligned}
$

$\text { Hence, } \mathrm{y} \text { given by eq. (i) is a solution of } y^{\prime}-2 x-2=0 \text {. }$

Ex 9.2 Question 3.

$y=\cos x+\mathrm{C}: y^{\prime}+\sin x=0$

Answer.

Given: $y=\cos x+\mathrm{C}$ $\qquad$ (i)

To prove: $\mathrm{y}$ is a solution of the differential equation $y^{\prime}+\sin x=0$ $\qquad$
Proof: From eq. (i),
$
y^{\prime}=-\sin x
$
L.H.S. of eq. (ii),
$
y^{\prime}+\sin x=-\sin x+\sin x=0=\text { R.H.S. }
$

Hence, $\mathrm{y}$ given by eq. (i) is a solution of $y^{\prime}+\sin x=0$.

Ex 9.2 Question 4.

$y=\sqrt{1+x^2}: y^{\prime}=\frac{x y}{1+x^2}$

Answer.

Given: $y=\sqrt{1+x^2}$ $\qquad$ (i)

To prove:y is a solution of the differential equation $y^{\prime}=\frac{x y}{1+x^2}$ $\qquad$
Proof: From eq. (i),
$
\begin{aligned}
& y^{\prime}=\frac{d}{d x} \sqrt{1+x^2}=\frac{d}{d x}\left(1+x^2\right)^{1 / 2} \\
& =\frac{1}{2}\left(1+x^2\right)^{-1 / 2} \frac{d}{d x}\left(1+x^2\right)=\frac{1}{2}\left(1+x^2\right)^{-1 / 2} \cdot 2 x \\
& =\frac{x}{\sqrt{1+x^2}} \ldots \ldots \ldots \text { (iii) }
\end{aligned}
$

Now R.H.S. of eq. (ii)

$
\begin{aligned}
& =\frac{x y}{1+x^2} \\
& =\frac{x}{1+x^2} \sqrt{1+x^2} \text { [From eq. (i)] } \\
& =\frac{x}{\sqrt{1+x^2}}=y^{\prime} \\
& \therefore \text { L.H.S. = R.H.S }
\end{aligned}
$

Hence, $y$ given by eq. (i) is a solution of $y^{\prime}=\frac{x y}{1+x^2}$.

Ex 9.2 Question 5.

$y=\mathrm{A} x: x y^{\prime}=y(x \neq 0)$

Answer.

Given: $y=\mathrm{A} x$ ..(i)

To prove:y given by eq. (i) is a solution of differential equation $x y^{\prime}=y(x \neq 0)$
Proof: From eq. (i)
$
y^{\prime}=\mathrm{A}(1)=\mathrm{A}
$
L.H.S. of eq. (ii)
$=x y^{\prime}=x \mathrm{~A}=\mathrm{A} x=y=$ R.H.S. of eq. (ii)
$\therefore \mathrm{y}$ given by eq. (i) is a solution of differential equation $x y^{\prime}=y(x \neq 0)$.

Ex 9.2 Question 6.

$y=x \sin x: x y^{\prime}=y+x \sqrt{x^2-y^2}(x \neq 0$ and $x>y$ or $x<-y)$

Answer.

Given: $y=x \sin x$ ...(i) $\qquad$
To prove:y given by eq. (i) is a solution of differential equation $x y^{\prime}=y+x \sqrt{x^2-y^2}$..(ii)
Proof: From eq. (i),

$
\begin{aligned}
& \frac{d y}{d x}\left(=y^{\prime}\right)=x \frac{d}{d x} \sin x+\sin x \frac{d}{d x} x \\
& =x \cos x+\sin x
\end{aligned}
$
L.H.S. of eq. (ii)
$
=x y^{\prime}=x(x \cos x+\sin x)=x^2 \cos x+x \sin x
$
R.H.S. of eq. (ii)
$
\begin{aligned}
& =y+x \sqrt{x^2-y^2} \\
& =x \sin x+x \sqrt{x^2-x^2 \sin ^2 x} \text { [From eq. (i)] } \\
& =x \sin x+x \sqrt{x^2\left(1-\sin ^2 x\right)} \\
& =x \sin x+x \sqrt{x^2 \cos ^2 x} \\
& =x \sin x+x \cdot x \cos x \\
& =x \sin x+x^2 \cos x \\
& =x^2 \cos x+x \sin x \\
& \therefore \text { L.H.S. }=\text { R.H.S }
\end{aligned}
$

Hence, $\mathrm{y}$ given by eq. (i) is a solution of $x y^{\prime}=y+x \sqrt{x^2-y^2}$.

In each of the questions 7 to 10, verify that the given functions (explicit or implicit) is a solution of the corresponding differential equation:
Ex 9.2 Question 7.

$x y=\log y+\mathrm{C}: y^{\prime}=\frac{y^2}{1-x y}(x y \neq 1)$

Answer.

