MCQ 2011 Mark
Differential equation having solution $y=A x+B^3$ is of order
Answer(b) : Given solution contanty two arbitrary constant.
$\therefore \quad$ Order differential equation is 2 .
View full question & answer→MCQ 2021 Mark
The solution of $x \frac{d y}{d x}+y=e^x$ is
AnswerCorrect option: A. $y=\frac{e^x}{x}+\frac{c}{x}$
$x \frac{d y}{d x}+y=e^x$
$\frac{d y}{d x}+\frac{y}{x}=\frac{e^x}{x}$
It is a linear differential equation with
$\text { I.F. }=e^{\int \frac{1}{x} d x}=e^{\log x}=x$
Now, solution is $y \cdot x=\int \frac{e^x}{x} \cdot x d x+c$
$\Rightarrow y x=e^x+c$
$\Rightarrow y=\frac{e^x}{x}+\frac{c}{x}$
View full question & answer→MCQ 2031 Mark
The integrating factor of the differential equation $\frac{d y}{d x}+y=\frac{1+y}{x}$ is
- A
$\frac{x}{e^x}$
- ✓
$\frac{e^x}{x}$
- C
$x e^x$
- D
$e^x$
AnswerCorrect option: B. $\frac{e^x}{x}$
$\frac{d y}{d x}+y=\frac{1+y}{x}$
$\Rightarrow \frac{d y}{d x}+y=\frac{1}{x}+\frac{y}{x}$
$\Rightarrow \frac{d y}{d x}+y-\frac{y}{x}=\frac{1}{x}$
$\Rightarrow \frac{d y}{d x}+\left(1-\frac{1}{x}\right) y=\frac{1}{x}$
$\therefore \text { I.F. }=e^{\int\left(1-\frac{1}{x}\right) d x}$
$=e^{x-\log x}=e^x e^{-\log x}$
$=e^x e^{\log (x)^{-1}}$
$=e^x x^{-1}$
$=\frac{e^x}{x}$
View full question & answer→MCQ 2041 Mark
The solution of the differential equation $\frac{d y}{d x}=\frac{1+y^2}{1+x^2}$ is
AnswerCorrect option: B. $\tan ^{-1} y-\tan ^{-1} x=c$
(b) : $\frac{d y}{d x}=\frac{1+y^2}{1+x^2} \Rightarrow \frac{d y}{1+y^2}=\frac{d x}{1+x^2}$
On integrating both sides, we get
$
\tan ^{-1} y=\tan ^{-1} x+c \Rightarrow \tan ^{-1} y-\tan ^{-1} x=c
$
View full question & answer→MCQ 2051 Mark
Integrating factor of the differential equation $\frac{d y}{d x}+y \tan x-\sec x=0$ is
- A
$\cos x$
- ✓
$\sec x$
- C
$e^{\cos x}$
- D
$e^{\sec x}$
AnswerCorrect option: B. $\sec x$
(b) : $\frac{d y}{d x}+y \tan x-\sec x=0$
$\therefore \quad$ I.F. $=e^{\int \tan x d x}=e^{\log \sec x}=\sec x$
View full question & answer→MCQ 2061 Mark
Which of the following is a second order differential equation?
- A
$\left(y^{\prime}\right)^2+x=y^2$
- ✓
$y^{\prime} y^{\prime \prime}+y=\sin x$
- C
$y^{\prime \prime \prime}+\left(y^{\prime \prime}\right)^2+y=0$
- D
$y^{\prime}=y^2$
AnswerCorrect option: B. $y^{\prime} y^{\prime \prime}+y=\sin x$
(b) : (a) $\left(\frac{d y}{d x}\right)^2+x=y^2$; order $=1$
(b) $\left(\frac{d y}{d x}\right)\left(\frac{d^2 y}{d x^2}\right)+y=\sin x ;$ order $=2$
(c) $\frac{d^3 y}{d x^3}+\left(\frac{d^2 y}{d x^2}\right)^2+y=0 ;$ order $=3$
(d) $\frac{d y}{d x}=y^2 ;$ order $=1$
View full question & answer→MCQ 2071 Mark
The number of solutions of $\frac{d y}{d x}=\frac{y+1}{x-1}$, when $y(1)=2$ is
Answer$\frac{d y}{d x}=\frac{y+1}{x-1}$
$\Rightarrow \frac{d y}{y+1}=\frac{d x}{x-1}$
On integrating both sides, we get
$\log (y+1)+\log c=\log (x-1)$
$\Rightarrow(y+1) c=(x-1)$
Now, $y(1)=2$
$\Rightarrow 3 c=0$
$\Rightarrow c=0$
$\therefore x-1=0$
$\Rightarrow x=1$
Hence, only one solution exists.
