MCQ 1011 Mark
The number of arbitrary constants in the particular solution of a differential equation of third order are:
AnswerThe number of arbitrary constants in a particular solution of a differential equation of any order is zero (0) as a particular solution is a solution which contains no arbitrary constant. Therefore, option (d) is correct.
View full question & answer→MCQ 1021 Mark
Find the general solution of: $\frac{\text{dy}}{\text{dx}}=\text{y}\sin\text{x:}$
AnswerConcept:
$\int\frac{\text{dx}}{\text{x}}=\log\text{x}+\text{c}$
$\int\sin{\text{x}}{\text{ dx}}=-\cos\text{x}+\text{c}$
Calculation:
Given: $\frac{\text{dx}}{\text{dy}}=\text{y}\sin\text{x}$
$\Rightarrow\frac{\text{dx}}{\text{dy}}=\sin\text{x}\text{ dx}$
Integrating both sides, we get
$\Rightarrow\int\frac{\text{dy}}{\text{y}}=\int\sin\text{x}\text{ dx}$
$\Rightarrow\log\text{y}=-\cos\text{x}+\text{c}$
$\Rightarrow\log\text{y}+\cos\text{x}-\text{c}=0$
View full question & answer→MCQ 1031 Mark
Which of the following is a homogeneous differnetial equation?
- A
$(4x + 6y + 5)dy - (3y + 2x + 4)dx = 0$
- B
$xy dx - (x^3 + y^3)dy = 0$
- C
$(x^3 + 2y^2)dx + 2xy dy = 0$
- ✓
$y^2dx + (x^2 - xy - y^2) = 0$
AnswerCorrect option: D. $y^2dx + (x^2 - xy - y^2) = 0$
A differential equation is said to be homogenous if all the in the terms in the equation have equal degree and it can be written in the from $\frac{\text{dy}}{\text{dx}}=\frac{\text{f}(\text{x,}\text{y})}{\text{g}(\text{x,}\text{y})}.$
In $(a), (b)$ and $(c),$ the degree of all the terms is not equal.
But in the equation $y^2 dx + (x^2- xy - y^2)dy = 0,$ the degree of all the terms is $2.$
Thus, $(d)$ constant a homogeneous differential equation.
View full question & answer→MCQ 1041 Mark
The general solution of differention eqution of the type $\frac{\text{dx}}{\text{dy}}+\text{P}_{1}\text{x}=\text{Q}_{1}$ is:
- A
$\text{ye}^{\int\text{P}_{1}\text{dy}}=\int \left\{\text{Q}_{1}\text{e}^{\int\text{P}_{1}\text{dy}}\right\}\text{dy}+\text{C}$
- B
$\text{ye}^{\int\text{P}_{1}\text{dy}}=\int \left\{\text{Q}_{1}\text{e}^{\int\text{P}_{1}\text{dx}}\right\}\text{dx}+\text{C}$
- ✓
$\text{xe}^{\int\text{P}_{1}\text{dy}}=\int \left\{\text{Q}_{1}\text{e}^{\int\text{P}_{1}\text{dx}}\right\}\text{dx}+\text{C}$
- D
$\text{xe}^{\int\text{P}_{1}\text{dx}}=\int \left\{\text{Q}_{1}\text{e}^{\int\text{P}_{1}\text{dx}}\right\}\text{dx}+\text{C}$
AnswerCorrect option: C. $\text{xe}^{\int\text{P}_{1}\text{dy}}=\int \left\{\text{Q}_{1}\text{e}^{\int\text{P}_{1}\text{dx}}\right\}\text{dx}+\text{C}$
We have,
$\frac{\text{dx}}{\text{dy}}+\text{P}_{1}\text{x}=\text{Q}_{1}$
Comparing with the equation $\frac{\text{dx}}{\text{dy}}+\text{P}\text{x}=\text{Q}$ we get,
$\text{P}=\text{P}_{1}, \text{Q}=\text{Q}_{1}$
The solution of the equation $\frac{\text{dx}}{\text{dy}}+\text{P}\text{x}=\text{Q}$ is given by
$\text{xe}^{\int\text{P}_{1}\text{dy}}=\int \left\{\text{Q}_{1}\text{e}^{\int\text{P}_{1}\text{dy}}\right\}\text{dy}+\text{C}\ ...(\text{i})$
Putting the value of P and Q in (i),
$\text{xe}^{\int\text{P}_{1}\text{dy}}=\int \left\{\text{Q}_{1}\text{e}^{\int\text{P}_{1}\text{dy}}\right\}\text{dy}+\text{C}$
View full question & answer→MCQ 1051 Mark
If $\frac{\text{dy}}{\text{dx}}=\frac{\text{x}+\text{y}}{\text{x}},\text{y}(1)=1,$ then $y =$
- A
$x +$ In $x$
- B
$x^2+ x$ In $x$
- C
$xe^{x-1}$
- ✓
$x + x$ In $x$
AnswerCorrect option: D. $x + x$ In $x$
View full question & answer→MCQ 1061 Mark
What is the degree of differential equation $(y\ ’’’)^2 + (y\ ’’)^3 + (y\ ’)^4 + y^5 = 0?$
AnswerThe degree is the power raised to the highest order derivative. Therefore, in the given differential equation, $(y\ ’’’)^2 + (y\ ’’)^3 + (y\ ’)^4 + y^5 = 0$, the degree will be power raised to $y\ ’’’.$
View full question & answer→MCQ 1071 Mark
If P and q are the order and degree of the differention $\text{y}\frac{\text{dy}}{\text{dx}}+\text{x}^{3}\frac{\text{d}^{2}\text{y}}{\text{dx}^{3}}+\text{xy}=\cos\text{x}$ then:
AnswerWe have,$\text{y}\frac{\text{dy}}{\text{dx}}+\text{x}^{3}\frac{\text{d}^{2}\text{y}}{\text{dx}^{3}}+\text{xy}=\cos\text{x}$
The highest order is $\frac{\text{d}^{2}\text{y}}{\text{dz}^{2}}$ and it's degree is 1.
So, the order is 2 and the degree is 1.
$\text{p}=2, \text{q}=1$
Clearly, $\text{p}>\text{q}$
View full question & answer→MCQ 1081 Mark
Choose the correct answer from the given four options.The solution of the differential equation $\frac{\text{dy}}{\text{dx}}+\frac{2\text{xy}}{1+\text{x}^2}=\frac{1}{(1+\text{x}^2)^2}$ is:
- ✓
$\text{y}(1+\text{x}^2)=\text{C}+\tan^{-1}\text{x}$
- B
$\frac{\text{y}}{1+\text{x}^2}=\text{C}+\tan^{-1}\text{x}$
- C
$\text{y}\log(1+\text{x}^2)=\text{C}+\tan^{-1}\text{x}$
- D
$\text{y}(1+\text{x}^2)=\text{C}+\sin^{-1}\text{x}$
AnswerCorrect option: A. $\text{y}(1+\text{x}^2)=\text{C}+\tan^{-1}\text{x}$
Given is, $\frac{\text{dy}}{\text{dx}}+\frac{2\text{xy}}{1+\text{x}^2}=\frac{1}{(1+\text{x}^2)^2}$
Here, $\text{P}=\frac{2\text{x}}{1+\text{x}^2}$ and $\text{Q}=\frac{1}{(1+\text{x}^2)^2}$
This ia a linerar differential equation.
