MCQ 1511 Mark
The number of arbitrary constants in the particular solution of a differential equation of third order is:
AnswerThe solution free from arbitrary constants i.e., the solution obtained from the general solution by giving particular values to the arbitrary constants is called a particular solution of the differential equation.
View full question & answer→MCQ 1521 Mark
The solution of $\text{x}^{2}+\text{y}^{2}\frac{\text{dy}}{\text{dx}}=4$ is:
- A
$x^2+y^2=12 x+C$
- B
$x^2+y^2=3 x+C$
- C
$x^3+y^3=3 x+c$
- ✓
$x^3+y^3=12 x+C$
AnswerCorrect option: D. $x^3+y^3=12 x+C$
We have,
$\text{x}^{2}+\text{y}^{2}\frac{\text{dy}}{\text{dx}}=4$
$\Rightarrow\text{y}^{2}\frac{\text{dy}}{\text{dx}}=4-\text{x}^{2}$
$\Rightarrow\text{y}^{2}\frac{\text{dy}}{\text{dx}}=(4-\text{x}^{2})\text{dx}$
Integrating both sides, we get
$\int\text{y}^{2}\frac{\text{dy}}{\text{dx}}=\int(4-\text{x}^{2})\text{dx}$
$\Rightarrow \frac{\text{y}^{3}}{3}=4\text{x}-\frac{\text{x}^{3}}{3}+\text{D} $
$\Rightarrow \text{y}^{3}=12\text{x}-\text{x}^{3}+3\text{D}$
$\Rightarrow \text{x}^{3}+\text{y}^{3}=12\text{x}+\text{C}$
View full question & answer→MCQ 1531 Mark
Which of the following transformation reduce the differential quation into the form $\frac{\text{du}}{\text{dx}}+\text{P}(\text{x})\text{u}=\text{Q}(\text{x})$ into the from $\frac{\text{dz}}{\text{dx}}+\frac{\text{z}}{\text{x}}\log\text{z}=\frac{\text{z}}{\text{x}^{2}}(\log\text{z})^{2}$
- A
$\text{u}=\log\text{x}$
- B
$\text{u}=\text{e}^{\text{z}}$
- ✓
$\text{u}=(\log\text{z})^{-1}$
- D
$\text{u}=(\log\text{z})^{2}$
AnswerCorrect option: C. $\text{u}=(\log\text{z})^{-1}$
We have,
$\frac{\text{dz}}{\text{dx}}+\frac{\text{z}}{\text{x}}\log\text{z}=\frac{\text{z}}{\text{x}^{2}}(\log\text{z})^{2}\ ...(\text{i})$
Let $\text{u}=(\log\text{z})^{-1}$
$\frac{\text{du}}{\text{dx}}=-\frac{1}{(\log)^{2}}\times\frac{1}{\text{z}}\times\frac{\text{dz}}{\text{dx}}$
$\frac{\text{du}}{\text{dx}}=-\text{z}(\log\text{z})^{2}\frac{\text{du}}{\text{dx}}$
Substituting the value of the equation $(i),$
$-\text{z}(\log\text{z})^{2}\frac{\text{du}}{\text{dx}}+\frac{\text{z}}{\text{x}}\log\text{z}=\frac{\text{z}}{\text{x}^{2}}(\log\text{z}^{2})$
$\frac{\text{du}}{\text{dx}}-\frac{1}{\text{x}}\frac{1}{\log\text{z}}=-\frac{1}{\text{x}^{2}}$
$\frac{\text{du}}{\text{dx}}-\frac{1}{\text{x}}({\log\text{z}})^{-1}=-\frac{1}{\text{x}^{2}}$
$\frac{\text{du}}{\text{dx}}-\frac{1}{\text{x}}(\text{u})=-\frac{1}{\text{x}^{2}}$
It can be written as,
$\frac{\text{du}}{\text{dx}}+\text{P}(\text{x})\text{u}=\text{Q}(\text{x})$
Where, $\text{p}(\text{x})=-\frac{1}{\text{x}}$