View full question & answer→MCQ 2081 Mark
Which of the following is the integrating factor of $x d y / d x-y=x^4-3 x$ ?
- A
$x$
- B
$\log x$
- ✓
$\frac{1}{x}$
- D
$-x$
AnswerCorrect option: C. $\frac{1}{x}$
$ x \frac{d y}{d x}-y=x^4-3 x$
$\Rightarrow \frac{d y}{d x}-\frac{y}{x}=\frac{x^4-3 x}{x}$
$\Rightarrow \frac{d y}{d x}-\frac{1}{x} \cdot y=x^3-3$
$\therefore \text { I.F. }=e^{-\int \frac{1}{x} d x}$
$=e^{-\log x}$
$=e^{\log (x)^{-1}}$
$=x^{-1}$
$=\frac{1}{x}$
View full question & answer→MCQ 2091 Mark
The order and degree respectively of the differential equation $\frac{d^2 y}{d x^2}+\left(\frac{d y}{d x}\right)^{\frac{1}{4}}+x^{\frac{1}{5}}=0$, are
Answer(a): $\frac{d^2 y}{d x^2}+\left(\frac{d y}{d x}\right)^{\frac{1}{4}}+x^{\frac{1}{5}}=0$
Clearly, order of given differential equation is 2 and degree is not defined.
View full question & answer→MCQ 2101 Mark
If $\frac{d y}{d x}=\frac{x+y}{x}, y(1)=1$, then $y=$
- A
$x+\ln x$
- B
$x^2+x \ln x$
- C
$x e^{x-1}$
- ✓
$x+x \ln x$
AnswerCorrect option: D. $x+x \ln x$
It is a homogeneous equation.
Substitute $y=v x$
$\Rightarrow \frac{d y}{d x}=x \frac{d v}{d x}+v$
Now, given equation becomes
$x \frac{d v}{d x}+v=1+v$
$\Rightarrow d v=\frac{d x}{x}$
On integrating both sides, we get
$v=\ln x+c$
$\Rightarrow \frac{y}{x}=\ln x+c$
$\because y(1)=1$
$\Rightarrow x=1, y=1$
$\Rightarrow c=1$
$\therefore y=x+x \ln x$
View full question & answer→MCQ 2111 Mark
Which of the following is the general solution of the differential equation $\frac{d y}{d x}=\frac{y}{x}$ ?
