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$:\ \frac{\text{dy}}{\text{dx}}=\text{y}\sin\text{x:}$
- A
$y + \log \sin x + c = 0$
- B
$\log y - \cos x - c = 0$
- ✓
$\log y + \cos x - c = 0$
- D
AnswerCorrect option: C. $\log y + \cos x - c = 0$
Concept:
$\int\frac{\text{dx}}{\text{x}}=\log\text{x}+\text{c}$
$\int\sin{\text{x}}{\text{ dx}}=-\cos\text{x}+\text{c}$
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
$\Rightarrow\int\frac{\text{dy}}{\text{y}}=\int\sin\text{x}\text{ dx}$
$\Rightarrow\log\text{y}=-\cos\text{x}+\text{c}$
$\Rightarrow\log\text{y}+\cos\text{x}-\text{c}=0$
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Which of the following is a homogeneous differnetial equation?
- A
$(4 x+6 y+5) d y-(3 y+2 x+4) d x=0$
- B
$xy \sim dx-\left(x^3+y^3\right) d y=0$
- C
$\left(x^3+2 y^2\right) d x+2 x y \sim d y=0$
- ✓
$y^2 d x+\left(x^2-x y-y^2\right)=0$
AnswerCorrect option: D. $y^2 d x+\left(x^2-x y-y^2\right)=0$
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 $y^2 d x+\left(x^2-x y-y^2\right) d y=0,$ the degree of all the terms is $2.$
Thus$, (d)$ constant a homogeneous differential equation.
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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 $\left(y^{\prime \prime \prime}\right)^2+\left(y^{\prime \prime}\right)^3+\left(y^{\prime}\right)^4+y^5=0 ?$
AnswerThe degree is the power raised to the highest order derivative. Therefore, in the given differential equation, $\left(y^{\prime \prime \prime}\right)^2+\left(y^{\prime \prime}\right)^3+\left(y^{\prime}\right)^4+y^5=0 ?$, 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:
- A
$\text{p}<\text{q}$
- B
$\text{p}=\text{q}$
- ✓
$\text{p}>\text{q}$
- D
AnswerCorrect option: C. $\text{p}>\text{q}$
We 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
$\text{y}.(1+\text{x}^2)=\int(1+\text{x}^2)\frac{1}{(1+\text{x}^2)}+\text{C}$
$\Rightarrow\text{y}(1+\text{x}^2)=\int\frac{1}{(1+\text{x}^2)}\text{dx}+\text{C}$
$\Rightarrow\text{y}(1+\text{x}^2)=\tan^{-1}\text{x}+\text{C}$
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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
$\text{y}\ ''+\text{y}\ '=0$
- B
$\text{y}\ ''-\omega^{2}\text{y}=0$
- ✓
$\text{y}\ ''=-\omega^{2}\text{y}=0$
- D
$\text{y}\ ''+\text{y}=0$
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. $\text{y}-\text{x}=\text{k}(1+\text{xy})$
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 $n^{\text {th }}$ 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
- A
$\text{m}=3, \text{n}=3$
- ✓
$\text{m}=3, \text{n}=2$
- C
- D
$\text{m}=3,\text{n}=1$
AnswerCorrect option: B. $\text{m}=3, \text{n}=2$
We 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 $
$\Rightarrow y = x$
Solution of differential equation
$\text{x dy - y x = 0}$
reperesnts straight line passing through origin
View full question & answer→MCQ 1191 Mark
The degree and the order of the differential equation$:\ \frac{\text{d}^2\text{y}}{\text{dx}^2}=\sqrt{1}+\Big(\frac{\text{dy}}{\text{dx}}\Big)^3\text{ are:}$
- A
$2$ and $3.$
- B
$3$ and $2.$
- ✓
$2$ and $2.$
- D
$3$ and $3.$
AnswerCorrect option: C. $2$ and $2.$
Concept:
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:
- ✓
$2$ and $4$
- B
$2$ and $2$
- C
$2$ and $3$
- D
$3$ and $3$
AnswerCorrect option: A. $2$ and $4$
We 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 $\bigg(\frac{\text{d}^2\text{y}}{\text{dx}^2}\bigg)^3 + \bigg(\frac{\text{dy}}{\text{dx}}\bigg)^2+\text{sin} \bigg(\frac{\text{dy}}{\text{dx}}\bigg) + 1 =0$ 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 $\text{y}_3)$
View full question & answer→MCQ 1311 Mark
What is the solution of the differential equation$:\ \text{In}\Big(\frac{\text{dx}}{\text{dy}}\Big)-\text{a}=0?$
- ✓
$y=x e^a+c$
- B
$x=y e^a+c$
- C
$\text{y = In x + c}$
- D
$\text{x = In y + c}$
AnswerCorrect option: A. $y=x e^a+c$
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
$x^{a b c}$
- ✓
$1$
- D
Answer$x \frac{b-c}{b c} \times \frac{c-a}{a c} \times \frac{a-b}{a c} \times \frac{b-c}{b c}+\frac{c-a}{c a}+\frac{a-b}{a c}$