$\text{q}(\text{x})=-\frac{1}{\text{x}^{2}}$
View full question & answer→MCQ 1541 Mark
Choose the correct answer from the given four options.The general solution of $\frac{\text{dy}}{\text{dx}}=2\text{x}\ \text{e}^{\text{x}^{2}-\text{y}}$ is:
- A
$\text{e}^{\text{x}^{2}-\text{y}}=\text{C}$
- B
$\text{e}^{-\text{y}}+\text{e}^{\text{x}^{2}}=\text{C}$
- ✓
$\text{e}^{\text{y}}=\text{e}^{\text{x}^{2}}+\text{C}$
- D
$\text{e}^{\text{x}^{2}+\text{y}}=\text{C}$
AnswerCorrect option: C. $\text{e}^{\text{y}}=\text{e}^{\text{x}^{2}}+\text{C}$
We have $\frac{\text{dy}}{\text{dx}}=2\text{x}\ \text{e}^{\text{x}^{2}-\text{y}}$
$\Rightarrow\text{e}^{\text{y}}=\frac{\text{dy}}{\text{dx}}=2\text{x}\ \text{e}^{\text{x}^{2}}$
$\Rightarrow\int\text{e}^{\text{y}}\text{dy}=2\int\text{x}\text{e}^{\text{x}^{2}}\text{dx}$
Put $\text{x}^2=\text{t}$ in $\text{R.H.S.}$ integral we get
$2\text{x}\text{dx}=\text{dt}$
$\Rightarrow\int\text{e}^{\text{y}}\text{dy}=\int\text{e}^{\text{t}}\text{dt}$
$\Rightarrow\text{e}^{\text{y}}=\text{e}^{\text{t}}+\text{C}$
$\Rightarrow\text{e}^{\text{y}}=\text{e}^{\text{x}^2}+\text{C}$
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Choose the correct answer from the given four option.Solution of the differential equation $\tan\text{y}\sec^2\text{xdx} + \tan\text{x }\sec^2\text{ydy}=0$ is:
- A
$\tan\text{x}+\tan\text{y}=\text{k}$
- B
$\tan\text{x}-\tan\text{y}=\text{k}$
- C
$\frac{\tan\text{x}}{\tan\text{y}}=\text{k}$
- ✓
$\tan\text{x}.\tan\text{y}=\text{k}$
AnswerCorrect option: D. $\tan\text{x}.\tan\text{y}=\text{k}$
We have, $\tan\text{y}\sec^2\text{xdx} + \tan\text{x }\sec^2\text{ydy}=0$
$\Rightarrow\tan\text{y}\sec^2\text{xdx} =- \tan\text{x }\sec^2\text{ydy}$
$\Rightarrow\frac{\sec^2\text{x}}{\tan\text{x}}\text{dx}=-\frac{\sec^2\text{y}}{\tan\text{y}}\text{dy}$
$\Rightarrow\int\frac{\sec^2\text{x}}{\tan\text{x}}\text{dx}=-\int\frac{\sec^2\text{y}}{\tan\text{y}}\text{dy}$
$\Rightarrow\log\tan\text{x}=-\log\tan\text{y}+\log\text{k}$
$\Rightarrow\log\tan\text{x}+\log\tan\text{y}=\log\text{k}$
$\Rightarrow\log(\tan\text{x}\tan\text{y})=\log\text{k}$
$\Rightarrow\tan\text{x}\tan\text{y}=\text{k}$
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The general solution of differention eqution $\frac{\text{y}\ \text{dx}-\text{x}\ \text{dy}}{\text{y}}=0$ is:
- A
$xy = C$
- B
$x=C y^2$
- ✓
$y = Cx$
- D
$y=C x^2$
AnswerCorrect option: C. $y = Cx$
We have,
$\frac{\text{y}\ \text{dx}-\text{x}\ \text{dy}}{\text{y}}=0$
$\Rightarrow \text{y}\ \text{dx}=\text{x}\ \text{dy}$
$\Rightarrow \frac{1}{\text{y}}\ \text{dy}=\frac{1}{\text{x}}\ \text{dx}$
Integrating both sides, we get,