- A
$y=\frac{k}{x}$
- B
$y=k \log x$
- ✓
$y=k x$
- D
$\log y=k x$
AnswerCorrect option: C. $y=k x$
(c) : $\frac{d y}{d x}=\frac{y}{x} \Rightarrow \frac{d x}{x}=\frac{d y}{y}$
$
\Rightarrow \log x=\log y+\log C \Rightarrow x=y C \Rightarrow y=\frac{1}{C} x \Rightarrow y=k x \text {. }
$
View full question & answer→MCQ 2121 Mark
The general solution of the differential equation $\frac{d y}{d x}=\frac{x^2}{y^2}$ is
- ✓
$x^3-y^3=c$
- B
$x^3+y^3=c$
- C
$x^2+y^2=c$
- D
$x^2-y^2=c$
AnswerCorrect option: A. $x^3-y^3=c$
$\frac{d y}{d x}=\frac{x^2}{y^2}$
$\Rightarrow y^2 d y=x^2 d x$
$\Rightarrow \int y^2 d y=\int x^2 d x$
$\Rightarrow \frac{y^3}{3}=\frac{x^3}{3}+C$
$\Rightarrow x^3-y^3=-3 C=c \text { (say). }$
View full question & answer→MCQ 2131 Mark
If $\frac{d y}{d x}=y \sin 2 x, y(0)=1$, then solution is
- ✓
$y=e^{\sin ^2 x}$
- B
$y=\sin ^2 x$
- C
$y=\cos ^2 x$
- D
$y=e^{\cos ^2 x}$
AnswerCorrect option: A. $y=e^{\sin ^2 x}$
(a) : We have, $\frac{d y}{d x}=y \sin 2 x$
$
\Rightarrow \quad \frac{d y}{y}=\sin 2 x d x \Rightarrow \log y=-\frac{\cos 2 x}{2}+C
$
Since $y(0)=1 \Rightarrow x=0, y=1 \Rightarrow C=1 / 2$
$
\therefore \quad \log y=\frac{1}{2}(1-\cos 2 x) \Rightarrow \log y=\sin ^2 x \Rightarrow y=e^{\sin ^2 x}
$
View full question & answer→MCQ 2141 Mark
The general solution of the differential equation $x\left(1-y^2\right) d x+y\left(1-x^2\right) d y=0$ is
- A
$\left(1-x^2\right)\left(1-y^2\right)=0$
- ✓
$\left(1-x^2\right)\left(1-y^2\right)=C$
- C
$\left(1-x^2\right)=C\left(1-y^2\right)$
- D
$\left(1+y^4\right)=C\left(1-x^2\right)$
AnswerCorrect option: B. $\left(1-x^2\right)\left(1-y^2\right)=C$
(b) : Given differential equation is
$
\begin{aligned}
x\left(1-y^2\right) d x+y\left(1-x^2\right) d y & =0 \\
\Rightarrow\left(\frac{x}{1-x^2}\right) d x+\left(\frac{y}{1-y^2}\right) d y & =0
\end{aligned}
$
On integrating, we get
$
\begin{aligned}
& \frac{1}{2} \log \left(1-x^2\right)-\frac{1}{2} \log \left(1-y^2\right)=k \\
\Rightarrow \quad & \log \left(1-x^2\right)\left(1-y^2\right)=-2 k \Rightarrow\left(1-x^2\right)\left(1-y^2\right)=e^{-2 k}=C
\end{aligned}
$
View full question & answer→MCQ 2151 Mark
If $y^{\prime}=y+1, y(0)=1$, then $y(\ln 2)=$
Answer(c) : $y^{\prime}=y+1 \Rightarrow \frac{d y}{y+1}=d x$
$\Rightarrow \ln (y+1)=x+C$ (Integrating both sides)
Now, $y(0)=1 \Rightarrow C=\ln 2$
$\therefore \ln \left(\frac{y+1}{2}\right)=x \Rightarrow y+1=2 e^x \Rightarrow y=2 e^x-1$
So, $y(\ln 2)=2 e^{\ln 2}-1=4-1=3$
View full question & answer→MCQ 2161 Mark
The differential equation $\frac{d y}{d x}=\frac{\sqrt{1-y^2}}{y}$ determines a family of circle with
- A
variable radii and fixed centre $(0,1)$
- B
variable radii and fixed centre $(0,-1)$
- ✓
fixed radius 1 and variable centre on $x$-axis
- D
fixed radius 1 and variable centre on $y$-axis
AnswerCorrect option: C. fixed radius 1 and variable centre on $x$-axis
(c) : We have, $\frac{y d y}{\sqrt{1-y^2}}=d x$
On integration, we get $-\sqrt{1-y^2}=x+C$
$
\Rightarrow 1-y^2=(x+C)^2 \Rightarrow(x+C)^2+y^2=1
$
which is a circle of radius 1 and centre on the $x$-axis.
View full question & answer→MCQ 2171 Mark
For the differential equation $x \frac{d y}{d x}+2 y=x y \frac{d y}{d x}$, which of the following is true?