$x^0=1$
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 $\frac{\text{dy}}{\text{dx}}+\text{y}\cot\text{x}=\operatorname{cosec}\ \text{x}$ 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. $\text{y}\sin\text{x}=\text{x}+\text{C}$
$\frac{\text{dy}}{\text{dx}}+\text{y}\cot\text{x}=\operatorname{cosec}\ \text{x}$
Comparting with $\frac{\text{dy}}{\text{dx}}+\text{Py}=\text{Q}$ we get
$\text{P}=\cot\text{x}$
$\text{Q}=\operatorname{cosec} \ \text{x}$
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
$\Rightarrow \text{y}\sin\text{x}=\int\sin\text{x}\times\operatorname{cosec}\ \text{x}\text{dx}+\text{C}$
$\Rightarrow \text{y}\sin\text{x}=\text{x}+\text{C}$
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 $a x^2+2 h x y+b y^2+c=0$
Dividing by aa, we get
$x ^2+\left(\frac{2 h}{ a }\right) ax +\left(\frac{ b }{ a }\right) y ^2+\left(\frac{ c }{ a }\right)=0$
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 $x d y+y d y=x^2 y d y-y^2 x d x$, is:
- ✓
$x^2-1=C\left(1+y^2\right)$
- B
$x^2+1=C\left(1+y^2\right)$
- C
$x^3-1=C\left(1+y^3\right)$
- D
$x^3+1=C\left(1-y^3\right)$
AnswerCorrect option: A. $x^2-1=C\left(1+y^2\right)$
a. $x^2-1=C\left(1+y^2\right)$
Solution:
We have,
$x d x+y d y=x^2 d y-y^2 x d x$
$\Rightarrow\left(x+x y^2\right) d x=\left(x^2 y-y\right) d y$
$\Rightarrow \frac{x}{\left(x^2-1\right)} d x=\frac{y}{(1+x)^2} d y$
$\Rightarrow \frac{2 x}{2\left(x^2-1\right)} d x=\frac{2 y}{2(1+y)^2} d y$
Integrating both sides, we get
$\frac{1}{2} \int \frac{2 y }{(1+ y )^2} dy =\frac{1}{2} \int \frac{2 x }{(1+ x )^2} dx$
$\Rightarrow \log \left|\left(1+ y ^2\right)\right|=\frac{1}{2} \log \left|\left( x ^2-1\right)\right|-\frac{1}{2} \log |C|$
$\Rightarrow \log \left|\left(1+ y ^2\right)\right|=\log \left|\left( x ^2-1\right)\right|-\log |C|$
$\Rightarrow \log \left|\left(1+ y ^2\right)\right|=\log \left|\frac{ x ^2-1}{ C }\right|$
$\Rightarrow 1+ y ^2=\frac{ x ^2-1}{ C }$
$\Rightarrow C \left(1+ y ^2\right)= x ^2-1$
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
$\text{e}^\text{x}\ \text{dy}+(\text{y e}^\text{x}+2\text{x})\text{dx}=0$
or$\ \ \text{e}^\text{x}\frac{\text{dy}}{\text{dx}}+\text{y e}^\text{x}+2\text{x}=0\ \ $or$\ \ \frac{\text{dy}}{\text{dx}}+\text{y}+\frac{2\text{x}}{\text{e}^\text{x}}=0$ or $\ \ \frac{\text{dy}}{\text{dx}}+\text{y}=-2\text{x e}^{-\text{x}}$
Comparing it with $\frac{\text{dy}}{\text{dx}}+\text{Py}=\text{Q},$
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 $\text{w.r.t.x,}$ 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,
- A
$\text{a}=\text{b}$
- ✓
$\text{a}=-\text{b}$
- C
$\text{a}=-2\text{b}$
- D
$\text{a}=2\text{b}$
AnswerCorrect option: B. $\text{a}=-\text{b}$
b. $a=-b$
Solution:
We have,
$\frac{d y}{d x}=\frac{a x+g}{b y+f}$
$\Rightarrow(b y+f) d y=(a x+g) d x$
Intergrating both sides, we get
$\Rightarrow \int( by + f ) dy =\int( ax + g ) dx$
$\Rightarrow b \frac{ y ^2}{2}+ fy = a \frac{ x ^2}{2}+ gx + C$
$\Rightarrow b \frac{ y ^2}{2}+ fy - a \frac{ x ^2}{2}-g x = C$
$\Rightarrow b y^2+2 f y-a x^2-2 g x-2 C=0$
The above equation resprasents a circle.
Therefore, the coffrcients of $x^2$ and $y^2$ must be equal.
$- a = b$
$\Rightarrow a =- b$
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
$\text{x}=\log\sec\text{y}+\text{C}$
$\Rightarrow\text{x}-\text{C}=\log\sec\text{y}$
$\Rightarrow\sec\text{y}=\text{e}^{\text{x}-\text{c}}$
$\Rightarrow\frac{1}{\cos\text{y}}=\frac{\text{e}^\text{x}}{\text{e}^\text{c}}$
$\Rightarrow\text{e}^{\text{x}}\cos\text{y}=\text{k}$
$[$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 ?
- A
$1$ only
- ✓
$2$ only
- C
Both $1$ and $2$
- D
AnswerCorrect option: B. $2$ only
$\frac{\text{d}^2\text{y}}{\text{d}\text{x}^2}+\cos\Big(\frac{\text{dx}}{\text{dy}}\Big)=0:$
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 $\text{w.r.t. x},$ 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 $\text{y}(\text{x}+\text{y}^{3})\text{dx}=\text{x}(\text{y}^{3}-\text{x})\text{dy}$ 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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