$\int\frac{1}{\text{y}}\ \text{dy}=\int\frac{1}{\text{x}}\ \text{dx}$
$\Rightarrow \log\text{y}=\log\text{x}+\text{D}$
$\Rightarrow \log\text{y}-\log\text{x}=\text{C}$
$\Rightarrow \log\big(\frac{\text{y}}{2}\big)=\log\text{C}$
$\Rightarrow \frac{\text{y}}{2}=\text{C}$
$\Rightarrow \text{y}=\text{Cx}$
View full question & answer→MCQ 1571 Mark
The differential equation $\text{x}\frac{\text{dy}}{\text{dx}}-\text{y}=\text{x}^{2}$ has the general solution:
- A
$y-x^3=2 C x$
- ✓
$2 y-x^3=C x$
- C
$2 y+x^2=2 C x$
- D
$y+x^2=2 C x$
AnswerCorrect option: B. $2 y-x^3=C x$
We have,
$\text{x}\frac{\text{dy}}{\text{dx}}-\text{y}=\text{x}^{2}$
$\Rightarrow \frac{\text{dy}}{\text{dx}}-\frac{1}{\text{x}}\text{y}=\text{x}^{2}$
Comparing with we get,
$\text{P}=-\frac{1}{\text{x}}$
$\text{Q}=\text{x}^{2}$
Now,
$\text{I.F}=\text{e}^{-\int\frac{1}{\text{x}}\text{dx}}$
$=\text{e}^{-\log|\text{x}|}$
$=\text{e}^{\log|\frac{1}{\text{x}}|}$
$=\frac{1}{\text{x}}$
$\text{y}\times\text{I.F}=\int\text{x}^{2}\times\text{I.F}\text{dx}+\text{C}$
$\Rightarrow \text{y}\frac{1}{\text{x}}=\int\text{x}^{2}\times\frac{1}{\text{x}}\text{dx}+\text{C}$
$\Rightarrow \text{y}\frac{1}{\text{x}}=\int\text{x}^{2}\text{dx}+\text{C}$
$\Rightarrow \text{y}\frac{1}{\text{x}}=\frac{\text{x}^{2}}{2}+\text{C}$
$\Rightarrow 2\text{y}-\text{x}^{3}=\text{Cx}$
View full question & answer→MCQ 1581 Mark
The degree of the differential equation $\big(\frac{\text{d}^{2}\text{y}}{\text{dx}^{2}}\big)^{3}+\big(\frac{\text{dy}}{\text{dx}}\big)^{2}+\sin\big(\frac{\text{dy}}{\text{dx}}\big)+1=0$ is:
AnswerWe have,
$\big(\frac{\text{d}^{2}\text{y}}{\text{dx}^{2}}\big)^{3}+\big(\frac{\text{dy}}{\text{dx}}\big)^{2}+\sin\big(\frac{\text{dy}}{\text{dx}}\big)+1=0$
The highest order derivative in this equation is $\frac{\text{d}^{2}\text{y}}{\text{d}^{2}\text{x}}.$
But the equation cannot be as a polynomial in differential coeffcient.
Hence, the degree is not defined.
View full question & answer→MCQ 1591 Mark
Which of the following differential equations has y = x as one of its particular solution?
- A
$\frac{\text{d}^2\text{y}}{\text{dx}^2}-\text{x}^2\frac{\text{dy}}{\text{dx}}+\text{xy}=\text{x}$
- B
$\frac{\text{d}^2\text{y}}{\text{dx}^2}+\text{x}\frac{\text{dy}}{\text{dx}}+\text{xy}=\text{x}$
- ✓
$\frac{\text{d}^2\text{y}}{\text{dx}^2}-\text{x}^2\frac{\text{dy}}{\text{dx}}+\text{xy}=0$
- D
$\frac{\text{d}^2\text{y}}{\text{dx}^2}+\text{x}\frac{\text{dy}}{\text{dx}}+\text{xy}=0$
AnswerCorrect option: C. $\frac{\text{d}^2\text{y}}{\text{dx}^2}-\text{x}^2\frac{\text{dy}}{\text{dx}}+\text{xy}=0$
The given equation of curve is y = x.