AnswerCorrect option: A. Order is 1 and degree is 1
(a) : Given, $x \frac{d y}{d x}(1-y)+2 y=0$
$
\Rightarrow\left(\frac{1-y}{y}\right) d y+2 \frac{d x}{x}=0 \Rightarrow \frac{1}{y} d y-d y+2 \frac{d x}{x}=0
$
On integrating, we get
$
\Rightarrow \ln y-y+2 \ln x=C \Rightarrow \ln \left(y x^2\right)=C+y
$
View full question & answer→MCQ 2181 Mark
Given the differential equation
$
\frac{d y}{d x}=\frac{6 x^2}{2 y+\cos y} ; y(1)=\pi \text {. }
$
Which of the following statements is correct?
- A
General solution is $y^2-\sin y=-2 x^3+C$.
- ✓
General solution is $y^2+\sin y=2 x^3+C$.
- C
Particular solution is $y^2+\sin y=2 x^3+\pi^2+2$.
- D
The value of the integration constant is $\pi^2+2$.
AnswerCorrect option: B. General solution is $y^2+\sin y=2 x^3+C$.
(b) : We have, $\frac{d y}{d x}=\frac{6 x^2}{2 y+\cos y}$
$
\begin{array}{ll}
\Rightarrow & \int(2 y+\cos y) d y=\int 6 x^2 d x \\
\Rightarrow & y^2+\sin y=2 x^3+C \\
\because & y(1)=\pi \therefore C=\pi^2-2 \\
\therefore & \text { Solution is } y^2+\sin y=2 x^3+\pi^2-2 \\
\Rightarrow & y^2+\sin y=2 x^3+C \text {, where } C=\pi^2-2
\end{array}
$
View full question & answer→MCQ 2191 Mark
Solution of the differential equation $\frac{e^x-e^{-x}}{e^x+e^{-x}}=\frac{d x-d y}{d x+d y}$, is
- A
$2 y e^{2 x}=C \cdot e^{2 x}+1$
- ✓
$2 y e^{2 x}=C \cdot e^{2 x}-1$
- C
$y e^{2 x}=C \cdot e^{2 x}+2$
- D
AnswerCorrect option: B. $2 y e^{2 x}=C \cdot e^{2 x}-1$
Given equation $\frac{e^x-e^{-x}}{e^x+e^{-x}}=\frac{d x-d y}{d x+d y}$
Applying componendo and dividendo, we get
$\frac{d y}{d x}=\frac{e^{-x}}{e^x}$
$\Rightarrow d y=e^{-2 x} d x$
$\Rightarrow 2 y=-e^{-2 x}+C ($ Integrating both sides $)$
$\Rightarrow 2 y e^{2 x}=C \cdot e^{2 x}-1$
View full question & answer→MCQ 2201 Mark
The solution of the differential equation $\frac{d y}{d x}=\frac{x^2+1}{2 x y}$ satisfying $y(1)=1$, is
AnswerCorrect option: A. $y^2=\frac{x^2}{2}+\ln |x|+\frac{1}{2}$
(a) : Given, $\frac{d y}{d x}=\frac{x^2+1}{2 x y}$
$
\Rightarrow 2 y d y=\frac{\left(x^2+1\right)}{x} d x \Rightarrow 2 y d y=\frac{(x+1)}{x} d x
$
Integrating on both sides, we get
$
\begin{aligned}
& y^2=\frac{x^2}{2}+\ln |x|+C \\
\Rightarrow \quad & y^2=\frac{x^2}{2}+C
\end{aligned}
$
When $x=1, y=1$
$
\therefore \quad C=\frac{1}{2} \quad \therefore \quad y^2=\frac{x^2}{2}+\ln |x|+\frac{1}{2}
$
View full question & answer→MCQ 2211 Mark
The degree of the differential equation $\left(\frac{d^2 y}{d x^2}\right)^{2 / 3}+4-3 \frac{d y}{d x}=0$ is
Answer(a) : $\left(\frac{d^2 y}{d x^2}\right)^{2 / 3}=3 \frac{d y}{d x}-4 \Rightarrow\left(\frac{d^2 y}{d x^2}\right)^2=\left(3 \frac{d y}{d x}-4\right)^3$
$\therefore \quad$ Degree of the differential equation is 2 .
