Exercise 4.6 - Chapter 4 - Differential Equations - 12th Maths Guide Guide Samacheer Kalvi Solutions

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Ex 4.6
Choose the correct answer.
Question 1.
The degree of the differential equation $\frac{d^4 y}{d x^4}-\left(\frac{d^2 y}{d x^2}\right)^4+\frac{d y}{d x}=3$ is
(a) 1
(b) 2
(c) 3
(d) 4
Answer:
(a) 1
Hint:
Since the power of $\frac{d^4 y}{d x^4}$ is 1
Question 2.
The order and degree of the differential equation $\sqrt{\frac{d^2 y}{d x^2}}=\sqrt{\frac{d y}{d x}+5}$ are respectively
(a) 2 and 3
(b) 3 and 2
(c) 2 and 1
(d) 2 and 2
Answer:
(c) 2 and 1
Hint:
Squaring both sides, we get $\frac{d^2 y}{d x^2}=\frac{d y}{d x}+5$
So order $=2$, degree $=1$
Question 3.
The order and degree of the differential equation $\left(\frac{d^2 y}{d x^2}\right)^{\frac{3}{2}}-\sqrt{\left(\frac{d y}{d x}\right)}-4=0$ are respectively.
(a) 2 and 6
(b) 3 and 6
(c) 1 and 4
(d) 2 and 4
Answer:
(a) 2 and 6
Hint:

$
\left(\frac{d^2 y}{d x^2}\right)^{\frac{3}{2}}=\sqrt{\frac{d y}{d x}}+4
$
Squaring both sides
$
\begin{aligned}
& \left(\frac{d^2 y}{d x^2}\right)^3=\frac{d y}{d x}+16+8 \sqrt{\frac{d y}{d x}} \\
& \left(\frac{d^2 y}{d x^2}\right)^3-\frac{d y}{d x}-16=8 \sqrt{\frac{d y}{d x}}
\end{aligned}
$
Again squaring both sides,
$
\begin{aligned}
& {\left[\left(\frac{d^2 y}{d x^2}\right)^3-\frac{d y}{d x}-16\right]^2=64 \frac{d y}{d x}} \\
& \Rightarrow \text { Order }=2 \text { degree }=6
\end{aligned}
$
Question 4.
The differential equation $\left(\frac{d x}{d y}\right)^3+2 y^{\frac{1}{2}}=x$ is
(a) of order 2 and degree 1
(b) of order 1 and degree 3
(c) of order 1 and degree 6
(d) of order 1 and degree 2
Answer:
(b) of order 1 and degree 3
Question 5.
The differential equation formed by eliminating $a$ and $b$ from $y=a e^{\mathrm{x}}+b e^{-\mathrm{x}}$ is
(a) $\frac{d^2 y}{d x^2}-y=0$
(b) $\frac{d^2 y}{d x^2}-\frac{d y}{d x}=0$
(c) $\frac{d^2 y}{d x^2}=0$
(d) $\frac{d^2 y}{d x^2}-x=0$
Answer:
(a) $\frac{d^2 y}{d x^2}-y=0$
Hint:

$
\begin{aligned}
& y=a e^x+b e^{-x} ; y^{\prime}=a e^x-b e^{-x} \\
& y^{\prime \prime}=a e^x+b e^{-x} \Rightarrow y^{\prime \prime}-y^{\prime}=0
\end{aligned}
$
Question 6.
If $y=c x+c-c^3$ then its differential equation is
(a) $y=x \frac{d y}{d x}+\frac{d y}{d x}-\left(\frac{d y}{d x}\right)^3$
(b) $y+\left(\frac{d y}{d x}\right)^3=x \frac{d y}{d x}-\frac{d y}{d x}$
(c) $\frac{d y}{d x}+y=\left(\frac{d y}{d x}\right)^3-x \frac{d y}{d x}$
(d) $\frac{d^3 y}{d x^3}=0$
Answer:
(a) $y=x \frac{d y}{d x}+\frac{d y}{d x}-\left(\frac{d y}{d x}\right)^3$
Question 7.
The integrating factor of the differential equation $\frac{d x}{d y}+\mathrm{Px}=\mathrm{Q}$ is
(a) $e^{\int P d x}$
(b) $\int \mathrm{Pdx}$
(c) $\int$ Pdy
(d) $\mathrm{e}^{\int \mathrm{Pdy}}$
Answer:
(d) $e^{\int P d y}$
Question 8.
The complementary function of $\left(D^2+4\right) y=e^{2 x}$ is
(a) $(\mathrm{Ax}+\mathrm{B}) \mathrm{e}^{2 \mathrm{x}}$
(b) $(\mathrm{Ax}+\mathrm{B}) \mathrm{e}^{-2 \mathrm{x}}$
(c) $A \cos 2 x+B \sin 2 x$
(d) $\mathrm{Ax}^{-2 \mathrm{x}}+\mathrm{Be}^{2 \mathrm{x}}$
Answer:
(c) $A \cos 2 x+B \sin 2 x$
Hint:

$
\begin{aligned}
& \text { A. } E=m^2+4=0 \Rightarrow m=\pm 2 i \\
& \text { C.F }=e^{0 x}(A \cos 2 x+B \sin 2 x)
\end{aligned}
$
Question 9.
The differential equation of $\mathrm{y}=\mathrm{mx}+\mathrm{c}$ is ( $\mathrm{m}$ and $\mathrm{c}$ are arbitrary constants).
(a) $\frac{d^2 y}{d x^2}=0$
(b) $\mathrm{y}=\mathrm{x} \frac{d y}{d x}$
(c) $x d y+y d x=0$
(d) $y d x-x d y=0$
Answer:
(a) $\frac{d^2 y}{d x^2}=0$
Question 10.
The particular integral of the differential equation $\frac{d^2 y}{d x^2}-8 \frac{d y}{d x}+16 y=2 e^{4 x}$ is
(a) $\frac{x^2 e^{4 x}}{2 !}$
(b) $\frac{e^{4 x}}{2 !}$
(c) $x^2 e^{4 x}$
(d) $x e^{4 x}$
Answer:
(c) $x^2 e^{4 x}$
Hint:
Replace D by 4. PI $=\frac{1}{16+16-32} 2 e^{4 x}$
$
\therefore \quad \mathrm{PI}=x \frac{1}{2 \mathrm{D}-8} 2 e^{4 x}
$
Again $2(4)-8=0$
So
$
\mathrm{PI}=x^2 \frac{1}{2}\left(2 e^{4 x}\right)=x^2 e^{4 x}
$
Question 11.
Solution of $\frac{d x}{d y}+\mathrm{px}=0$
(a) $x=c e^{p y}$
(b) $x=c e^{-p y}$

(c) $x=p y+c$
(d) $x=c y$
Answer:
(b) $x=c e^{-p y}$
Hint:
$
\begin{aligned}
\frac{d x}{d y}+p x & =0 \Rightarrow \frac{d x}{d y}=-p x \\
\frac{d x}{x} & =-p d y \\
\log x & =-p y+c_1 \\
x & =e^{-p y} e^{c_1}=c e^{-p y}
\end{aligned}
$
Question 12 .
If $\sec ^2 \mathrm{x}$ is an integrating factor of the differential equation $\frac{d y}{d x}+\mathrm{Py}=\mathrm{Q}$ then $\mathrm{P}=$
(a) $2 \tan x$
(b) $\sec x$
(c) $\cos ^2 x$
(d) $\tan ^2 x$
Answer:
(a) $2 \tan x$
Hint:
$
\begin{aligned}
\text { I.F } & =e^{\int p d x}=\sec ^2 x \\
\Rightarrow \int \mathrm{P} d x & =\log \sec ^2 x=2 \log \sec x \\
\mathrm{P}=\frac{d}{d x} 2 \log \sec x & =\frac{2 \sec x \tan x}{\sec x}=2 \tan x
\end{aligned}
$

Question 13.
The integrating factor of $\mathrm{x} \frac{d y}{d x}-\mathrm{y}=\mathrm{x}^2$ is
(a) $\frac{-1}{x}$
(b) $\frac{1}{x}$
(c) $\log x$
(d) $\mathrm{x}$
Answer:
(b) $\frac{1}{x}$
Hint:
$
\begin{array}{r}
x \frac{d y}{d x}-y=x^2 \Rightarrow \frac{d y}{d x}-\frac{y}{x}=x \\
\text { IF }=e^{\int \frac{-1}{x} d x}=e^{-\log x}=\frac{1}{x}
\end{array}
$
Question 14.
The solution of the differential equation $\frac{d y}{d x}+\mathrm{Py}=\mathrm{Q}$ where $\mathrm{P}$ and $\mathrm{Q}$ are the function of $\mathrm{x}$ is
(a) $y=\int \mathrm{Q} e^{\int p d x} d x+c$
(b) $y=\int \mathrm{Q} e^{-\int p d x} d x+c$
(c) $y e^{\int p d x}=\int \mathrm{Q} e^{\int p d x} d x+c$
(d) $y e^{\int p d x}=\int \mathrm{Q} e^{-\int p d x} d x+c$
Answer:
(c) $y e^{\int p d x}=\int Q e^{\int p d x} d x+c$