$\therefore\text{I.F.}=\text{e}^{\int\frac{2\text{x}}{1+\text{x}^2}\text{dx}}$
Put $1+\text{x}^2=\text{t}\Rightarrow2\text{x}\text{ dx}=\text{dt}$
$\therefore\text{I.F.}=\text{e}^{\int\frac{\text{dt}}{\text{t}}}=\text{e}^{\log\text{t}}$
$=\text{e}^{\log(1+\text{x}^2)}=1+\text{x}^2$
Thus, the general solution is
$\text{y}.(1+\text{x}^2)=\int(1+\text{x}^2)\frac{1}{(1+\text{x}^2)}+\text{C}$
$\Rightarrow\text{y}(1+\text{x}^2)=\int\frac{1}{(1+\text{x}^2)}\text{dx}+\text{C}$
$\Rightarrow\text{y}(1+\text{x}^2)=\tan^{-1}\text{x}+\text{C}$
View full question & answer→MCQ 1091 Mark
The radius of a circle is increasing at the rate of $0.4 \ cm/ s.$ The rate of increasing of its circumference is:
- A
$0.4\pi\text{ cm}/\text{ s}.$
- ✓
$0.8\pi\text{ cm}/\text{ s}.$
- C
$0.8\text{ cm}/\text{ s}.$
- D
AnswerCorrect option: B. $0.8\pi\text{ cm}/\text{ s}.$
View full question & answer→MCQ 1101 Mark
The differential equation obtained on eliminating A and B from $\text{y}=\text{A}\cos\omega\text{t}+\text{B}\sin\omega\text{t}$ is:
- A
- B
$\text{y}''-\omega^{2}\text{y}=0$
- ✓
$\text{y}''=-\omega^{2}\text{y}=0$
- D
AnswerCorrect option: C. $\text{y}''=-\omega^{2}\text{y}=0$
We have,
$\text{y}=\text{A}\cos\omega\text{t}+\text{B}\sin\omega\text{t}\ ...(\text{i})$
Differentiating both sides of (i) with respect to x, we get
$\frac{\text{dy}}{\text{dt}}=-\text{A}\omega\sin\omega\text{t}+\text{B}\omega\cos\omega\text{t}\ ...(\text{ii})$
Differentiating both sides of (ii)
$\frac{\text{d}^{2}\text{y}}{\text{dt}^{2}}=-\text{A}\omega^{2}\cos\omega\text{t}+\text{B}\omega^{2}\sin\omega\text{t}$
$\Rightarrow \frac{\text{d}^{2}\text{y}}{\text{dt}^{2}}=-\omega^{2}(\text{A}\cos\omega\text{t}+\text{B}\sin\omega\text{t})$
$\Rightarrow \frac{\text{d}^{2}\text{y}}{\text{dt}^{2}}=-\omega^{2}\text{y}$
$\text{y}''=-\omega^{2}\text{y}$
View full question & answer→MCQ 1111 Mark
The differential equation of all parabolas whose axes are parallel to $y-$axis is:
- ✓
$\frac{\text{dy}}{\text{dx}}=-\frac{\text{c}^2}{\text{x}^2}$
- B
$\frac{\text{d}^2\text{x}}{\text{dy}^2}=\text{c}$
- C
$\frac{\text{d}^3\text{y}}{\text{dx}^3}+\frac{\text{d}^2\text{x}}{\text{dy}^2}=0$
- D
$\frac{\text{d}^2\text{y}}{\text{dx}^2}+2\frac{\text{dy}}{\text{dx}}=0$
AnswerCorrect option: A. $\frac{\text{dy}}{\text{dx}}=-\frac{\text{c}^2}{\text{x}^2}$
View full question & answer→MCQ 1121 Mark
If $\sin\text{x}\frac{\text{dy}}{\text{dx}}+\text{y}\cos\text{x}=\text{x}\sin\text{x},$ then $(\text{y}-1)\sin\text{x}=$
- A
$\text{c}-\text{x}\sin\text{x}$
- B
$\text{c}+\text{x}\cos\text{x}$
- ✓
$\text{c}-\text{x}\cos\text{x}$
- D
$\text{c}+\text{x}\sin\text{x}$
AnswerCorrect option: C. $\text{c}-\text{x}\cos\text{x}$
View full question & answer→MCQ 1131 Mark
Choose the correct answer from the given four option.
The solution of the differential equation $\frac{\text{d}\text{y}}{\text{d}\text{x}}=\frac{1+\text{y}^2}{1+\text{x}^2}$ is:
- A
$\text{y}=\tan^{-1}\text{x}$
- ✓
$\text{y}-\text{x}=\text{k}(1+\text{xy})$
- C
$\text{x}=\tan^{-1}\text{y}$
- D
$\tan(\text{xy})=\text{k}$
AnswerCorrect option: B. $\text{y}-\text{x}=\text{k}(1+\text{xy})$
Given that, $\frac{\text{d}\text{y}}{\text{d}\text{x}}=\frac{1+\text{y}^2}{1+\text{x}^2}$
$\Rightarrow\frac{\text{dy}}{1+\text{y}^2}=\frac{\text{dx}}{1+\text{x}^2}$
On integrating both sides, we get
$\tan^{-1}\text{y}=\tan^{-1}\text{x}+\text{C}$
$\Rightarrow\tan^{-1}\text{y}-\tan^{-1}\text{x}=\text{C}$
$\Rightarrow\tan^{-1}\Big(\frac{\text{y}-\text{x}}{1+\text{xy}}\Big)=\text{C}$
$\Rightarrow\frac{\text{y}-\text{x}}{1+\text{xy}}=\tan\text{C}$
$\Rightarrow\text{y}-\text{x}=\tan\text{C}(1+\text{xy})$
$\Rightarrow\text{y}-\text{x}=\text{K}(1+\text{xy})$
Where, $\text{k}=\tan\text{C}$
View full question & answer→MCQ 1141 Mark
The number of arbitrary constants in the general solution of a differential equation of fourth order are:
AnswerThe number of arbitrary constants $(c_1, c_2, c_3,$ etc$.)$ in the general solution of a differential equation of $n^{th}$ order is $n.$
View full question & answer→MCQ 1151 Mark
The order of the differential equartion $\sqrt{1-\text{x}^{4}}+\sqrt{1-\text{y}^{4}}=\text{a}(\text{x}^{2}-\text{y}^{2})$ is:
AnswerThe order of a differention depends on the number of constent in it.
Since $\sqrt{1-\text{x}^{4}}+\sqrt{1-\text{y}^{4}}=\text{a}(\text{x}^{2}-\text{y}^{2})$ constant only 1 constant, the order of the differential equation is 1.
View full question & answer→MCQ 1161 Mark
If m and n are the order and degree of the differential equation $(\text{y}_{2})^{5}+\frac{4(\text{y}_{2})^{3}}{\text{y}^{3}}+\text{y}^{3}=\text{y}_{3}=\text{x}^{2}-1$, then
AnswerWe have,
$(\text{y}_{2})^{5}+\frac{4(\text{y}_{2})^{3}}{\text{y}^{3}}+\text{y}_{3}=\text{x}^{2}-1$
$\text{y}_{3}(\text{y}_{2})^{5}+{4(\text{y}_{2})^{3}}+(\text{y}_{3})^{2}=\text{y}_{3}(\text{x}^{2}-1)$
The highest order is $y_3$ and its highest in this equation is $2$.
