Question 11 Mark
The general solution of the differential equation $x d y+y d x=0$ is:
AnswerWe have, $x d y+y d x=0$
$\Rightarrow x d y=-y d x$
$\Rightarrow \int \frac{d y}{y}=-\int \frac{d x}{x}$
$\Rightarrow \log y=-\log x+\log c$
$\Rightarrow y=\frac{c}{x}$
$\Rightarrow x y=\text { constant }$
View full question & answer→Question 21 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→Question 31 Mark
The integrating factor of the differential equation $\left(x+2 y^2\right) \frac{d y}{d x}=y(y > 0)$ is :
Answer$\text {We have, }\left(x+2 y^2\right) \frac{d y}{d x}=y$
$\Rightarrow \frac{x+2 y^2}{y}=\frac{d x}{d y}$
$\Rightarrow \frac{d x}{d y}-\frac{x}{y}=2 y$
$\therefore \text { I.F. }=e^{\int \frac{1}{y} d y}=e^{-\log y}=\frac{1}{y}$
View full question & answer→Question 41 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$\text {Let } F(x, y)=\cos x-\sin \frac{y}{x}$
$\Rightarrow F(\lambda x, \lambda y)=\cos \lambda x-\sin \frac{\lambda y}{\lambda x}=\cos \lambda x-\sin \frac{y}{x}$
$ \neq \lambda\left(\cos x-\sin \frac{y}{x}\right)$
$\therefore \cos x-\sin \frac{y}{x}$ is not homogeneous.
View full question & answer→Question 51 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→Question 61 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→Question 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$\text {}\left(1+\left(\frac{d y}{d x}\right)^2\right)^3=\left(\frac{d^2 y}{d x^2}\right)^2$
$\Rightarrow \text { degree }=2$
View full question & answer→Question 81 Mark
The general solution of the differential equation $y d x-x d y=0$; (Given $x, y>0)$, is of the form
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→Question 91 Mark
In which of the following differential equations is the degree equal to its order?
Answer$x^2\left(\frac{d y}{d x}\right)^4+\sin y-\left(\frac{d^2 y}{d x^2}\right)^2=0$
View full question & answer→Question 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?
View full question & answer→Question 111 Mark
The integrating factor of the differential equation $\left(3 x^2+y\right) \frac{d x}{d y}=x$ is
AnswerGiven, $\left(3 x^2+y\right) \frac{d x}{d y}=x$
$\Rightarrow x \frac{d y}{d x}=3 x^2+y$
$\Rightarrow \frac{d y}{d x}=3 x+\frac{y}{x}$
$\Rightarrow \frac{d y}{d x}-\frac{1}{x} \cdot y=3 x$
$\text { I.F. }=e^{\int \frac{1}{x} d x}=e^{-\ln x}=\frac{1}{x}$
View full question & answer→Question 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→Question 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→Question 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→Question 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 $\int \frac{d y}{y+1}=\int \frac{d x}{x-1}$
$\Rightarrow \log (y+1)=\log (x-1)-\log C$
$\Rightarrow \log (y+1)+\log C=\log (x-1)$
$\Rightarrow C=\frac{x-1}{y+1}$
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→Question 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→Question 171 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→Question 181 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$
$\Rightarrow d y =\left(\frac{2}{x}+x\right) d x$
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→Question 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→Question 201 Mark
The integrating factor of the differential equation $\left(x+3 y^2\right) \frac{d y}{d x}=y$ is
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→Question 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→Question 221 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?
Answer(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→Question 231 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
Answer$(b):$ 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
$\frac{d y}{d x}=\frac{e^{-x}}{e^x}$
$\Rightarrow d y=e^{-2 x} d x$
$\Rightarrow 2 y=-e^{-2 x}+C ($Integrating both sides$)$
$\Rightarrow 2 y e^{2 x}=C \cdot e^{2 x}-1$
View full question & answer→Question 241 Mark
The differential equation $\frac{d y}{d x}=\frac{\sqrt{1-y^2}}{y}$ determines a family of circle with
Answer(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→Question 251 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→Question 261 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?
