MCQ 11 Mark
The integrating factor of the differential equation $\left(x+2 y^2\right) \frac{d y}{d x}=y(y>0)$ is :
- A
$\frac{1}{x}$
- B
$x$
- C
$y$
- D
$\frac{1}{y}$
Answer$\begin{array}{l}\text {We have, }\left(x+2 y^2\right) \frac{d y}{d x}=y \\ \Rightarrow \quad \frac{x+2 y^2}{y}=\frac{d x}{d y} \Rightarrow \frac{d x}{d y}-\frac{x}{y}=2 y \\ \therefore \quad \text { I.F. }=e^{\int \frac{1}{y} d y}=e^{-\log y}=\frac{1}{y}\end{array}$
View full question & answer→MCQ 21 Mark
The order of the following differential equation $\frac{d^3 y}{d x^3}+x\left(\frac{d y}{d x}\right)^5=4 \log \left(\frac{d^4 y}{d x^4}\right)$ is:
View full question & answer→MCQ 31 Mark
The degree of the differential equation $\left(y^{\prime \prime}\right)^2+\left(y^{\prime}\right)^3$ $=x \sin \left(y^{\prime}\right)$ is :
AnswerSince, the given differential equation is not a polynomial in $\frac{d y}{d x}$.
$\therefore \quad$ Its degree is not defined.
View full question & answer→MCQ 41 Mark
The general solution of the differential equation $x d y+y d x=0$ is:
- A
$x y=c$
- B
$x+y=c$
- C
$x^2+y^2=c^2$
- D
$\log y=\log x+c$
AnswerWe have, $x d y+y d x=0$
$
\begin{array}{l}
\Rightarrow x d y=-y d x \Rightarrow \int \frac{d y}{y}=-\int \frac{d x}{x} \Rightarrow \log y=-\log x+\log c \Rightarrow y=\frac{c}{x} \\
\Rightarrow x y=\text { constant }
\end{array}
$
View full question & answer→MCQ 51 Mark
The degree and order of differential equation $y^{\prime \prime 2}+\log \left(y^{\prime}\right)=x^5$ respectively are:
AnswerWe have, $y^{\prime \prime 2}+\log \left(y^{\prime}\right)=x^5$
As the highest order derivative is $y^{\prime \prime}$. So, order $=2$
But, the given differential equation is not a polynomial.
Therefore, its degree is not defined.
View full question & answer→MCQ 61 Mark
The differential equation $\frac{d y}{d x}=F(x, y)$ will not be a homogeneous differential equation, if $F(x, y)$ is:
Answer$
\begin{array}{l}\text {Let } F(x, y)=\cos x-\sin \frac{y}{x} \\
\Rightarrow F(\lambda x, \lambda y)=\cos \lambda x-\sin \frac{\lambda y}{\lambda x}=\cos \lambda x-\sin \frac{y}{x} \\
\quad \neq \lambda\left(\cos x-\sin \frac{y}{x}\right)
\end{array}
$$\therefore \quad \cos x-\sin \frac{y}{x}$ is not homogeneous.
View full question & answer→MCQ 71 Mark
The degree of the differential equation $\left[1+\left(\frac{d y}{d x}\right)^2\right]^3=\left(\frac{d^2 y}{d x^2}\right)^2$ is
Answer$\begin{array}{l}\text {}\left(1+\left(\frac{d y}{d x}\right)^2\right)^3=\left(\frac{d^2 y}{d x^2}\right)^2 \\ \Rightarrow \text { degree }=2\end{array}$
View full question & answer→MCQ 81 Mark
The general solution of the differential equation $y d x-x d y=0$; (Given $x, y>0)$, is of the form
- A
$x y=c$
- B
$x=c y^2$
- C
$y=c x$
- D
$y=c x^2$
AnswerGiven, $y d x-x d y=0 \Rightarrow y d x=x d y \Rightarrow \frac{d y}{y}=\frac{d x}{x}$;
Integrating both sides, we get $\int \frac{d y}{y}=\int \frac{d x}{x}$
$\log _{ e }|y|=\log _{ e }|x|+\log _{ e }|c|$
Since $x, y, c>0$, we write $\log _{ c } y=\log _{ e } x+\log _{ e } c \Rightarrow y=c x$.