Hence, $m = 3, n = 2$.
View full question & answer→MCQ 1171 Mark
Choose the correct answer from the given four options.Solution of the differential equation $\frac{\text{dy}}{\text{dx}}+\frac{\text{y}}{\text{x}}=\sin\text{x}$ is:
- ✓
$\text{x}(\text{y}+\cos\text{x})=\sin\text{x}+\text{c}$
- B
$\text{x}(\text{y}-\cos\text{x})=\sin\text{x}+\text{c}$
- C
$\text{x}\text{y}\cos\text{x}=\sin\text{x}+\text{c}$
- D
$\text{x}(\text{y}+\cos\text{x})=\cos\text{x}+\text{c}$
AnswerCorrect option: A. $\text{x}(\text{y}+\cos\text{x})=\sin\text{x}+\text{c}$
Given differential equation is
$\frac{\text{dy}}{\text{dx}}+\text{y}\frac{1}{\text{x}}=\sin\text{x}$
Which is liner differential equations.
Here, $\text{P}=\frac{1}{\text{x}}$ and $\text{Q}=\sin\text{x}$
$\therefore\text{I.F.}=\text{e}^{\int\frac{1}{\text{x}}\text{dx}}$
$=\text{e}^{\log\text{x}}=\text{x}$
The general solution is
$\text{yx}=\int\text{x}\sin\text{xdx}+\text{c}\ .....(\text{i})$
Take $\text{I}=\int\text{x}\sin\text{xdx}$
$-\text{x}\cos\text{x}-\int-\cos\text{xdx}$
$=-\text{x}\cos\text{x}+\sin\text{x}$
Put the value of 1 in Eq. (i), we get
$\text{xy}=-\text{x}\cos\text{x}+\sin\text{x}+\text{c}$
$\Rightarrow\text{x}(\text{y}+\cos\text{x})=\sin\text{x}+\text{c}$
View full question & answer→MCQ 1181 Mark
Solution of differential equation x dy - yx = 0 represents:
- A
- ✓
straight line passing through origin
- C
parabola whose vertex is at origin
- D
circle whose center is at origin
AnswerCorrect option: B. straight line passing through origin
$=\text{x}\text{ dx}-\text{y}\text{ dx}=0$
$\Rightarrow\frac{\text{dy}}{\text{y}}=\frac{\text{dx}}{\text{x}}$
Integrating both sides
y = ln x ⇒ y = x
Solution of differential equation
x dy - y x = 0
reperesnts straight line passing through origin
View full question & answer→MCQ 1191 Mark
The degree and the order of the differential equation:
$\frac{\text{d}^2\text{y}}{\text{dx}^2}=\sqrt{1}+\Big(\frac{\text{dy}}{\text{dx}}\Big)^3\text{ are:}$
AnswerConcept:
Order of a differential equation is the highest order of derivative that occurs in the differential equation.
Degree of a differential equation is the highest power of the highest order derivative that occurs in the equation, after all the derivatives are converted into rational and radical free form.
Calculation:
Getting rid of the radicals by raising both the sides to power 3 will give us:
$\frac{\text{d}^2\text{y}}{\text{dx}^2}=\sqrt{1}+\Big(\frac{\text{dy}}{\text{dx}}\Big)^3$
$\Rightarrow\frac{\text{d}^2\text{y}}{\text{dx}^2}=1+\Big(\frac{\text{dy}}{\text{dx}}\Big)^3$
The highest derivative in it is $\frac{\text{d}^2\text{y}}{\text{dx}^2},$ therefore its order is 2.
And, the highest power of this derivative in the equation is also 2, therefore its degree is also 2.
View full question & answer→MCQ 1201 Mark
Choose the correct answer from the given four option.
The order and degree of the differential equation $\frac{\text{d}^2\text{y}}{\text{d}\text{x}^2}+\Big(\frac{\text{d}\text{y}}{\text{d}\text{x}}\Big)^{\frac{1}{4}}+\text{x}^{\frac{1}{5}}$ respectively, are:
AnswerWe have, $\frac{\text{d}^2\text{y}}{\text{d}\text{x}^2}+\Big(\frac{\text{d}\text{y}}{\text{d}\text{x}}\Big)^{\frac{1}{4}}+\text{x}^{\frac{1}{5}}$
$\Rightarrow\Big(\frac{\text{d}\text{y}}{\text{d}\text{x}}\Big)^\frac{1}{4}=-\Big(\text{x}^{\frac{1}{5}}+\frac{\text{d}^2\text{y}}{\text{d}\text{x}^2}\Big)$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\Big(\text{x}^{\frac{1}{5}}+\frac{\text{d}^2\text{y}}{\text{d}\text{x}^2}\Big)^4$
$\therefore$ order = -2, Degree = 4
View full question & answer→MCQ 1211 Mark
Choose the correct answer from the given four option.
The solution of $\frac{\text{d}\text{y}}{\text{d}\text{x}}+\text{y}=\text{e}^{\text{-x}},\text{ y}(0)$is:
- A
$\text{y}=\text{e}^{\text{x}}(\text{x}-1)$
- ✓
$\text{y}=\text{x}\text{e}^{-\text{x}}$
- C
$\text{y}=\text{x}\text{e}^{\text{x}}+1$
- D
$\text{y}=(\text{x}+1)\text{e}^{-\text{x}}$
AnswerCorrect option: B. $\text{y}=\text{x}\text{e}^{-\text{x}}$
We have, $\frac{\text{d}\text{y}}{\text{d}\text{x}}+\text{y}=\text{e}^{\text{-x}}$
This is a linear differential equation.
On comparing it with $\frac{\text{d}\text{y}}{\text{d}\text{x}}+\text{P}\text{y}=\text{Q},$we get
$\text{P}=1,\text{Q}=\text{e}^{-\text{x}}$
$\text{I.F.}=\text{e}^{\int\text{Pdx}}=\text{e}^{{\int\text{dx}}}$
So, the general solution is,
$\text{y}.\text{e}^{\text{x}}=\int\text{e}^{-\text{x}}\text{e}^{\text{x}}\text{dx}+\text{C}$
$\Rightarrow\text{y}.\text{e}^{\text{x}}=\int\text{dx}+\text{C}$
$\Rightarrow\text{y}.\text{e}^{\text{x}}=\text{x}+\text{C}$
Given that when x = 0 and y = 0
$\Rightarrow0 = 0 + \text{C}$
$\Rightarrow\text{C}=0$
Eq. (i) becomes $\text{y}.\text{e}^{\text{x}}=\text{x}$
$\Rightarrow\text{y}=\text{x}\text{e}^{-\text{x}}$
View full question & answer→MCQ 1221 Mark
The degree of the differential equation
$\bigg(\frac{\text{d}^2\text{y}}{\text{dx}^2}\bigg)^3 + \bigg(\frac{\text{dy}}{\text{dx}}\bigg)^2+\text{sin} \bigg(\frac{\text{dy}}{\text{dx}}\bigg) + 1 =0 \ \text{is}$
AnswerThe given differential equation is $\bigg(\frac{\text{d}^2\text{y}}{\text{dx}^2}\bigg)^3 + \bigg(\frac{\text{dy}}{\text{dx}}\bigg)^2+\text{sin} \bigg(\frac{\text{dy}}{\text{dx}}\bigg) + 1 =0 \ $
Since the differential equation is not a polynomial equation in its derivatives.