Question 15.
The differential equation formed by eliminating A and $B$ from $y=e^{-2 x}(A \cos x+B \sin x)$ is
(a) $y_2-4 y_1+5 y=0$
(b) $y_2+4 y_1-5 y=0$
(c) $\mathrm{y}_2-4 \mathrm{y}_1-5 \mathrm{y}=0$
(d) $y_2+4 y_1+5 y=0$
Answer:
(d) $y_2+4 y_1+5 y=0$
Hint:
$
\begin{aligned}
& y=e^{-2 x}(\mathrm{~A} \cos x+\mathrm{B} \sin x) \\
& y^{\prime}=e^{-2 x}(-\mathrm{A} \sin x+\mathrm{B} \cos x)+(\mathrm{A} \cos x+\mathrm{B} \sin x)(-2) e^{-2 x} \\
& y^{\prime}=e^{-2 x}[-\mathrm{A} \sin x+\mathrm{B} \cos x]-2 y \\
& y^{\prime \prime}=e^{-2 x}[-\mathrm{A} \cos x-\mathrm{B} \sin x]+[-\mathrm{A} \sin x+\mathrm{B} \cos x]\left(-2 e^{-2 x}\right)-2 y^{\prime} \\
& y^{\prime \prime}=-y-2\left(y^{\prime}+2 y\right)-2 y^{\prime} \\
& y^{\prime \prime}=-y-2 y^{\prime}-4 y-2 y^{\prime} \Rightarrow y^{\prime \prime}+4 y^{\prime}+5 y=0
\end{aligned}
$
Question 16.
The particular integral of the differential equation $f(D) y=e^{a x}$ where $f(D)=(D-a)^2$
(a) $\frac{x^2}{2} e^{2 x}$
(b) $\mathrm{xe}^{\mathrm{ax}}$
(c) $\frac{x}{2} e^{2 x}$
(d) $\mathrm{x}^2 \mathrm{e}^{2 \mathrm{x}}$
Answer:

(a) $\frac{x^2}{2} e^{2 x}$

Question 17.
The differential equation of $x^2+y^2=a^2$ is
(a) $x d y+y d x=0$
(b) $y d x-x d y=0$
(c) $x d x-y d x=0$
(d) $x d x+y d y=0$
Answer:
(d) $x d x+y d y=0$
Hint:
$
\begin{aligned}
& \mathrm{x}^2+\mathrm{y}^2=\mathrm{a}^2 \\
& \Rightarrow 2 \mathrm{x}+2 \mathrm{y} \frac{d y}{d x}=0 \\
& \Rightarrow \mathrm{x} \mathrm{dx}+\mathrm{y} \mathrm{dy}=0
\end{aligned}
$
Question 18.
The complementary function of $\frac{d^2 y}{d x^2}-\frac{d y}{d x}=0$ is
(a) $\mathrm{A}+\mathrm{B} \mathrm{e}^{\mathrm{x}}$
(b) $(A+B) e^x$
(c) $(\mathrm{Ax}+\mathrm{B}) \mathrm{e}^{\mathrm{x}}$
(d) $\left(A e^x+B\right)$
Answer:
(a) $\mathrm{A}+\mathrm{B} \mathrm{e}^{\mathrm{x}}$
Hint:
A.E is $\mathrm{m}^2-\mathrm{m}=0$
$\Rightarrow \mathrm{m}(\mathrm{m}-1)=0$
$\Rightarrow \mathrm{m}=0,1$
$\mathrm{CF}$ is $\mathrm{Ae}^{0 \mathrm{x}}+\mathrm{Be}^{\mathrm{x}}=\mathrm{A}+\mathrm{Be}^{\mathrm{x}}$
Question 19.
The P.I of $\left(3 \mathrm{D}^2+\mathrm{D}-14\right) \mathrm{y}=13 \mathrm{e}^{2 \mathrm{x}}$ is
(a) $\frac{x}{2} \mathrm{e}^{2 \mathrm{x}}$
(b) $\mathrm{xe}^{2 \mathrm{x}}$
(c) $\frac{x^2}{2} \mathrm{e}^{2 \mathrm{x}}$