$(1 / 2)$
View full question & answer→MCQ 91 Mark
In which of the following differential equations is the degree equal to its order?
- A
$x^3\left(\frac{d y}{d x}\right)-\frac{d^3 y}{d x^3}=0$
- B
$\left(\frac{d^3 y}{d x^3}\right)^3+\sin \left(\frac{d y}{d x}\right)=0$
- ✓
$x^2\left(\frac{d y}{d x}\right)^4+\sin y-\left(\frac{d^2 y}{d x^2}\right)^2=0$
- D
$\left(\frac{d y}{d x}\right)^3+x\left(\frac{d^2 y}{d x^2}\right)-y^3\left(\frac{d^3 y}{d x^3}\right)+y=0$
AnswerCorrect option: C. $x^2\left(\frac{d y}{d x}\right)^4+\sin y-\left(\frac{d^2 y}{d x^2}\right)^2=0$
$x^2\left(\frac{d y}{d x}\right)^4+\sin y-\left(\frac{d^2 y}{d x^2}\right)^2=0$
View full question & answer→MCQ 101 Mark
Kapila is trying to find the general solution of the following differential equations.
(i) $x e^{\frac{x}{y}} d x-y e^{\frac{3 x}{y}} d y=0$
(ii) $(2 x+1) \frac{d y}{d x}=3-2 y$
(iii) $\frac{d y}{d x}=\sin x-\cos y$
Which of the above become variable separable by substituting $y=b . x$, where $b$ is a variable?
- ✓
- B
- C
all - (i), (ii) and (iii)
- D
View full question & answer→MCQ 111 Mark
The integrating factor of the differential equation $\left(3 x^2+y\right) \frac{d x}{d y}=x$ is
- A
$\frac{1}{x}$
- B
$\frac{1}{x^2}$
- C
$\frac{2}{x}$
- D
$-\frac{1}{x}$
AnswerGiven, $\left(3 x^2+y\right) \frac{d x}{d y}=x \Rightarrow x \frac{d y}{d x}=3 x^2+y$
$
\begin{array}{l}
\Rightarrow \frac{d y}{d x}=3 x+\frac{y}{x} \Rightarrow \frac{d y}{d x}-\frac{1}{x} \cdot y=3 x \\
\text { I.F. }=e^{\int \frac{1}{x} d x}=e^{-\ln x}=\frac{1}{x}
\end{array}
$
View full question & answer→MCQ 121 Mark
The general solution of the differential equation $x d y-\left(1+x^2\right) d x=d x$ is :
AnswerGiven differential equation is
$x d y-\left(1+x^2\right) d x=d x$
$\Rightarrow x d y=d x+\left(1+x^2\right) d x$
$=\left(2+x^2\right) d x$
$\Rightarrow \quad d y=\left(\frac{2}{x}+x\right) d x$
Integrating both sides, we get
$\int d y=\int\left(\frac{2}{x}+x\right) d x$
$\Rightarrow y=2 \log x+\frac{x^2}{2}+C$
View full question & answer→MCQ 131 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
AnswerWe 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 141 Mark
The sum of the order and the degree of the differential equation $\frac{d^2 y}{d x^2}+\left(\frac{d y}{d x}\right)^3=\sin y$ is:
AnswerGiven differential equation is
$
\frac{d^2 y}{d x^2}+\left(\frac{d y}{d x}\right)^3=\sin y
$
Since, the highest order derivative is 2 and its power is 1 .