$\therefore$ its degree is not defined.
View full question & answer→MCQ 1231 Mark
Which of the following equation is a linear differential equation of order 3?
[Note: The original question asks for linear equation, but it should be linear differential equation]:
- ✓
$\frac{\text{d}^3\text{y}}{\text{dx}^3}+\frac{\text{d}^2\text{y}\text{ dx}}{\text{dx}^2\text{dx}}+\text{y}=\text{x}$
- B
$\frac{\text{d}^3\text{y}}{\text{dx}^3}+\frac{\text{d}^2\text{y}}{\text{dx}^2}+\text{y}^2=\text{x}^2$
- C
$\text{x}\frac{\text{d}^3\text{y}}{\text{dx}^3}+\frac{\text{d}^2\text{y}}{\text{dx}^2}=\text{e}^\text{x}$
- D
$\frac{\text{d}^2\text{y}}{\text{dx}^2}+\frac{\text{dy}}{\text{dx}}=\log\text{x}$
AnswerCorrect option: A. $\frac{\text{d}^3\text{y}}{\text{dx}^3}+\frac{\text{d}^2\text{y}\text{ dx}}{\text{dx}^2\text{dx}}+\text{y}=\text{x}$
Linear equation is an equation between two variables that gives a straight line and order of linear equation is highest order derivative in linear equation.Since, in option A equation is linear equation in which highest order is 3.
View full question & answer→MCQ 1241 Mark
The degree of differential equation $[1+(\frac{\text{dy}}{\text{dx}})^2]^{\frac{1}{2}}=\frac{\text{d}^2\text{y}}{\text{dx}^2}$ is:
View full question & answer→MCQ 1251 Mark
The general solution of the dofferential equation $\frac{\text{dy}}{\text{dx}}=\text{e}^{\text{x}+\text{y}}$ is:
- ✓
$\text{e}^{\text{x}}+\text{e}^{-\text{y}}=\text{C}$
- B
$\text{e}^{\text{x}}+\text{e}^{\text{y}}=\text{C}$
- C
$\text{e}^{-\text{x}}+\text{e}^{\text{y}}=\text{C}$
- D
$\text{e}^{-\text{x}}+\text{e}^{-\text{y}}=\text{C}$
AnswerCorrect option: A. $\text{e}^{\text{x}}+\text{e}^{-\text{y}}=\text{C}$
We have,$\frac{\text{dy}}{\text{dx}}=\text{e}^{\text{x}+\text{y}}$
$\Rightarrow \frac{\text{dy}}{\text{dx}}=\text{e}^{\text{x}}\times\text{e}^{\text{y}}$
$\Rightarrow \text{e}^{-\text{y}}\text{dy}=\text{e}^{\text{x}}\text{dx}$
Integrating both sides, we get
$\int\text{e}^{-\text{y}}\text{dy}=\int\text{e}^{\text{x}}\text{dx}$
$\Rightarrow \text{e}^{-\text{y}}=\text{e}^{\text{x}}+\text{D}$
$\Rightarrow \text{e}^{\text{x}}+\text{e}^{-\text{y}}=-\text{D}$
$\Rightarrow \text{e}^{\text{x}}+\text{e}^{-\text{y}}=-\text{C}$
View full question & answer→MCQ 1261 Mark
Which of the following differentials equation has $\text{y}=\text{C}_{1}\text{e}^{\text{x}}+\text{C}_{2}\text{e}^{-\text{x}}$ as the general solution?
- A
$\frac{\text{d}^{2}\text{y}}{\text{dx}^{2}}+\text{y}=0$
- ✓
$\frac{\text{d}^{2}\text{y}}{\text{dx}^{2}}-\text{y}=0$
- C
$\frac{\text{d}^{2}\text{y}}{\text{dx}^{2}}+\text{1}=0$
- D
$\frac{\text{d}^{2}\text{y}}{\text{dx}^{2}}-\text{1}=0$
AnswerCorrect option: B. $\frac{\text{d}^{2}\text{y}}{\text{dx}^{2}}-\text{y}=0$
We have,
$\text{y}=\text{C}_{1}\text{e}^{\text{x}}+\text{C}_{2}\text{e}^{-\text{x}}\ ...(\text{i})$
Differentiating both sides of (i) with we get,
$\frac{\text{dy}}{\text{dx}}=\text{C}_{1}\text{e}^{\text{x}}+\text{C}_{2}\text{e}^{-\text{x}}\ ...(\text{ii})$
Differentiating both sides of (ii) with we get,
$\frac{\text{d}^{2}\text{y}}{\text{dx}^{2}}=\text{C}_{1}\text{e}^{\text{x}}+\text{C}_{2}\text{e}^{-\text{x}}$
$\Rightarrow \frac{\text{d}^{2}\text{y}}{\text{dx}^{2}}=\text{y}$
$\Rightarrow \frac{\text{d}^{2}\text{y}}{\text{dx}^{2}}-\text{y}=0$
View full question & answer→MCQ 1271 Mark
If $(\text{x}+\text{y})^2\frac{\text{dy}}{\text{dx}}=\text{a}^2,\text{y}=0$ when $x = 0,$ then $y = a$ if $\frac{\text{x}}{\text{a}}=$
- A
$1$
- B
$\tan1$
- C
$\tan1+1$
- ✓
$\tan1-1$
AnswerCorrect option: D. $\tan1-1$
View full question & answer→MCQ 1281 Mark
If $(\text{x}+2\text{y}^3)\frac{\text{dy}}{\text{dx}}=\text{y},$ then:
- A
$\frac{\text{x}}{\text{y}}+\text{y}^2=\text{c}$
- B
$\frac{\text{y}}{\text{x}}+\text{x}^2=\text{c}$
- ✓
$\frac{\text{x}}{\text{y}}-\text{y}^2=\text{c}$
- D
$\frac{\text{y}}{\text{x}}-\text{x}^2=\text{c}$
AnswerCorrect option: C. $\frac{\text{x}}{\text{y}}-\text{y}^2=\text{c}$
View full question & answer→MCQ 1291 Mark
The solution of the differention equation $(1+\text{x}^{2})\frac{\text{dy}}{\text{dx}}+1+\text{y}^{2}=0$ is:
- A
$\tan^{-1}\text{x}-\tan^{-1}\text{y}=\tan^{-1}\text{C}$
- B
$\tan^{-1}\text{y}-\tan^{-1}\text{x}=\tan^{-1}\text{C}$
- C
$\tan^{-1}\text{y}\pm\tan^{-1}\text{x}=\tan^{-1}\text{C}$
- ✓
$\tan^{-1}\text{y}+\tan^{-1}\text{x}=\tan^{-1}\text{C}$
AnswerCorrect option: D. $\tan^{-1}\text{y}+\tan^{-1}\text{x}=\tan^{-1}\text{C}$
We have,
$(1+\text{x}^{2})\frac{\text{dy}}{\text{dx}}+1+\text{y}^{2}=0$
$\Rightarrow (1+\text{x}^{2})\frac{\text{dy}}{\text{dx}}=-(1+\text{y}^{2})$
$\Rightarrow \frac{1}{(1+\text{y}^{2})}\text{dy}=-\frac{1}{(1+\text{x}^{2})}\text{dx}$
Integrating both sides, we get
$\int\frac{1}{(1+\text{y}^{2})}\text{dy}=-\int\frac{1}{(1+\text{x}^{2})}\text{dx}$
$\Rightarrow \tan^{-1}\text{y}=-\tan^{-1}\text{x}+\tan^{-1}\text{C}$
$\Rightarrow \tan^{-1}\text{y}+\tan^{-1}\text{x}=\tan^{-1}\text{C}$
View full question & answer→MCQ 1301 Mark
The degree of the differential equation $\text{y}_3^\frac{2}{3}+2+3\text{y}_2+\text{y}_1=0$ is:
Answer$=\text{y}_3^\frac{2}{3}+2+3\text{y}_2+\text{y}_1=0$
$\Rightarrow\text{y}_3^\frac{2}{3}=-(2+3\text{y}+\text{y}_1)$
$\Rightarrow\text{y}^\frac{2}{3}=-(2+3\text{y}_2+\text{y}_1)^2$
cubing both sides,
Hence degree of given differential equation is $2 ($power on $y_3)$
View full question & answer→MCQ 1311 Mark
What is the solution of the differential equation$:\text{In}\Big(\frac{\text{dx}}{\text{dy}}\Big)-\text{a}=0?$
- ✓
$y = xe^a + c$
- B
$x = ye^a + c$
- C
$\text{y = In x + c}$
- D
$\text{x = In y + c}$
AnswerCorrect option: A. $y = xe^a + c$
Calculation:
Given: $\text{In}\Big(\frac{\text{dy}}{\text{dx}}\Big)-\text{a}=0$
$\Rightarrow\text{In}\Big(\frac{\text{dy}}{\text{dx}}\Big)=\text{a}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\text{e}^\text{a}$
$\Rightarrow\int\frac{\text{dy}}{\text{dx}}=\int\text{e}^\text{a}$
On integrating both sides, we get
$\Rightarrow\text{y}=\text{xe}^\text{a}+\text{c}$
View full question & answer→MCQ 1321 Mark
The solution of the differential equation $(\text{x}^{2}+1)\frac{\text{dy}}{\text{dx}}+(\text{y}^{2}+1)=0$ is:
- A
$\text{y}=2+\text{x}^{2}$
- B
$\text{y}=\frac{1+\text{x}}{1-\text{x}}$
- C
$\text{y}=\text{x}(\text{x}-1)$
- ✓
$\text{y}=\frac{1+\text{y}}{1-\text{y}}$
AnswerCorrect option: D. $\text{y}=\frac{1+\text{y}}{1-\text{y}}$
We have,
$(\text{x}^{2}+1)\frac{\text{dy}}{\text{dx}}=-(\text{y}^{2}+1)=0$
$\Rightarrow (\text{x}^{2}+1)\frac{\text{dy}}{\text{dx}}=-(\text{y}^{2}+1)$
$\Rightarrow \frac{1}{(\text{y}^{2}+1)}\text{dy}=-\frac{1}{(\text{x}^{2}+1)}\text{dx}$
Intergrating both sides, we get
$\Rightarrow \int\frac{1}{(\text{y}^{2}+1)}\text{dy}=-\int\frac{1}{(\text{x}^{2}+1)}\text{dx}$
$\Rightarrow \tan^{-1}\text{y}=-\tan^{-1}\text{x}+\tan^{-1}\text{C}$
$\Rightarrow \tan^{-1}\text{y}+\tan^{-1}\text{x}=\tan^{-1}\text{C}$
$\Rightarrow \tan^{-1}\Big(\frac{\text{x}+\text{y}}{1-\text{xy}}\Big) =\tan^{-1}\text{C}$
$\Rightarrow \frac{\text{x}+\text{y}}{1+\text{xy}}=\text{C}$
Disclaimer : The initial value given, So the find will be C = 1, So
$\Rightarrow \text{x}+\text{y}=1-\text{xy}$
$\Rightarrow \text{y}+\text{xy}=1-\text{x}$
$\Rightarrow \text{y}(1+\text{x})=1-\text{x}$
$\Rightarrow \text{y}=\frac{1-\text{x}}{1+\text{x}}$
View full question & answer→MCQ 1331 Mark
$\text{x}\frac{\text{b}-\text{c}}{\text{b}\text{c}}\ \text{x}\frac{\text{c}-\text{a}}{\text{c}\text{a}}\text{x}\frac{\text{a}-\text{b}}{\text{a}\text{c}}=$
- A
$a^{a+b+c}$
- B
$x^{abc}$
- ✓
$1$
- D
$0$
Answer$\text{x}\frac{\text{b}-\text{c}}{\text{b}\text{c}}\text{x}\frac{\text{c}-\text{a}}{\text{ac}}\text{x}\frac{\text{a}-\text{b}}{\text{ac}}\text{x}\frac{\text{b}-\text{c}}{\text{bc}}+\frac{\text{c}-\text{a}}{\text{ca}}+\frac{\text{a}-\text{b}}{\text{ac}}$
$\text{x}^0=1$
View full question & answer→MCQ 1341 Mark
Choose the correct answer from the given four option.
Solution of $\frac{\text{d}\text{y}}{\text{d}\text{x}}-\text{y}=1,\text{ y}(0)=1$is given by:
- A
$\text{xy}=-\text{e}^\text{x}$
- B
$\text{xy}=-\text{e}^{-\text{x}}$
- C
$\text{xy}=-1$
- ✓
$\text{y}=2\text{e}^\text{x}-1$
AnswerCorrect option: D. $\text{y}=2\text{e}^\text{x}-1$
Given is, $\frac{\text{d}\text{y}}{\text{d}\text{x}}-\text{y}=1$
$\Rightarrow\frac{\text{d}\text{y}}{\text{d}\text{x}}=\text{y}+1$
$\Rightarrow\frac{\text{d}\text{y}}{1+\text{y}}=\text{dx}$
On integrating both sides, we get
$\log(1+\text{y})=\text{x}+\text{C}\ ......(\text{i})$
When x = 0 and y = 1, then
$\log2=0+\text{C}$
$\Rightarrow\text{C}=\log2$
The required solution is
$\log(1+\text{y})=\text{x}+\log2$
$\Rightarrow\log\Big(\frac{1+\text{y}}{2}\Big)=\text{x}$
$\Rightarrow\frac{1+\text{y}}{2}=\text{e}^\text{x}$
$\Rightarrow1+\text{y}=2\text{e}^\text{x}$
$\Rightarrow\text{y}=2\text{e}^\text{x}-1$
View full question & answer→MCQ 1351 Mark
The general solution of the differntial equation $\frac{\text{dy}}{\text{dx}}+\text{y}\cot\text{x}=\text{coses}\ \text{x}$ is:
- A
$\text{x}+\text{y}\sin\text{x}=\text{C}$
- B
$\text{x}+\text{y}\cos\text{x}=\text{C}$
- C
$\text{y}+\text{x}(\sin\text{x}+\cos\text{x})=\text{C}$
- ✓
$\text{y}\sin\text{x}=\text{x}+\text{C}$
AnswerCorrect option: D. $\text{y}\sin\text{x}=\text{x}+\text{C}$
$\frac{\text{dy}}{\text{dx}}+\text{y}\cot\text{x}=\text{coses}\ \text{x}$
Comparting with $\frac{\text{dy}}{\text{dx}}+\text{Py}=\text{Q}$ we get
$\text{P}=\cot\text{x}$
$\text{Q}=\text{coses} \ \text{x}$
Now,
$\text{I.F}=\text{e}^{\int\cot\text{x}\text{dx}}$
$=\text{e}^{\log(\sin\text{x})}$
$=\sin\text{x}$
So, the solution is given by
$\Rightarrow \text{y}\sin\text{x}=\int\sin\text{x}\times\text{cosec}\ \text{x}\text{dx}+\text{C}$
$\Rightarrow \text{y}\sin\text{x}=\text{x}+\text{C}$
View full question & answer→MCQ 1361 Mark
The differential equation of all conics with centreat origin is of order:
AnswerThe general equation of all conics with center at origin can be written as
$ax^2+ 2hxy + by^2+ c = 0$
Dividing by aa, we get
$\text{x}^2+\Big(\frac{2\text{h}}{\text{a}}\Big)\text{ax}+\Big(\frac{\text{b}}{\text{a}}\Big)\text{y}^2+\Big(\frac{\text{c}}{\text{a}}\Big)=0$
Since, it has three arbitrary constants.