Differentiting with respect to x, we get:
$\frac{\text{dy}}{\text{dx}}=1 \ ...(1)$
Again, differentiating with respect to x, we get:
$\frac{\text{d}^2\text{y}}{\text{dx}^2}=0 \ ...(2)$
Now, on substituting the values of $\text{y}, \frac{\text{d}^2\text{y}}{\text{dx}^2},\ \text{and} \ \frac{\text{dy}}{\text{dx}}$ from equation (1) and (2) in each of the given alternatives, we find that only the differential equation given in alternative C is correct.
$\frac{\text{d}^2\text{y}}{\text{dx}^2}-\text{x}^2\frac{\text{dy}}{\text{dx}}+ \text{xy}=0-\text{x}^2\cdot1+\text{x} \cdot \text{x}$
$= -\text{x}^2+\text{x}^2$
$=0$
Hence, the correct answer is C.
View full question & answer→MCQ 1601 Mark
What is integrating factor of $\frac{\text{dy}}{\text{dx}}+\text{y}\sec\text{x}=\tan\text{x}?$
- ✓
$\sec\text{x}+\tan\text{x}$
- B
$\log(\sec\text{x}+\tan\text{x})$
- C
$\text{e}^{\sec\text{x}}$
- D
$\sec{\text{x}}$
AnswerCorrect option: A. $\sec\text{x}+\tan\text{x}$
We have,
$\frac{\text{dy}}{\text{dx}}+\text{y}\sec\text{x}=\tan\text{x}$
Comparing with We get,
$\text{P}=\sec{\text{x}}, \text{Q}=\tan{\text{x}}$
Now,
$\text{I.F}=\text{e}^{\int\sec\text{x}\text{dx}}$
$=\text{e}^{\log(\sec\text{x}+\tan\text{x})}$
$=\sec\text{x}+\tan\text{x}$
View full question & answer→MCQ 1611 Mark
Choose the correct answer from the given four options.The general solution of the differential equation $\frac{\text{dy}}{\text{dx}}\ \text{e}^{\frac{\text{x}^2}{2}}+\text{xy}$ is:
- A
$\text{y}=\text{c}\text{e}^{\frac{-\text{x}^2}{2}}$
- B
$\text{y}=\text{c}\text{e}^{\frac{\text{x}^2}{2}}$
- ✓
$\text{y}=(\text{x}+\text{c})\text{e}^{\frac{\text{x}^2}{2}}$
- D
$\text{y}=(\text{c}-\text{x})\text{e}^{\frac{\text{x}^2}{2}}$
AnswerCorrect option: C. $\text{y}=(\text{x}+\text{c})\text{e}^{\frac{\text{x}^2}{2}}$
Given that, $\frac{\text{dy}}{\text{dx}}\ \text{e}^{\frac{\text{x}^2}{2}}+\text{xy}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}-\text{xy}=\text{e}^{\frac{\text{x}^2}{2}}$
It is a linear differential equation.