(d) $13 x e^{2 x}$
Answer:
(b) $x^{2 x}$
Hint:
$
\mathrm{PI}=\frac{1}{3 \mathrm{D}^2+\mathrm{D}-14} 13 e^{2 x}
$
Replace D by $2 . \quad 3 \mathrm{D}^2+\mathrm{D}-14=0$
$
\therefore \quad \text { PI }=x \frac{1}{6 \mathrm{D}+1} 13 e^{2 x}
$
Replace D by 2. $\mathrm{PI}=x e^{2 x}$
Question 20.
The general solution of the differential equation $\frac{d y}{d x}=\cos \mathrm{x}$ is
(a) $y=\sin \mathrm{x}+1$
(b) $y=\sin \mathrm{x}-2$
(c) $y=\cos x+c, c$ is an arbitrary constant
(d) $y=\sin x+c, c$ is an arbitrary constant
Answer:
(d) $y=\sin x+c, c$ is an arbitrary constant
Question 21.
A homogeneous differential equation of the form $\frac{d y}{d x}=f\left(\frac{y}{x}\right)$ can be solved by making substitution
(a) $y=v x$
(b) $\mathrm{v}=\mathrm{yx}$
(c) $x=v y$
(d) $x=v$
Answer:
(a) $y=v x$
Question 22.
A homogeneous differential equation of the form $\frac{d x}{d y}=f\left(\frac{x}{y}\right)$ can be solved by making substitution,
(a) $x=v y$
(b) $y=v x$
(c) $y=v$
(d) $x=v$
Answer:
(a) $x=v y$

Question 23.
The variable separable form of $\frac{d y}{d x}=\frac{y(x-y)}{x(x+y)}$ by taking $y=v x$ and $\frac{d y}{d x}=v+x \frac{d v}{d x}$ is
(a) $\frac{2 v^2}{1+v} d v=\frac{d x}{x}$
(b) $\frac{2 v^2}{1+v} d v=-\frac{d x}{x}$
(c) $\frac{2 v^2}{1-v} d v=\frac{d x}{x}$
(d) $\frac{1+v}{2 v^2} d v=-\frac{d x}{x}$
Answer:
(d) $\frac{1+v}{2 v^2} d v=-\frac{d x}{x}$
Hint:
$
\begin{aligned}
v+x \frac{d v}{d x} & =\frac{v x(x-v x)}{x(x+v x)}=\frac{v(1-v)}{1+v}=\frac{v-v^2}{1+v} \\
\frac{x d v}{d x}=\frac{v-v^2}{1+v}-v & =\frac{v-v^2-v-v^2}{1+v}=\frac{-2 v^2}{1+v} \\
\Rightarrow \quad \frac{1+v}{2 v^2} d v & =\frac{-d x}{x}
\end{aligned}
$
Question 24.
Which of the following is the homogeneous differential equation?
(a) $(3 x-5) d x=(4 y-1) d y$
(b) $x y d x-\left(x^3+y^3\right) d y=0$
(c) $y^2 d x+\left(x^2-x y-y^2\right) d y=0$
(d) $\left(x^2+y\right) d x=\left(y^2+x\right) d y$
Answer:
(c) $y^2 d x+\left(x^2-x y-y^2\right) d y=0$
Question 25.
The solution of the differential equation $\frac{d y}{d x}=\frac{y}{x}+\frac{f\left(\frac{y}{x}\right)}{f^{\prime}\left(\frac{y}{x}\right)}$ is
(a) $\mathrm{f}\left(\frac{y}{x}\right)=\mathrm{kx}$
(b) $\mathrm{xf}\left(\frac{y}{x}\right)=\mathrm{k}$
(c) $\mathrm{f}\left(\frac{y}{x}\right)=\mathrm{ky}$
(d) $y \mathrm{f}\left(\frac{y}{x}\right)=\mathrm{k}$
Answer:
(a) $f\left(\frac{y}{x}\right)=\mathrm{kx}$

Hint:
$
\begin{aligned}
& \frac{d y}{d x}=\frac{y}{x}+\frac{f\left(\frac{y}{x}\right)}{f^{\prime}\left(\frac{y}{x}\right)} \\
& \text { Put } y=v x \text { and } \frac{d y}{d x}=v+x \frac{d v}{d x} \\
& \Rightarrow \quad \frac{x d v}{d x}+v=\frac{v x}{x}+\frac{f\left(\frac{v x}{x}\right)}{f^{\prime}\left(\frac{v x}{x}\right)} \\
& \frac{x d v}{d x}+v=v+\frac{f(v)}{f^{\prime}(v)} \\
& \frac{x d v}{d x}=\frac{f(v)}{f^{\prime}(v)} \\
& \int \frac{f^{\prime}(v)}{f(v)} d v=\int \frac{d x}{x} \\
& \log f(v)=\log x+\log k \\
& f(v)=k x \quad \text { (or }) f\left(\frac{y}{x}\right)=k x
\end{aligned}
$