Exercise 10.9 - Chapter 10 Ordinary Differential Equations 12th Maths Guide Samacheer Kalvi Solutions - SaraNextGen [2024-2025]

Updated By SaraNextGen

On April 24, 2024, 11:35 AM

Ex $10.9$
Choose the correct or the most suitable answer from the given four alternatives:
Question 1.
The order and degree of the differential equation $\frac{d^2 y}{d x^2}+\left(\frac{d y}{d x}\right)^{1 / 3}+x^{1 / 4}=0 \quad$ are respectively.
(a) 2,3
(b) 3,3
(c) 2,6
(d) 2,4
Solution:
(a) 2,3
Hint:
$
\begin{aligned}
\frac{d^2 y}{d x^2}+\left(\frac{d y}{d x}\right)^{\frac{1}{3}}+x^{\frac{1}{4}} & =0 \\
\frac{d^2 y}{d x^2}+x^{\frac{1}{4}} & =-\left(\frac{d y}{d x}\right)^{\frac{1}{3}}
\end{aligned}
$
Raising the power both sides to ' 3 '
$
\left[\frac{d^2 y}{d x^2}+x^{\frac{1}{4}}\right]^3=-\left(\frac{d y}{d x}\right)
$
Order $=2$,
degree $=3$
Question 2.
The differential equation representing the family of curves $y=A \cos (x+B)$, where $A$ and $B$ are parameters, is .......
(a) $\frac{d^2 y}{d x^2}-y=0$
(b) $\frac{d^2 y}{d x^2}+y=0$
(c) $\frac{d^2 y}{d x^2}=0$
(d) $\frac{d^2 x}{d y^2}=0$
Solution:
(b) $\frac{d^2 y}{d x^2}+y=0$

Hint:
$
\begin{aligned}
y & =\mathrm{A} \cos (x+\mathrm{B}) \\
\frac{d y}{d x} & =-\mathrm{A} \sin (x+\mathrm{B}) \\
\frac{d^2 y}{d x^2} & =-\mathrm{A} \cos (x+\mathrm{B})=-y \\
\frac{d^2 y}{d x^2}+y & =0
\end{aligned}
$
Question 3.
The order and degree of the differential equation $\sqrt{\sin x}(d x+d y)=\sqrt{\cos x}(d x-d y)$ is .....

(a) 1,2
(b) 2,2
(c) 1,1
(d) 2,1
Solution:
(c) 1, 1
Hint:
$
\sqrt{\sin x}(d x+d y)=\sqrt{\cos x}(d x-d y)
$
Since, the first order derivatives are involved, Order $=1$ and degree 1 .
Question 4.
The order of the differential equation of all circles with centre at $(h, k)$ and radius ' $a$ ' is
(a) 2
(b) 3
(c) 4
(d) 1
Solution:
(a) 2
Hint:
Equation of circle is $(\mathrm{x}-\mathrm{h})^2+(\mathrm{y}-\mathrm{k})^2=\mathrm{a}^2$
Equation is to be differentiated twice as two parameters are given.
$\therefore$ Order $=2$
Question 5.
The differential equation of the family of curves $\mathrm{y}=\mathrm{Ae}^{\mathrm{x}}+\mathrm{Be}^{-\mathrm{x}}$, where $\mathrm{A}$ and $\mathrm{B}$ are arbitrary constants is 

(a) $\frac{d^2 y}{d x^2}+y=0$
(b) $\frac{d^2 y}{d x^2}-y=0$
(c) $\frac{d y}{d x}+y=0$
(d) $\frac{d y}{d x}-y=0$
Solution:
(b) $
\frac{d^2 y}{d x^2}-y=0
$
Hint:
$
\begin{aligned}
y & =\mathrm{A} e^x+\mathrm{B} e^{-x} \\
\frac{d y}{d x} & =\mathrm{A} e^x-\mathrm{B} e^{-x} \\
\frac{d^2 y}{d x^2} & =\mathrm{A} e^x+\mathrm{B} e^{-x}=y \Rightarrow \frac{d^2 y}{d x^2}-y=0
\end{aligned}
$
Question 6.
The general solution of the differential equation $\frac{d y}{d x}=\frac{y}{x}$ is .....
(a) $x y=k$
(b) $y=k \log x$
(c) $y=\mathrm{kx}$
(d) $\log y=k x$
Solution:
(c) $y=k x$