Here, $\text{P}=-\text{x},\text{ Q}=\text{e}^{\frac{\text{x}^2}{2}}$
$\therefore\text{I.F.}=\text{e}^{\int-\text{xdx}}=\text{e}^{\frac{-\text{x}^2}{2}}$
The general solution is
$\text{y}\text{e}^{\frac{-\text{x}^2}{2}}=\int\text{e}^{\frac{-\text{x}^2}{2}}.\text{e}^{\frac{-\text{x}^2}{2}}\text{dx}+\text{c}$
$\Rightarrow\text{y}{\text{e}}^{\frac{-\text{x}^2}{2}}=\int1\text{dx}+\text{c}$
$\Rightarrow\text{y}\text{e}^{\frac{-\text{x}^2}{2}}=\text{x}+\text{c}$
$\Rightarrow\text{y}=\text{x}\text{e}^{\frac{\text{x}^2}{2}}+\text{c}\text{e}^{\frac{\text{x}^2}{2}}$
$\Rightarrow\text{y}=(\text{x}+\text{c})\text{e}^{\frac{\text{x}^2}{2}}$
View full question & answer→MCQ 1621 Mark
Choose the correct answer from the given four option.The differential equation for $\text{y}=\text{A}\cos\alpha\text{x}+\text{B}\sin\alpha\text{x},$ where $A$ and $B$ are arbitrary constants is:
- A
$\frac{\text{d}^2\text{y}}{\text{d}\text{x}^2}-\alpha^2\text{y}=0$
- ✓
$\frac{\text{d}^2\text{y}}{\text{d}\text{x}^2}+\alpha^2\text{y}=0$
- C
$\frac{\text{d}^2\text{y}}{\text{d}\text{x}^2}+\alpha\text{y}=0$
- D
$\frac{\text{d}^2\text{y}}{\text{d}\text{x}^2}-\alpha\text{y}=0$
AnswerCorrect option: B. $\frac{\text{d}^2\text{y}}{\text{d}\text{x}^2}+\alpha^2\text{y}=0$
Given, $\text{y}=\text{A}\cos\alpha\text{x}+\text{B}\sin\alpha\text{x}$
$\Rightarrow\frac{\text{d}\text{y}}{\text{d}\text{x}}=-\alpha\text{A}\sin\alpha\text{x}+\alpha\text{B}\cos\alpha\text{x}$
Again, differentiating both sides $\text{w.r.t. x,}$ we get
$\frac{\text{d}^2\text{y}}{\text{d}\text{x}^2}=-\text{A}\alpha^2\cos\alpha\text{x}-\alpha^2\text{B}\sin\alpha\text{x}$
$\Rightarrow\frac{\text{d}^2\text{y}}{\text{d}\text{x}^2}=-\alpha^2(\text{A}\cos\alpha\text{x}-\text{B}\sin\alpha\text{x})$
$\Rightarrow\frac{\text{d}^2\text{y}}{\text{d}\text{x}^2}=-\alpha^2\text{y}$
$\Rightarrow\frac{\text{d}^2\text{y}}{\text{d}\text{x}^2}+\alpha^2\text{y}=0$
View full question & answer→MCQ 1631 Mark
Order of $\Big(\frac{\text{dy}}{\text{dx}}+3\text{x}\Big)^\frac{3}{2}=\text{x}+\frac{3\text{dy}}{\text{dx}}$ is:
AnswerOrder$:-$ The number of the highest derivative in a differential equation, that is
$=\frac{\text{dy}}{\text{dx}}=1$
View full question & answer→MCQ 1641 Mark
The general solution of the differntial equation $\frac{\text{dy}}{\text{dx}}=\frac{\text{y}}{\text{x}}$ is:
AnswerCorrect option: B. $\text{y}=\text{kx}$
We have,
$\frac{\text{dy}}{\text{dx}}=\frac{\text{y}}{\text{x}}$