So, the differential equation is of order $3.$
View full question & answer→MCQ 1371 Mark
Integration factor of differential equation $\frac{\text{dy}}{\text{dx}}+\text{py}=\text{Q},$ where $P$ and $IQ$ are functions of $x$ is:
View full question & answer→MCQ 1381 Mark
The solution of the differential equation $xdy + ydy = x^2y dy - y^2x dx$, is:
AnswerWe have,
$\text{x}\ \text{dx}+\text{y}\ \text{dy}=\text{x}^{2}\ \text{dy}-\text{y}^{2}\text{x}\ \text{dx}$
$\Rightarrow (\text{x}+\text{xy}^{2})\text{dx}=(\text{x}^{2}\text{y}-\text{y})\text{dy}$
$\Rightarrow \frac{\text{x}}{(\text{x}^{2}-1)}\text{dx}=\frac{\text{y}}{(1+\text{x})^{2}}\text{dy}$
$\Rightarrow \frac{2\text{x}}{2(\text{x}^{2}-1)}\text{dx}=\frac{2\text{y}}{2(1+\text{y})^{2}}\text{dy}$
Integrating both sides, we get
$\frac{1}{2}\int\frac{2\text{y}}{(1+\text{y})^{2}}\text{dy}=\frac{1}{2}\int \frac{2\text{x}}{(1+\text{x})^{2}}\text{dx}$
$\Rightarrow \log|(1+\text{y}^{2})|=\frac{1}{2}\log|(\text{x}^{2}-1)|-\frac{1}{2}\log|\text{C}|$
$\Rightarrow \log|(1+\text{y}^{2})|=\log|(\text{x}^{2}-1)|-\log|\text{C}|$
$\Rightarrow \log|(1+\text{y}^{2})|=\log|\frac{\text{x}^{2}-1}{\text{C}}|$
$\Rightarrow 1+\text{y}^{2}=\frac{\text{x}^{2}-1}{\text{C}}$
$\Rightarrow \text{C}(1+\text{y}^{2})=\text{x}^{2}-1$
View full question & answer→MCQ 1391 Mark
The general solution of the differential equation $\text{e}^\text{x}\ \text{dy}+(\text{y e}^\text{x}+2\text{x})\text{dx}=0\ \text{is}$
- A
$\text{x e}^\text{y}+\text{x}^2=\text{C}$
- B
$\text{x e}^\text{y}+\text{y}^2=\text{C}$
- ✓
$\text{y e}^\text{x}+\text{x}^{2}=\text{C}$
- D
$\text{y e}^\text{y}+\text{x}^2=\text{C}$
AnswerCorrect option: C. $\text{y e}^\text{x}+\text{x}^{2}=\text{C}$
The given differential equation is
$\text{e}^\text{x}\ \text{dy}+(\text{y e}^\text{x}+2\text{x})\text{dx}=0$
$\text{or}\ \ \text{e}^\text{x}\frac{\text{dy}}{\text{dx}}+\text{y e}^\text{x}+2\text{x}=0\ \ \text{or}\ \ \frac{\text{dy}}{\text{dx}}+\text{y}+\frac{2\text{x}}{\text{e}^\text{x}}=0$ $\text{or}\ \ \frac{\text{dy}}{\text{dx}}+\text{y}=-2\text{x e}^{-\text{x}}$
$\text{Comparing it with }\frac{\text{dy}}{\text{dx}}+\text{Py}=\text{Q},$ $\text{we get},\ \text{P}=1,\ \text{Q}=-2\text{x e}^{-\text{x}}$
$\therefore\ \ \int\text{P dx}=\int1\ \text{dx}=\text{'x},\ \text{e}^{\int\text{P dx}}=\text{e}^\text{x}$
Solution of given differential equation is
$\text{y e}^{\int\text{P dx}}=\int\text{Q e}^{\int\text{P dx}}+\text{C}$ $\text{or}\ \ \text{y e}^\text{x}=\int(-2\text{x e}^\text{x}).\text{e}^{-\text{x}}\ \text{dx}+\text{C}$
$\text{or}\ \ \text{y e}^\text{x}=-2\int\text{x dx}+\text{C}\ \ \text{or}\ \ \text{y e}^\text{x}=-\text{x}^2+\text{C}$
$\text{or}\ \ \text{y e}^\text{x}+\text{x}^2=\text{C}$
$\therefore$ (C) is correct answer.
View full question & answer→MCQ 1401 Mark
Choose the correct answer from the given four options.The differential equation of the family of curves $\text{y}^2=4\text{a}(\text{x}+\text{a})$ is:
- A
$\text{y}^2=4\frac{\text{dy}}{\text{dx}}\Big(\text{x}+\frac{\text{dy}}{\text{dx}}\Big)$
- B
$2\text{y}\frac{\text{dy}}{\text{dx}}=4\text{a}$
- C
$\text{y}\frac{\text{d}^2\text{y}}{\text{d}\text{x}^2}+\Big(\frac{\text{dy}}{\text{dx}}\Big)^2=0$
- ✓
$2\text{x}\frac{\text{dy}}{\text{dx}}+\text{y}\Big(\frac{\text{dy}}{\text{dx}}\Big)^2-\text{y}=0$
AnswerCorrect option: D. $2\text{x}\frac{\text{dy}}{\text{dx}}+\text{y}\Big(\frac{\text{dy}}{\text{dx}}\Big)^2-\text{y}=0$
We have equation of the curve as
$\text{y}^2=4\text{a}(\text{x}+\text{a})\ .....(\text{i})$
On differentiating both sides w.r.t.x, we gat
$2\text{y}\frac{\text{dy}}{\text{dx}}=4\text{a}$
$\Rightarrow\frac{1}{2}\text{y}\frac{\text{dy}}{\text{dx}}=\text{a}$
On putting the value of a in Eq. (i), we get
$\text{y}^2=2\text{y}\frac{\text{dy}}{\text{dx}}\Big(\text{x}+\frac{1}{2}\text{y}\frac{\text{dy}}{\text{dx}}\Big)$
$\Rightarrow\text{y}^2=2\text{xy}\frac{\text{dy}}{\text{dx}}+\text{y}^2\Big(\frac{\text{dy}}{\text{dx}}\Big)^2$
$\Rightarrow2\text{x}\frac{\text{dy}}{\text{dx}}+\text{y}\Big(\frac{\text{dy}}{\text{dx}}\Big)^2-\text{y}=0$
View full question & answer→MCQ 1411 Mark
The solution of the differential equation $\frac{\text{dy}}{\text{dx}}=\frac{\text{ax}+\text{g}}{\text{by}+\text{f}}$ represents a circle when,
AnswerWe have,
$\frac{\text{dy}}{\text{dx}}=\frac{\text{ax}+\text{g}}{\text{by}+\text{f}}$
$\Rightarrow (\text{by}+\text{f})\text{dy}=(\text{ax}+\text{g})\text{dx}$
Intergrating both sides, we get
$\Rightarrow \int(\text{by}+\text{f})\text{dy}=\int(\text{ax}+\text{g})\text{dx}$
$\Rightarrow \text{b}\frac{\text{y}^{2}}{2}+\text{fy}=\text{a}\frac{\text{x}^{2}}{2}+\text{gx}+\text{C}$
$\Rightarrow \text{b}\frac{\text{y}^{2}}{2}+\text{fy}-\text{a}\frac{\text{x}^{2}}{2}-\text{gx}=\text{C}$
$\Rightarrow \text{b}\text{y}^{2}+2\text{fy}-\text{a}\text{x}^{2}-2\text{gx}-2\text{C}=0$
The above equation resprasents a circle.