So, order $=2$
and degree $=1$
$\therefore \quad$ Required sum $=2+1=3$
View full question & answer→MCQ 151 Mark
The number of solutions of the differential equation $\frac{d y}{d x}=\frac{y+1}{x-1}$, when $y(1)=2$, is
AnswerGiven that; $\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 $\int \frac{d y}{y+1}=\int \frac{d x}{x-1}$
$
\begin{array}{l}
\Rightarrow \log (y+1)=\log (x-1)-\log C \\
\Rightarrow \log (y+1)+\log C=\log (x-1) \Rightarrow C=\frac{x-1}{y+1}
\end{array}
$
Now, $y(1)=2 \Rightarrow C=\frac{1-1}{2+1}=0$
$\therefore \quad$ Required solution is $x-1=0$
Hence, only one solution exist.
View full question & answer→MCQ 161 Mark
The difference of the order and the degree of the differential equation $\left(\frac{d^2 y}{d x^2}\right)^2+\left(\frac{d y}{d x}\right)^3+x^4=0$ is:
AnswerSince, $\left(\frac{d^2 y}{d x^2}\right)^2+\left(\frac{d y}{d x}\right)^3+x^4=0$
Order $=2$ and Degree $=2$
$\therefore \quad$ Required difference $=2-2=0$
View full question & answer→MCQ 171 Mark
The general solution of the differential equation $x d y-\left(1+x^2\right) d x=d x$ is :
AnswerGiven differential equation is
$
\begin{aligned}
x d y-\left(1+x^2\right) d x & =d x \\
\Rightarrow \quad x d y & =d x+\left(1+x^2\right) d x \\
& =\left(2+x^2\right) d x \\
\Rightarrow \quad d y & =\left(\frac{2}{x}+x\right) d x
\end{aligned}
$
Integrating both sides, we get
$
\begin{array}{l}
\int d y=\int\left(\frac{2}{x}+x\right) d x \\
\Rightarrow \quad y=2 \log x+\frac{x^2}{2}+C
\end{array}$
View full question & answer→MCQ 181 Mark
The sum of the order and the degree of the differential equation $\frac{d}{d x}\left(\left(\frac{d y}{d x}\right)^3\right)$ is
Answer[There is error in question, the given differential equation should be $\frac{d}{d x}\left(\frac{d y}{d x}\right)^3=0$.]
The given differential equation is,
$
\begin{aligned}
& \frac{d}{d x}\left(\left(\frac{d y}{d x}\right)^3\right)=0 \Rightarrow 3\left(\frac{d y}{d x}\right)^2\left(\frac{d^2 y}{d x^2}\right)=0 \\
\therefore \quad & \text { Order }=2 \text { and degree }=1 . \text { So, required sum }=2+1=3
\end{aligned}
$
View full question & answer→MCQ 191 Mark
If $m$ and $n$, respectively, are the order and the degree of the differential equation $\frac{d}{d x}\left[\left(\frac{d y}{d x}\right)\right]^4=0$, then $m+n=$
AnswerThe given differential equation is $\frac{d}{d x}\left[\left(\frac{d y}{d x}\right)\right]^4=0$$
\Rightarrow 4\left(\frac{d y}{d x}\right)^3 \frac{d^2 y}{d x^2}=0
$Here, $m=2$ and $n=1$
Hence, $m+n=3$
View full question & answer→MCQ 201 Mark
The integrating factor of the differential equation $\left(x+3 y^2\right) \frac{d y}{d x}=y$ is
- A
$y$
- B
$-y$
- C
$\frac{1}{ v }$
- D
$-\frac{1}{y}$
AnswerWe 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 211 Mark
The number of arbitrary constants in the particular solution of a differential equation of second order is (are)
AnswerIn the particular solution of a differential equation of any order, there is no arbitrary constant because in the particular solution of any differential equation, we remove all the arbitrary constant by substituting some particular values.