Answer(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→Question 271 Mark
The integrating factor of the differential equation $\frac{d y}{d x}+y=\frac{1+y}{x}$ is
Answer$\text { (b) : } \frac{d y}{d x}+y=\frac{1+y}{x}$
$\Rightarrow \frac{d y}{d x}+y=\frac{1}{x}+\frac{y}{x}$
$\Rightarrow \frac{d y}{d x}+y-\frac{y}{x}=\frac{1}{x}$
$\Rightarrow \frac{d y}{d x}+\left(1-\frac{1}{x}\right) y=\frac{1}{x}$
$\therefore \text { I.F. }=e^{\int\left(1-\frac{1}{x}\right) d x}=e^{x-\log x}=e^x e^{-\log x}=e^x e^{\log (x)^{-1}}$
$=e^x x^{-1}=\frac{e^x}{x}$
View full question & answer→Question 281 Mark
Which of the following is a second order differential equation?
Answer(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→Question 291 Mark
The general solution of the differential equation $\frac{d y}{d x}=\frac{x^2}{y^2}$ is
Answer$(a) : \frac{d y}{d x}=\frac{x^2}{y^2}$
$\Rightarrow y^2 d y=x^2 d x$
$\Rightarrow \int y^2 d y=\int x^2 d x$
$\Rightarrow \frac{y^3}{3}=\frac{x^3}{3}+C$
$\Rightarrow x^3-y^3=-3 C=c \text { (say). }$
View full question & answer→Question 301 Mark
Integrating factor of differential equation $\frac{d y}{d x}+y \tan x-\sec x=0$ is
Answer(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→Question 311 Mark
The differential equation whose solution is $y=A e^{3 x}+B e^{-3 x}$ is given by
Answer$(c) :$ 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 $\text{w.r.t. x}$, 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→Question 321 Mark
The solution of $x \frac{d y}{d x}+y=e^x$ is
Answer$\text { (a) : } x \frac{d y}{d x}+y=e^x$
$\frac{d y}{d x}+\frac{y}{x}=\frac{e^x}{x}$
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→Question 331 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→Question 341 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→Question 351 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→Question 361 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→Question 371 Mark
The integrating factor of the differential equation $x \frac{d y}{d x}+2 y=x^2$ is
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→Question 381 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→Question 391 Mark
The integrating factor of the differential equation $x \frac{d y}{d x}-y=\log x$ is
Answer$(a):$ 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
$\frac{d y}{d x}+P y=Q$
$\therefore \text { I.F. }=e^{\int-\frac{1}{x} d x}=e^{-\log x}=\frac{1}{x}$
View full question & answer→Question 401 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→Question 411 Mark
Which of the following is the integrating factor of $x d y / d x-y=x^4-3 x$ ?
Answer$\text { (c) : } x \frac{d y}{d x}-y=x^4-3 x$
$\Rightarrow \frac{d y}{d x}-\frac{y}{x}=\frac{x^4-3 x}{x}$
$\Rightarrow \frac{d y}{d x}-\frac{1}{x} \cdot y=x^3-3$
$\therefore \text { I.F. }=e^{-\int \frac{1}{x} d x}=e^{-\log x}=e^{\log (x)^{-1}}=x^{-1}=\frac{1}{x}$
View full question & answer→Question 421 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→Question 431 Mark
If $\frac{d y}{d x}=\frac{x+y}{x}, y(1)=1$, then $y=$
Answer$(d) :$ 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
$v=\ln x+c$
$\Rightarrow \frac{y}{x}=\ln x+c$
$\because y(1)=1$
$\Rightarrow x=1, y=1$
$\Rightarrow c=1$
$\therefore y=x+x \ln x$
View full question & answer→Question 441 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→Question 451 Mark
Integrating factor of the differential equation $\frac{d y}{d x}+y \tan x-\sec x=0$ is
Answer(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→Question 461 Mark
The integrating factor for solving the differential equation $x \frac{d y}{d x}-y=2 x^2$ is
Answer(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→Question 471 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→Question 481 Mark
Which of the following is the general solution of the differential equation $\frac{d y}{d x}=\frac{y}{x}$ ?
Answer(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→Question 491 Mark
The solution of the differential equation $\frac{d y}{d x}=\frac{1+y^2}{1+x^2}$ is
Answer(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→Question 501 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
$\log (y+1)+\log c=\log (x-1)$
$ \Rightarrow(y+1) c=(x-1)$
$\text { Now, } y(1)=2$
$ \Rightarrow 3 c=0$
$ \Rightarrow c=0$
$\therefore x-1=0 $
$\Rightarrow x=1$
Hence, only one solution exists.
View full question & answer→