Therefore, the coffrcients of $x^2$ and $y^2$ must be equal.
$-\text{a}=\text{b}$
$\Rightarrow \text{a}=-\text{b}$
View full question & answer→MCQ 1421 Mark
Choose the correct answer from the given four option.
The general solution of $\text{e}^{\text{x}}\cos\text{ydx}-\text{e}^\text{x}\sin\text{ydy}=0$ is:
- ✓
$\text{e}^{\text{x}}\cos\text{y}=\text{k}$
- B
$\text{e}^{\text{x}}\sin\text{y}=\text{k}$
- C
$\text{e}^{\text{x}}=\text{k}\cos\text{y}$
- D
$\text{e}^{\text{x}}=\text{k}\sin\text{y}$
AnswerCorrect option: A. $\text{e}^{\text{x}}\cos\text{y}=\text{k}$
Given is, $\text{e}^{\text{x}}\cos\text{ydx}-\text{e}^\text{x}\sin\text{ydy}=0$
$\Rightarrow\text{e}^{\text{x}}\cos\text{ydx}=\text{e}^\text{x}\sin\text{ydy}$
$\Rightarrow\frac{\text{d}\text{y}}{\text{d}\text{x}}=\tan\text{ydy}$
$\Rightarrow\text{dx}=\tan\text{ydy}$
On integrating both sides, we get
$\text{x}=\log\sec\text{y}+\text{C}$
$\Rightarrow\text{x}-\text{C}=\log\sec\text{y}$
$\Rightarrow\sec\text{y}=\text{e}^{\text{x}-\text{c}}$
$\Rightarrow\frac{1}{\cos\text{y}}=\frac{\text{e}^\text{x}}{\text{e}^\text{c}}$
$\Rightarrow\text{e}^{\text{x}}\cos\text{y}=\text{k}$ $[\text{where},\text{K}=\text{e}^\text{c}]$
View full question & answer→MCQ 1431 Mark
The solution of differential equation $\frac{\text{dy}}{\text{dx}}-3\text{y}=\sin2\text{x}$ is:
- A
$\text{y}=\text{e}^{-3\text{x}}\Big[\frac{\cos2\text{x}+3\sin2\text{x}}{13}\Big]+\text{c}$
- B
$\text{y}=\text{e}^{-3\text{x}}\Big[\frac{\cos2\text{x}-3\sin2\text{x}}{13}\Big]+\text{c}$
- ✓
$\text{y}^{-3\text{x}}=-\text{e}^{-3\text{x}}\Big[\frac{2\cos2\text{x}+3\sin2\text{x}}{13}\Big]+\text{c}$
- D
AnswerCorrect option: C. $\text{y}^{-3\text{x}}=-\text{e}^{-3\text{x}}\Big[\frac{2\cos2\text{x}+3\sin2\text{x}}{13}\Big]+\text{c}$
View full question & answer→MCQ 1441 Mark
Solution of differential equation $x.dy - y.dx = Q$ represents:
- A
- B
Parabola whose vertex is at the origin
- ✓
Straight line passing through the origin
- D
A circle whose centre is at the origin
AnswerCorrect option: C. Straight line passing through the origin
View full question & answer→MCQ 1451 Mark
Consider the following statements in respect of the differential equation $\frac{\text{d}^2\text{y}}{\text{d}\text{x}^2}+\cos\Big(\frac{\text{dx}}{\text{dy}}\Big)=0:$
1. The degree of the differential equation is not defined.
2. The order of the differential equation is 2.
Which of the above statements is/are correct ?
Answer$\frac{\text{d}^2\text{y}}{\text{d}\text{x}^2}+\cos\Big(\frac{\text{dx}}{\text{dy}}\Big)=0:$
The order of differential equation is 2.
Degree of a differential equation is power of highest order differential equation
$\therefore$ The degree of this differential equation is 1.
View full question & answer→MCQ 1461 Mark
Choose the correct answer from the given four option.
The differential equation of the family of curves $\text{x}^2+\text{y}^2-2\text{ay}=0,$ where a is arbitrary constant, is:
- ✓
$(\text{x}^{2}-\text{y}^{2})\ \frac{\text{dy}}{\text{dx}}=2\text{xy}$
- B
$2(\text{x}^{2}+\text{y}^{2})\ \frac{\text{dy}}{\text{dx}}=\text{xy}$
- C
$2(\text{x}^{2}-\text{y}^{2})\ \frac{\text{dy}}{\text{dx}}=\text{xy}$
- D
$(\text{x}^{2}+\text{y}^{2})\ \frac{\text{dy}}{\text{dx}}=2\text{xy}$
AnswerCorrect option: A. $(\text{x}^{2}-\text{y}^{2})\ \frac{\text{dy}}{\text{dx}}=2\text{xy}$
Given equation is, $\text{x}^2+\text{y}^2-2\text{ay}=0$
$\Rightarrow\frac{\text{x}^2+\text{y}^2}{\text{y}}=2\text{a}$
On differentiating both sides w.r.t. x, we get
$\frac{\text{y}\Big(2\text{x}+2\text{y}\frac{\text{dy}}{\text{dx}}\Big)-(\text{x}^2+\text{y}^2)\frac{\text{dy}}{\text{dx}}}{\text{y}^2}=0$
$\Rightarrow2\text{xy}+2{\text{y}^2\frac{\text{dy}}{\text{dx}}}-(\text{x}^2+\text{y}^2)\frac{\text{dy}}{\text{dx}}=0$
$\Rightarrow(2\text{y}^2-\text{x}^2-\text{y}^2)\frac{\text{dy}}{\text{dx}}=-2\text{xy}$
$\Rightarrow(\text{y}^2-\text{x}^2)\frac{\text{dy}}{\text{dx}}=-2\text{xy}$
$\Rightarrow(\text{x}^2-\text{y}^2)\frac{\text{dy}}{\text{dx}}=2\text{xy}$
View full question & answer→MCQ 1471 Mark
Choose the correct answer from the given four option.