$\Rightarrow\frac{\text{1}}{\text{y}}\text{dy}=\frac{\text{1}}{\text{x}}\text{dx}$
Integrating both sides, we get
$\int\frac{\text{1}}{\text{y}}\text{dy}=\int\frac{\text{1}}{\text{x}}\text{dx}$
$\log\text{y}=\log\text{x}+\log\text{k}$
$\log\text{y}-\log\text{x}=\log\text{k}$
$\log\frac{\text{y}}{\text{x}}=\log\text{k}$
$\Rightarrow\frac{\text{y}}{\text{x}}=\text{k}$
$\Rightarrow\text{y}=\text{k}{\text{x}}$
View full question & answer→MCQ 1651 Mark
Choose the correct answer from the given four option.The integrating factor of the differential equation $\frac{\text{d}\text{y}}{\text{d}\text{x}}+\text{y}=\frac{1+\text{y}}{\text{x}}$ is:
- A
$\frac{\text{x}}{\text{e}^{\text{x}}}$
- ✓
$\frac{\text{e}^{\text{x}}}{\text{x}}$
- C
${\text{x}}\text{e}^{\text{x}}$
- D
$\text{e}^{\text{x}}$
AnswerCorrect option: B. $\frac{\text{e}^{\text{x}}}{\text{x}}$
We have, $\frac{\text{d}\text{y}}{\text{d}\text{x}}+\text{y}=\frac{1+\text{y}}{\text{x}}$
$\Rightarrow\frac{\text{d}\text{y}}{\text{d}\text{x}}=\frac{1+\text{y}}{\text{x}}-\text{y}$
$\Rightarrow\frac{\text{d}\text{y}}{\text{d}\text{x}}=\frac{1+\text{y}-\text{xy}}{\text{x}}$
$\Rightarrow\frac{\text{d}\text{y}}{\text{d}\text{x}}=\frac{1}{\text{x}}+\frac{\text{y}(1-\text{x})}{\text{x}}$
$\Rightarrow\frac{\text{d}\text{y}}{\text{d}\text{x}}-\Big(\frac{1-\text{x}}{\text{x}}\Big)$
$\Rightarrow\text{y}=\frac{1}{\text{x}}$
Here, $\text{P}=\frac{-(1-\text{x})}{\text{x}},\text{Q}=\frac{1}{\text{x}}$
$\text{I.F.}=\text{e}^{\int\text{Pdx}}$
$=\text{e}^{-\int\frac{1+\text{x}}{\text{x}}\text{dx}}=\text{e}^{\frac{\text{x}-1}{\text{x}}}$
$=\text{e}^{{\int\Big(1-\frac{1}{\text{x}}\Big)}\text{dx}}=\text{e}^{\int\text{x}-\log\text{x}}$
$=\text{e}^{\text{x}}.\text{e}^{\log\Big(\frac{1}{\text{x}}\Big)}=\text{e}^{\text{x}}.\frac{1}{\text{x}}$
View full question & answer→MCQ 1661 Mark
The differential equation for $\text{y}=\text{A}\cos\text{ax}+\text{B}\sin\text{ax}$ where $A$ and $B$ are arbitary constants is:
- A
$\frac{\text{d}^2\text{y}}{\text{dx}^2}-\text{a}^2\text{y}=0$
- ✓
$\frac{\text{d}^2\text{y}}{\text{dx}^2}+\text{a}^2\text{y}=0$
- C
$\frac{\text{d}^2\text{y}}{\text{dx}^2}+\text{ay}=0$
- D
$\frac{\text{d}^2\text{y}}{\text{dx}^2}-\text{ay}=0$
AnswerCorrect option: B. $\frac{\text{d}^2\text{y}}{\text{dx}^2}+\text{a}^2\text{y}=0$
View full question & answer→MCQ 1671 Mark
The solution of the differential equation $\frac{\text{dy}}{\text{dx}}+\frac{2\text{y}}{\text{x}}=0$ with $y(1) = 1$ is given by.