Question 7.
The solution of the differential equation (a) straight lines $2 x \frac{d y}{d x}-y=3$
(b) circles
(c) parabola
(d) ellipse
Solution:
(c) parabola Hint:
$
\begin{aligned}
2 x \frac{d y}{d x}-y & =3 \Rightarrow 2 x \frac{d y}{d x}=y+3 \\
2 \int \frac{d y}{y+3} & =\int \frac{d x}{x} \\
2 \log (y+3) & =\log x+\log c \Rightarrow \log (y+3)^2=\log c x
\end{aligned}
$
Question 8.
The solution of $\frac{d y}{d x}+p(x) y=0$ is .......
(a) $y=c e^{\int \mathrm{P} d x}$
(b) $y=c e^{-\int P d x}$
(c) $x=c e^{-\int \mathrm{P} d y}$
(d) $x=c e^{\int P d y}$
Solution:
(b) $y=c e^{-\int \mathbf{P} d x}$
Hint:

Solution of $\frac{d y}{d x}+p(x) y=0$, Here $\mathrm{Q}(x)=0$
$
\begin{aligned}
& y e^{\int P d x}=\int Q e^{\int P d x} d x+c \\
& y e^{\int P d x}=0+c \\
& y=c e^{-\int P d x}
\end{aligned}
$
Question 9.
The integrating factor of the differential equation $\frac{d y}{d x}+y=\frac{1+y}{x}$ is .......
(a) $\frac{x}{e^\lambda}$
(b) $\frac{e^x}{x}$
(c) $\lambda e^x$
$(d) e^x$
Solution:
(b) $\frac{e^z}{x}$
Hint:
$
\begin{gathered}
\frac{d y}{d x}+y=\frac{1+y}{x} \\
\frac{d y}{d x}+y-\frac{y}{x}=\frac{1}{x} \Rightarrow \frac{d y}{d x}+y\left(1-\frac{1}{x}\right)=\frac{1}{x} \\
\int \mathrm{P} d x=\int\left(1-\frac{1}{x}\right) d x=x-\log x \\
\text { I.F. }=e^{\int \mathrm{P} d x}=e^{x-\log x}=e^x e^{-\log x}=\frac{e^x}{x}
\end{gathered}
$
Question 10.
The integrating factor of the differential equation $\frac{d y}{d x}+\mathrm{P}(x) y=\mathrm{Q}(x)$ is x, then P(x)
(a) $x$
(b) $\frac{x^2}{2}$
(c) $\frac{1}{x}$
(d) $\frac{1}{x^2}$

Solution:

(c) $\frac{1}{x}$

Hint:
$
\begin{aligned}
& \frac{d y}{d x}+\mathrm{P}(x) y=\mathrm{Q}(x) \\
& \text { Given,I.F. }=e^{\int \mathrm{P} d x}=x \\
& \log e^{\int \mathrm{P} d x}=\log x \Rightarrow \int \mathrm{P} d x=\log x
\end{aligned}
$
we know $\int \frac{1}{x} d x=\log x$
$
\therefore \mathrm{P}=\frac{1}{x}
$
Question 11.
The degree of the dififerential equation (a) 2
$
y(x)=1+\frac{d y}{d x}+\frac{1}{1.2}\left(\frac{d y}{d x}\right)^2+\frac{1}{1.2 .3}\left(\frac{d y}{d x}\right)^3+\ldots
$
is ........
(b) 3
(c) 1
(d) 4
Solution:
(c) 1
Hint:
Degree $=1$

Question 12.
If $p$ and $q$ are the order and degree of the differential equation
when $
y \frac{d y}{d x}+x^3\left(\frac{d^2 y}{d x^2}\right)+x y=\cos x
$
(a) p $ (b) $p=q$
(c) $p>q$
(d) $\mathrm{p}$ exists and $\mathrm{q}$ does not exist
Solution:
(c) $p>q$
Hint:
$
y \frac{d y}{d x}+x^3\left(\frac{d^2 y}{d x^2}\right)+x y=\cos x
$
Order $p=2$, degree $q=1$
$
\therefore p>q \text { as } 2>1
$

Question 13.
The solution of the differential equation $\frac{d y}{d x}+\frac{1}{\sqrt{1-x^2}}=0$
(a) $y+\sin ^{-1} x=c$
(b) $x+\sin ^{-1} y=0$
(c) $y^2+2 \sin ^{-1} x=c$
(d) $x^2+2 \sin ^{-1} y=0$
Solution:
(a) $y+\sin ^{-1} x=c$
Hint:
$
\begin{aligned}
\frac{d y}{d x}+\frac{1}{\sqrt{1-x^2}} & =0 \\
\frac{d y}{d x} & =-\frac{1}{\sqrt{1-x^2}} \Rightarrow d y=-\int \frac{d x}{\sqrt{1-x^2}} \\
y & =-\sin ^{-1} x+c \Rightarrow y+\sin ^{-1} x=c
\end{aligned}
$