Integrating factor of the differential equation $\frac{\text{d}\text{y}}{\text{d}\text{x}}+\text{y}\tan\text{x}-\sec\text{x}=0$ is:
AnswerCorrect option: B. $\sec\text{x}$
Given that, $\frac{\text{d}\text{y}}{\text{d}\text{x}}+\text{y}\tan\text{x}-\sec\text{x}=0$
$\Rightarrow\frac{\text{d}\text{y}}{\text{d}\text{x}}+\text{y}\tan\text{x}=\sec\text{x}$
$\Big($It is a linear differential equation of form $\frac{\text{d}\text{y}}{\text{d}\text{x}}+\text{P}\text{y}=\text{Q}\Big)$
Here, $\text{P}=\tan\text{x},\text{Q}=\sec\text{x}$
$=\text{I.F.}=\text{e}^{\int\text{Pdx}}$
$=\text{e}^{\int\tan\text{x}\text{dx}}$
$=\text{e}^{(\log\sec\text{x})}$
$=\sec\text{x}$
View full question & answer→MCQ 1481 Mark
Solution of the differential equation $\frac{\text{dy}}{\text{dx}}+\frac{\text{y}}{\text{x}}=\sin\text{x}$ is:
- ✓
$\text{x}(\text{y}+\cos\text{x})=\sin\text{x}+\text{C}$
- B
$\text{x}(\text{y}-\cos\text{x})=\sin\text{x}+\text{C}$
- C
$\text{x}(\text{y}+\cos\text{x})=\cos\text{x}+\text{C}$
- D
AnswerCorrect option: A. $\text{x}(\text{y}+\cos\text{x})=\sin\text{x}+\text{C}$
We have,
$\frac{\text{dy}}{\text{dx}}+\frac{\text{y}}{\text{x}}=\sin\text{x}$
$\Rightarrow \frac{\text{dy}}{\text{dx}}+\frac{1}{\text{x}}\text{y}=\sin\text{x}\ ...(\text{i})$
Comparing with we get,
$\text{P}=\frac{1}{\text{x}}$
$\text{Q}=\sin\text{x}$
Now,
$\text{I.F}=\text{e}^{\int\frac{1}{\text{x}}\text{dx}}$
$=\text{e}^{\log|\text{x}|}$
$=\text{x}$
Therefore, intergration of (i) is given by,
$\text{y}\times\text{I.F}=\int\text{x}^{2}\times\text{I.F.}\ \text{dx}+\text{C}$
$\Rightarrow\text{yx}=\int\text{x}\ \sin\text{x}\ \text{dx}+\text{C}$
$\Rightarrow\text{yx}=\text{x}\int\sin\text{x}\ \text{dx}-\int\Big[\frac{\text{d}}{\text{dx}}(\text{x})\int\sin\text{x}\ \text{dx}\Big]\text{dx}+\text{C}$
$\Rightarrow\text{yx}=-\text{x}\cos\text{x}+\int\cos\text{x}\ \text{dx}+\text{C}$
$\Rightarrow\text{yx}+\text{x}\cos\text{x}=\sin\text{x}+\text{C}$
$\Rightarrow\text{x}(\text{y}+\cos\text{x})=\sin\text{x}+\text{C}$
View full question & answer→MCQ 1491 Mark
The equation of the curve aatisfying the differential $\text{y}(\text{x}+\text{y}^{3})\text{dx}=\text{x}(\text{y}^{3}-\text{x})$ dy and passing through the point (1, 1) is:
- A
$\text{y}^{3}-2\text{x}+3\text{x}^{2}\text{y}=0$
- B
$\text{y}^{3}+2\text{x}+3\text{x}^{2}\text{y}=0$
- ✓
$\text{y}^{3}+2\text{x}-3\text{x}^{2}\text{y}=0$
- D
AnswerCorrect option: C. $\text{y}^{3}+2\text{x}-3\text{x}^{2}\text{y}=0$
We have,$\text{y}(\text{x}+\text{y}^{3})\text{dx}=\text{x}(\text{y}^{3}-\text{x})\text{dy}$
$\Rightarrow (\text{xy}+\text{y}^{4})\text{dx}=(\text{xy}^{3}-\text{x}^{2})\text{dy}=0$
$\Rightarrow\text{xy}\ \text{dx}+\text{y}^{4}\text{dx}-\text{xy}^{3}\text{dy}+\text{x}^{2}\text{dy}=0$
$\Rightarrow \text{x}(\text{y}\text{dx}+\text{x}\text{dy})+\text{y}^{3}(\text{y}\text{dx}-\text{x}\text{dy})=0$
$\Rightarrow \text{xd}(\text{xy})+\text{x}^{2}\text{y}^{3}\frac{(\text{y}\text{dx}-\text{x}\text{dy})}{\text{x}^{2}}=0$
$\Rightarrow \frac{\text{d}(\text{xy})}{\text{x}^{2}\text{y}^{2}}-\frac{\text{y}}{\text{x}}\text{d}\Big(\frac{\text{y}}{\text{x}}\Big)=0$
$\Rightarrow \frac{\text{d}(\text{xy})}{\text{x}^{2}\text{y}^{2}}-\frac{\text{y}}{\text{x}}\text{d}\Big(\frac{\text{y}}{\text{x}}\Big)$
Integrating both sides we get,
$\Rightarrow \int\frac{\text{d}(\text{xy})}{\text{x}^{2}\text{y}^{2}}=\int\frac{\text{y}}{\text{x}}\text{d}\Big(\frac{\text{y}}{\text{x}}\Big)$
$\Rightarrow-\frac{1}{\text{xy}}=\frac{\Big(\frac{\text{y}}{\text{x}}\Big)^{2}}{2}-\text{C}$
$\Rightarrow-\frac{1}{\text{xy}}-\frac{1}{2}\Big(\frac{\text{y}^{2}}{\text{x}^{2}}\Big)+\text{C}=0$
$\Rightarrow \text{y}^{3}+2\text{x}+2\text{Cx}^{2}\text{y}=0$
It is given that the curve passes through (1, 1).
Hence,
$\text{y}^{3}+2\text{x}+2\text{Cx}^{2}\text{y}=0$
$(1)^{3}+2(1)+2\text{C}(1)(1)=0$
$1+2+2\text{C}=0$
$2\text{C}=-3$
$\text{C}=-\frac{3}{2}$
The required curve is,
$\text{y}^{3}+2\text{x}-3\text{x}^{2}\text{y}=0$
View full question & answer→MCQ 1501 Mark
Choose the correct answer from the given four option.
The differential equation $\text{y}\frac{\text{d}\text{y}}{\text{d}\text{x}}+\text{x}=\text{C}$ represents:
AnswerGiven that, $\text{y}\frac{\text{d}\text{y}}{\text{d}\text{x}}+\text{x}=\text{C}$
$\Rightarrow\text{y}\frac{\text{d}\text{y}}{\text{d}\text{x}}=\text{C}-\text{x}$
$\Rightarrow\text{ydy}=(\text{C}-\text{x})\text{dx}$
On integrating both sides, we get
$\int\text{ydy}=\int(\text{C}-\text{x})\text{dx}$
$\Rightarrow\frac{\text{y}^2}{2}=\text{Cx}-\frac{\text{x}^2}{2}+\text{k}$
$\Rightarrow\frac{\text{x}^2}{2}+\frac{\text{y}^2}{2}=\text{Cx}+\text{k}$
$\Rightarrow\frac{\text{x}^2}{2}+\frac{\text{y}^2}{2}-\text{Cx}=\text{k}$
which represent family of circles.
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