View full question & answer→MCQ 221 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 231 Mark
The integrating factor of the differential equation $\frac{d y}{d x}+y=\frac{1+y}{x}$ is
- A
$\frac{x}{e^x}$
- B
$\frac{e^x}{x}$
- C
$x e^x$
- D
$e^x$
Answer$
\begin{array}{l}
\text { (b) : } \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 \quad \frac{d y}{d x}+\left(1-\frac{1}{x}\right) y=\frac{1}{x} \\
\therefore \quad \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}
\end{array}
$
View full question & answer→MCQ 241 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 251 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 261 Mark
Order and degree of the differential equation $\left(1+\left(\frac{d y}{d x}\right)^3\right)^{\frac{7}{3}}=7 \frac{d^2 y}{d x^2}$ are respectively
Answer(b) : We have $\left(1+\left(\frac{d y}{d x}\right)^3\right)^{\frac{7}{3}}=7 \frac{d^2 y}{d x^2}$
$
\Rightarrow\left(1+\left(\frac{d y}{d x}\right)^3\right)^7=\left(7 \frac{d^2 y}{d x^2}\right)^3
$
$\therefore \quad$ Order is 2 and degree is 3 .
View full question & answer→MCQ 271 Mark
The solution of $x \frac{d y}{d x}+y=e^x$ is
Answer$
\begin{array}{l}
\text { (a) : } x \frac{d y}{d x}+y=e^x \\
\frac{d y}{d x}+\frac{y}{x}=\frac{e^x}{x}
\end{array}
$
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 281 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 291 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 301 Mark
The number of solutions of $\frac{d y}{d x}=\frac{y+1}{x-1}$, when $y(1)=2$ is
Answer(b) : $\frac{d y}{d x}=\frac{y+1}{x-1} \Rightarrow \frac{d y}{y+1}=\frac{d x}{x-1}$
On integrating both sides, we get
$
\begin{array}{l}
\log (y+1)+\log c=\log (x-1) \Rightarrow(y+1) c=(x-1) \\
\text { Now, } y(1)=2 \Rightarrow 3 c=0 \Rightarrow c=0 \\
\therefore \quad x-1=0 \Rightarrow x=1
\end{array}
$
Hence, only one solution exists.
View full question & answer→MCQ 311 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. $2 y e^{2 x}=C \cdot e^{2 x}-1$
(b): 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
$
\begin{array}{l}
\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 \\
\end{array}
$
View full question & answer→MCQ 321 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 331 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 341 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 351 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 361 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 371 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.
View full question & answer→MCQ 381 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 391 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 401 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. $\frac{1}{x}$
(a): 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
$
\begin{array}{l}
\frac{d y}{d x}+P y=Q \\
\therefore \quad \text { I.F. }=e^{\int-\frac{1}{x} d x}=e^{-\log x}=\frac{1}{x}
\end{array}
$
View full question & answer→MCQ 411 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 421 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. $x^3-y^3=c$
(a) : $\frac{d y}{d x}=\frac{x^2}{y^2} \Rightarrow y^2 d y=x^2 d x$
$
\begin{array}{l}
\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). }
\end{array}
$
View full question & answer→MCQ 431 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 441 Mark
The differential equation whose solution is $y=A e^{3 x}+B e^{-3 x}$ is given by
AnswerCorrect option: C. $y_2-9 y=0$
(c) : 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
$
\begin{array}{l}
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
\end{array}
$
View full question & answer→MCQ 451 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 461 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 471 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 481 Mark
Which of the following is the integrating factor of $x d y / d x-y=x^4-3 x$ ?
- A
$x$
- B
$\log x$
- C
$\frac{1}{x}$
- D
$-x$
Answer$
\begin{array}{l}
\text { (c) : } x \frac{d y}{d x}-y=x^4-3 x \\
\Rightarrow \quad \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 \quad \text { I.F. }=e^{-\int \frac{1}{x} d x}=e^{-\log x}=e^{\log (x)^{-1}}=x^{-1}=\frac{1}{x}
\end{array}
$
View full question & answer→MCQ 491 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 501 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. $x+x \ln x$
(d) : 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
$
\begin{array}{l}
v=\ln x+c \Rightarrow \frac{y}{x}=\ln x+c \\
\because \quad y(1)=1 \Rightarrow x=1, y=1 \Rightarrow c=1 \quad \therefore y=x+x \ln x
\end{array}
$
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