Question 14.
The solution of the differential equation $\frac{d y}{d x}=2 x y$ is
(a) $y=c e^{x^2}$
(b) $y=2 x^2+c$
(c) $y=c e^{-x^2}+c$
(d) $y=x^2+c$
Solution:
(a) $y=c e^{x^2}$
Hint:
$
\begin{aligned}
\frac{d y}{d x} & =2 x y \\
\int \frac{d y}{y} & =2 \int x d x \\
\log y & =2 \cdot \frac{x^2}{2}+c \Rightarrow \log y=x^2+c \\
y & =e^{x^2+c}=e^{x^2} \cdot e^c \Rightarrow y=c e^{x^2}
\end{aligned}
$
Question 15.
The general solution of the differential equation $\log \left(\frac{d y}{d x}\right)=x+y$
(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:
(b) $e^x+e^{-y}=c$

Hint:
$
\begin{aligned}
& \log \left(\frac{d y}{d x}\right)=x+y \\
& \frac{d y}{d x}=e^{x+y}=e^x \cdot e^y \\
& \int e^{-y} d y=\int e^x d x \\
& -e^{-y}=e^x+c \Rightarrow e^x+e^{-y}=c
\end{aligned}
$
Question 16.
The solution of $\frac{d y}{d x}=2^{y-x}$ is
Solution:
(c) $\frac{1}{2^z}-\frac{1}{2^y}=c$
Hint:
$
\begin{aligned}
\frac{d y}{d x} & =2^{y-x} \Rightarrow \frac{d y}{d x}=\frac{2^y}{2^x} \\
\int 2^{-y} d y & =\int 2^{-x} d x \\
\frac{-2^{-y}}{\log 2} & =\frac{-2^{-x}}{\log 2}+c
\end{aligned}
$
Simplifying $\frac{1}{2^x}-\frac{1}{2^y}=c \log 2 \Rightarrow \frac{1}{2^x}-\frac{1}{2^y}=c$ (constant)

Question 17.
$
\frac{d y}{d x}=\frac{y}{x}+\frac{\phi\left(\frac{y}{x}\right)}{\phi^{\prime}\left(\frac{y}{x}\right)}
$
The solution of the differential equation is ........
(a) $x \phi\left(\frac{y}{x}\right)=k$
(b) $\phi\left(\frac{y}{x}\right)=k x$
(c) $y \phi\left(\frac{y}{x}\right)=k$
(d) $\phi\left(\frac{y}{x}\right)=k y$
Solution:
(b) $\phi\left(\frac{y}{x}\right)=k x$

Hint:
$\frac{d y}{d x}=\frac{y}{x}+\frac{\phi\left(\frac{y}{x}\right)}{\phi^{\prime}\left(\frac{y}{x}\right)}$ is a Homogeneous differential equation.
$
\begin{aligned}
\frac{d y}{d x}-\frac{y}{x}+\phi^{\prime}\left(\frac{y}{x}\right) & =\phi\left(\frac{y}{x}\right) & \\
\phi\left(\frac{y}{x}\right) & =k x & {[\because k \text { constant }] }
\end{aligned}
$
Question 18.
If $\sin \mathrm{x}$ is the integrating factor of the linear differential equation $\frac{d y}{d x}+\mathrm{P} y=\mathrm{Q}$, then $\mathrm{P}$ is .....
(a) $\log \sin x$
(b) $\cos x$
(c) $\tan x$
(d) $\cot x$
Solution:
(d) $\cot x$
Hint:
$
\frac{d y}{d x}+\mathrm{P} y=\mathrm{Q}
$
$
\begin{aligned}
\text { I.F. }=e^{\int P d x} & =\sin x \\
\log e^{\int P d x} & =\log \sin x \Rightarrow \int P d x=\log \sin x
\end{aligned}
$
we know $\int \cot x d x=\log \sin x \Rightarrow \therefore \mathrm{P}=\cot x$
Question 19.
The number of arbitrary constants in the general solutions of order $n$ and $n+1$ are respectively
(a) $\mathrm{n}-1, \mathrm{n}$
(b) $\mathrm{n}, \mathrm{n}+1$
(c) $\mathrm{n}+1, \mathrm{n}+2$
(d) $\mathrm{n}+1, \mathrm{n}$
Solution:
(b) $n, n+1$