View full question & answer→MCQ 2221 Mark
The integrating factor of the differential equation $x \frac{d y}{d x}+2 y=x^2$ is
- A
$x$
- ✓
$x^2$
- C
$\frac{1}{x^2}$
- D
$x^3$
Answer(b): We have, $x \frac{d y}{d x}+2 y=x^2 \Rightarrow \frac{d y}{d x}+2 \frac{y}{x}=x$
$
\therefore \quad \text { I.F. }=e^{\int \frac{2}{x} d x}=e^{2 \log x}=e^{\log x^2}=x^2
$
View full question & answer→MCQ 2231 Mark
The degree of the differential equation $\left(\frac{d^2 y}{d x^2}\right)^2+\left(\frac{d y}{d x}\right)^2=x \sin \frac{d y}{d x}$ is
Answer(d) : Since, the given differential equation is not a poly-nomial in $\frac{d y}{d x}$. Therefore, its degree is not defined.
View full question & answer→MCQ 2241 Mark
The integrating factor of the differential equation $x \frac{d y}{d x}-y=\log x$ is
- ✓
$\frac{1}{x}$
- B
$x$
- C
$y$
- D
$\frac{1}{y}$
AnswerCorrect option: A. $\frac{1}{x}$
We have, $x \frac{d y}{d x}-y=\log x$
$\Rightarrow \frac{d y}{d x}-\frac{y}{x}=\frac{\log x}{x}$
Clearly, it is a linear differential equation of the form
$\frac{d y}{d x}+P y=Q$
$\therefore \text { I.F. }=e^{\int-\frac{1}{x} d x}$
$=e^{-\log x}$
$=\frac{1}{x}$
View full question & answer→MCQ 2251 Mark
The differential equation having solution as $y=17 e^x+a e^{-x}$ is
- A
$y^{\prime \prime}-x=0$
- ✓
$y^{\prime \prime}-y=0$
- C
$y^{\prime}-y=0$
- D
$y^{\prime}-x=0$
AnswerCorrect option: B. $y^{\prime \prime}-y=0$
(b): We have, $y=17 e^x+a e^{-x} \Rightarrow y^{\prime}=17 e^x-a e^{-x}$
$
\Rightarrow y^{\prime \prime}=17 e^x+a e^{-x} \Rightarrow y^{\prime \prime}=y \Rightarrow y^{\prime \prime}-y=0
$
View full question & answer→MCQ 2261 Mark
The integrating factor of the differential equation $\left(x+3 y^2\right) \frac{d y}{d x}=y$ is
- A
$y$
- B
$-y$
- ✓
$\frac{1}{y}$
- D
$-\frac{1}{y}$
AnswerCorrect option: C. $\frac{1}{y}$
(c) : We have, $\left(x+3 y^2\right) \frac{d y}{d x}=y$
$
\Rightarrow \frac{x+3 y^2}{y}=\frac{d x}{d y} \Rightarrow \frac{d x}{d y}-\frac{x}{y}=3 y
$
This is a linear differential equation.
$
\therefore \quad \text { I.F. }=e^{-\int \frac{d y}{y}}=e^{-\log y}=e^{\log y^{-1}}=\frac{1}{y}
$
View full question & answer→MCQ 2271 Mark
The integrating factor for solving the differential equation $x \frac{d y}{d x}-y=2 x^2$ is
- A
$e^{-y}$
- B
$e^{-e}$
- C
$x$
- ✓
$\frac{1}{x}$
AnswerCorrect option: D. $\frac{1}{x}$
(d) : We have, $x \frac{d y}{d x}-y=2 x^2$
i.e., $\frac{d y}{d x}-\frac{y}{x}=2 x \quad \therefore \quad$ I.F. $=e^{\int \frac{-1}{x} d x}=e^{-\ln x}=e^{\ln x^{-1}}=\frac{1}{x}$
$\therefore$ Integrating factor is $\frac{1}{x}$.