Given: $x y=\log y+\mathrm{C}$ (i)

To prove:y given by eq. (i) is a solution of differential equation $y^{\prime}=\frac{y^2}{1-x y}$
Proof: Differentiating both sides of eq. (i) w.r.t x, we have
$
\begin{aligned}
& x y^{\prime}+y(1)=\frac{1}{y} y^{\prime}+0 \\
& \Rightarrow x y^{\prime}-\frac{y^{\prime}}{y}=-y \\
& \Rightarrow y^{\prime}\left(x-\frac{1}{y}\right)=-y \\
& \Rightarrow y^{\prime}\left(\frac{x y-1}{y}\right)=-y \\
& \Rightarrow y^{\prime}(x y-1)=-y^2 \\
& \Rightarrow y^{\prime}=\frac{-y^2}{x y-1} \\
& \Rightarrow y^{\prime}=\frac{-y^2}{-(1-x y)}=\frac{y^2}{1-x y}
\end{aligned}
$

Hence, Function (implicit) given by eq. (i) is a solution of $y^{\prime}=\frac{y^2}{1-x y}$.

Ex 9.2 Question 8.

$y-\cos y=x:(y \sin y+\cos y+x) y^{\prime}=y$

Answer.

Given: $y-\cos y=x$ $\qquad$
To prove: y given by eq. (i) is a solution of differential equation
$
(y \sin y+\cos y+x) y^{\prime}=y
$
$\qquad$
Proof: Differentiating both sides of eq. (i) w.r.t x, we have

$
\begin{aligned}
& y^{\prime}+(\sin y) y^{\prime}=1 \\
& \Rightarrow y^{\prime}(1+\sin y)=1 \\
& \Rightarrow y^{\prime}=\frac{1}{1+\sin y} \ldots \ldots \ldots . .(\text { iii) }
\end{aligned}
$

Putting the value of $x$ from eq. (i) and value of $y^{\prime}$ from eq. (iii) in L.H.S. of eq. (ii),
$
\begin{aligned}
& (y \sin y+\cos y+x) y^{\prime} \\
& \Rightarrow(y \sin y+\cos y+y-\cos y) \frac{1}{1+\sin y} \\
& \Rightarrow(y \sin y+y) \frac{1}{1+\sin y} \\
& \Rightarrow y(\sin y+1) \frac{1}{1+\sin y}=y=\text { R.H.S. of (ii) }
\end{aligned}
$

Hence, Function given by eq. (i) is a solution of $(y \sin y+\cos y+x) y^{\prime}=y$.

Ex 9.2 Question 9.

$x+y=\tan ^{-1} y: y^2 y^{\prime}+y^2+1=0$

Answer.

Given: $x+y=\tan ^{-1} y$
To prove:y given by eq. (i) is a solution of differential equation $y^2 y^{\prime}+y^2+1=0$
Proof: Differentiating both sides of eq. (i) w.r.t x we have
$
\begin{aligned}
& 1+y^{\prime}=\frac{1}{1+y^2} y^{\prime} \\
& \Rightarrow\left(1+y^{\prime}\right)\left(1+y^2\right)=y^{\prime} \\
& \Rightarrow 1+y^2+y^{\prime}+y^{\prime} y^2=y^{\prime}
\end{aligned}
$

$
\Rightarrow y^2 y^{\prime}+y^2+1=0=\text { eq. }
$

Hence, Function given by eq. (i) is a solution of $y^2 y^{\prime}+y^2+1=0$

Ex 9.2 Question 10.

$y=\sqrt{a^2-x^2}, \epsilon(-a, a): x+y \frac{d y}{d x}=0(y \neq 0)$

Answer.

Given: $y=\sqrt{a^2-x^2}, x \in(-a, a)$
To prove:y given by eq. (i) is a solution of differential equation $x+y \frac{d y}{d x}=0$
Proof: From eq. (i),
$
\begin{aligned}
& \frac{d y}{d x}=\frac{1}{2 \sqrt{a^2-x^2}}(-2 x) \\
& =\frac{-x}{\sqrt{a^2-x^2}} \ldots \ldots . . . \text { (iii) }
\end{aligned}
$

Putting the values of $\mathrm{y}$ and $\frac{d y}{d x}$ from eq. (i) and (iii) in L.H.S. of eq. (ii),
$
\begin{aligned}
& =x+y \frac{d y}{d x} \\
& =x+\sqrt{a^2-x^2}\left(\frac{-x}{\sqrt{a^2-x^2}}\right) \\
& =x-x=0=\text { R.H.S. of eq. (ii) }
\end{aligned}
$

Hence, Function given by eq. (i) is a solution of $x+y \frac{d y}{d x}=0$.

Choose the correct answer:
Ex 9.2 Question 11.

The number of arbitrary constants in the general solution of a differential equation

of fourth order are:
(A) 0
(B) 2
(C) 3
(D) 4

Answer.

Option (D) is correct.
The number of arbitrary constants $\left(c_1, c_2, c_3\right.$, etc.) in the general solution of a differential equation of $n^{\text {th }}$ order is $n$.
Ex 9.2 Question 12.

The number of arbitrary constants in the particular solution of a differential equation of third order are:
(A) 3
(B) 2
(C) 1
(D) 0

Answer.

The number of arbitrary constants in a particular solution of a differential equation of any order is zero (0) as a particular solution is a solution which contains no arbitrary constant.

Therefore, option (D) is correct.