Question 20.
The number of arbitrary constants in the particular solution of a differential equation of third order is
(a) 3
(b) 2
(c) 1
(d) 0
Solution:
(d) 0
Question 21
Integrating factor of the differential equation
$
\frac{d y}{d x}=\frac{x+y+1}{x+1}
$
(a) $\frac{1}{x+1}$
(b) $x+1$
(c) $\frac{1}{\sqrt{x+1}}$
(d) $\sqrt{x+1}$
Solution:
(a) $\frac{1}{x+1}$
Hint:
$
\begin{aligned}
& \frac{d y}{d x}=\frac{x+y+1}{x+1} \\
& \frac{d y}{d x}=\frac{x+1}{x+1}+\frac{y}{x+1} \Rightarrow \frac{d y}{d x}-\frac{y}{x+1}=1 \\
& \int \mathrm{P} d x=-\int \frac{1}{x+1} d x=-\log (x+1)=\log \left(\frac{1}{x+1}\right) \\
& \text { I.F. }=e^{\int P d x}=e^{\log \left(\frac{1}{x+1}\right)}=\frac{1}{x+1}
\end{aligned}
$

Question 22.
The population $P$ in any year $t$ is such that the rate of increase in the population is proportional to the population. Then .....
(a) $\mathrm{P}=c e^{k t}$
(b) $\mathrm{P}=c e^{-k t}$
(c) $\mathrm{P}=c k t$
(d) $\mathrm{P}=c$
Solution:
(a) $\mathbf{P}=c e^{k t}$
Hint:
$
\text { Given } \begin{aligned}
\frac{d p}{d t} & \propto \mathrm{P} \Rightarrow \frac{d p}{d t}=k \mathrm{P} \\
\int \frac{d \mathrm{P}}{\mathrm{P}} & =k \int d t \\
\log \mathrm{P} & =k t+c \Rightarrow \mathrm{P}=e^{k t+c} \\
\mathrm{P} & =c e^{k t}
\end{aligned}
$
Question 23.
$\mathrm{P}$ is the amount of certain substance left in after time t. If the rate of evaporation of the substance is proportional to the amount remaining, then ......

(a) $P=c e^{k t}$
(b) $P=c e^{-k t}$
(c) $\mathrm{P}=\mathrm{ckt}$
(d) $\mathrm{Pt}=\mathrm{c}$
Solution:
(b) $\mathbf{P}=\mathrm{ce}^{-\mathrm{kt}}$
Hint:
Given $\frac{d p}{d t} \propto \mathrm{P}$
$
\begin{aligned}
\frac{d p}{d t} & =-k \mathrm{P} \\
\mathrm{P} & =c e^{-k t}
\end{aligned}
$
[decay constant]

Question 24.
If the solution of the differential equation $\frac{d y}{d x}=\frac{a x+3}{2 y+f}$ represents a circle, then the value of $a$ is............
(a) 2
(b) $-2$
(c) 1
(d) $-1$
Solution:
(b) $-2$
Hint:
$
\begin{aligned}
\frac{d y}{d x} & =\frac{a x+3}{2 y+f} \\
\int(2 y+f) d y & =\int(a x+3) d x \\
2 y^2+f y & =a x^2+3 x+c \Rightarrow a x^2-2 y^2+3 x-f y+c=0
\end{aligned}
$
In the equation of circle,
Co efficient of ' $x^{2 \prime}=$ Co efficient of ' $y^{2 \prime}$
$
a=-2
$

Question 25.
The slope at any point of a curve $\mathrm{y}=\mathrm{f}(\mathrm{x})$ is given by $\frac{d y}{d x}=3 x^2$ and it passes through $(-1,1)$. Then the equation of the curve is
(a) $y=x^3+2$
(b) $y=3 x^2+4$
(c) $y=3 x^3+4$
(d) $y=x^3+5$
Solution:
(a) $y=x^3+2$
Hint:
$
\begin{aligned}
& \frac{d y}{d x}=3 x^2 \Rightarrow \int d y=3 \int x^2 d x \\
& y=3 \frac{x^3}{3}+c \Rightarrow y=x^3+c
\end{aligned}
$
It passes through $(-1,1)$
$
\begin{aligned}
& \therefore 1=-1+c \\
& \Rightarrow c=2
\end{aligned}
$
$\therefore$ Equation of the curve is $y=x^3+2$