View full question & answer→MCQ 2281 Mark
Integrating factor of differential equation $\frac{d y}{d x}+y \tan x-\sec x=0$ is
- A
$\cos x$
- ✓
$\sec x$
- C
$e^{\cos x}$
- D
$e^{\sec x}$
AnswerCorrect option: B. $\sec x$
(b) : We have, $\frac{d y}{d x}+y \tan x-\sec x=0$
or $\frac{d y}{d x}+y \tan x=\sec x$
This is linear differential equation of the type $\frac{d y}{d x}+P y=Q$ with $P=\tan x, Q=\sec x$
$
\therefore \quad \text { I.F. }=e^{\int P d x}=e^{\int \tan x d x}=e^{(\log \sec x)}=\sec x
$
View full question & answer→MCQ 2291 Mark
The differential equation whose solution is $y=A e^{3 x}+B e^{-3 x}$ is given by
AnswerCorrect option: C. $y_2-9 y=0$
We have, $y=A e^{3 x}+B e^{-3 x}$
Differentiating $w.r.t.\ x$, we get
$y_1=3 A e^{3 x}-3 B e^{-3 x}$
Again differentiating $w.r.t.\ x$, we get
$y_2=9 A e^{3 x}+9 B e^{-3 x}$
$=9\left(A e^{3 x}+B e^{-3 x}\right)$
$=9 y$
$\Rightarrow y_2-9 y$
$=0$
View full question & answer→MCQ 2301 Mark
Let the differential equation is $\left[1+\left(\frac{d y}{d x}\right)^2\right]^{\frac{3}{2}}=\frac{d^2 y}{d x^2}$. Which of the following statements is/are true?
(i) Degree of the differential equation is 2 .
(ii) Order of the differential equation is 3 .
(iii) Order and degree of differential equation respectively are 2,2 .
Answer(d) : The given differential equation can be written as
$
\left[1+\left(\frac{d y}{d x}\right)^2\right]^3=\left(\frac{d^2 y}{d x^2}\right)^2
$
Clearly, order and degree of given differential equation are 2,2 respectively.
View full question & answer→MCQ 2311 Mark
The sum of the order and degree of the differential equation $1+\left(\frac{d y}{d x}\right)^4=7\left(\frac{d^2 y}{d x^2}\right)^3$ is
Answer(a) : Order $=2$, Degree $=3$
$\therefore$ Order + Degree $=2+3=5$
View full question & answer→MCQ 2321 Mark
The order and the degree of the differential equation $\left(1+3 \frac{d y}{d x}\right)^2=4 \frac{d^3 y}{d x^3}$ respectively, are
Answer(b) : We have, $\left(1+3 \frac{d y}{d x}\right)^2=4 \frac{d^3 y}{d x^3}$
Here, order $=3$ as highest order derivative is $\frac{d^3 y}{d x^3}$.
And degree $=1$, as power of highest order derivative i.e., $\frac{d^3 y}{d x^3}$ is 1 .
View full question & answer→MCQ 2331 Mark
The sum of the order and the degree of the differential equation $\frac{d}{d x}\left(\frac{d y}{d x}\right)^3$ is
Answer(b) : The given differential equation is,
$
\frac{d}{d x}\left(\frac{d y}{d x}\right)^3=0 \quad \Rightarrow \quad 3\left(\frac{d y}{d x}\right)^2\left(\frac{d^2 y}{d x^2}\right)=0
$
$\therefore \quad$ Order $=2$ and degree $=1$
So, required sum $=2+1=3$
View full question & answer→MCQ 2341 Mark
The order of the differential equation whose general solution is given by
$
y=\left(C_1+C_2\right) \cos \left(x+C_3\right)-C_4 e^{x+C_5}
$
where $C_1, C_2, C_3, C_4, C_5$ are arbitrary constants, is
Answer(c) : The given equation can be rewritten as
$
y=A \cos \left(x+C_3\right)-B e^x
$
where, $A=C_1+C_2$ and $B=C_4 e^{C_5}$
So, there are three independent variables, $\left(A, B, C_3\right)$.
Hence, the differential equation is of order 3 .
View full question & answer→MCQ 2351 Mark
The order of the differential equation whose solution is $y=a \cos x+b \sin x+c e^{-x}$ is
Answer(a): $y=a \cos x+b \sin x+c e^{-x}$
It is a third order differential equation, as it contains three arbitrary constants.
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