- ✓
$\text{y}=\frac{1}{\text{x}^{2}}$
- B
$\text{x}=\frac{1}{\text{y}^{2}}$
- C
$\text{x}=\frac{1}{\text{y}}$
- D
$\text{y}=\frac{1}{\text{x}}$
AnswerCorrect option: A. $\text{y}=\frac{1}{\text{x}^{2}}$
We have,
$\frac{\text{dy}}{\text{dx}}+\frac{2\text{y}}{\text{x}}=0$
$\Rightarrow \frac{\text{dy}}{\text{dx}}=\frac{-2\text{y}}{\text{x}}$
$\Rightarrow\frac{1}{2}\times\frac{1}{\text{y}}\text{dy}=\frac{-1}{\text{x}}\text{dx}$
Integrating both sides, we get
$\Rightarrow\frac{1}{2}\int\frac{1}{\text{y}}\text{dy}=-\int\frac{1}{\text{x}}\text{dx}$
$\Rightarrow\frac{1}{2}\log\text{y}=-\log{\text{x}}+\log\text{C}$
$\Rightarrow\log\text{y}^{\frac{1}{2}}+\log{\text{x}}=\log\text{C}$
$\Rightarrow \log(\sqrt{\text{yx}})=\log\text{C}$
$\Rightarrow\sqrt{\text{yx}}=\log\text{C}\ ...(\text{i})$
As $(i) y(1) = 1,$ we get
$1=\text{C}$
Putting the valur of $C$ in $(i)$
$\Rightarrow\sqrt{\text{yx}}=1$
$\Rightarrow\text{y}=\frac{1}{\text{x}^{2}}$
View full question & answer→MCQ 1681 Mark
Consider a differential equation of order $mm$ and degree $nn.$ Which one of the following pairs is not feasible?
- A
$(3,2)$
- ✓
$(2,\frac{3}{2})$
- C
$(2,4)$
- D
$(2,2)$
AnswerCorrect option: B. $(2,\frac{3}{2})$
Degree of the differential equation is always a positive integer.But in $B,$ degree is given $\frac{3}{2}$ which is not an integer.
So option $B$ is not the feasible pair.
View full question & answer→MCQ 1691 Mark
Integrating factor of the differential equation, $(1-\text{x}^2)\frac{\text{dy}}{\text{dx}}-\text{xy}=1$ is:
AnswerCorrect option: C. $\sqrt{1-\text{x}^2}$
View full question & answer→MCQ 1701 Mark
The degree of the differential equation$:\frac{\text{d}^2\text{y}}{\text{dx}^2}+3\Big(\frac{\text{dy}}{\text{dx}}\Big)^2=\text{x}^2\log\Big(\frac{\text{d}^2\text{y}}{\text{dx}^2}\Big)$
AnswerConcept:
Order: The order of a differential equation is the order of the highest derivative appearing in it.
Degree: The degree of a differential equation is the power of the highest derivative occurring in it, after the Equation has been expressed in a form free from radicals as far as the derivatives are concerned.
Calculation:
$\frac{\text{d}^2\text{y}}{\text{dx}^2}+3\Big(\frac{\text{dy}}{\text{dx}}\Big)^2=\text{x}^2\log\Big(\frac{\text{d}^2\text{y}}{\text{dx}^2}\Big)$
For the given differential equation the highest order derivative is $2.$
The given differential equation is not a polynomial equation because it involved a logarithmic term in its derivatives hence its degree is not defined.
View full question & answer→MCQ 1711 Mark
Choose the correct answer from the given four option.The degree of the differential equation $\Big[1+\Big(\frac{\text{d}\text{y}}{\text{d}\text{x}}\Big)^2\Big]^{\frac{3}{2}}=\frac{\text{d}^2\text{y}}{\text{d}\text{x}^2}$ is:
AnswerGiven is, $\Big[1+\Big(\frac{\text{d}\text{y}}{\text{d}\text{x}}\Big)^2\Big]^{\frac{3}{2}}=\frac{\text{d}^2\text{y}}{\text{d}\text{x}^2}$
On squaring both sides, we get
$\Big[1+\Big(\frac{\text{d}\text{y}}{\text{d}\text{x}}\Big)^2\Big]^{3}=\Big(\frac{\text{d}^2\text{y}}{\text{d}\text{x}^2}\Big)^2$
So, the degree of differential equation is $2.$
View full question & answer→MCQ 1721 Mark
Choose the correct answer from the given four option.Integrating factor of the differential equation $(1-\text{x}^2)\frac{\text{d}\text{y}}{\text{d}\text{x}}-\text{xy}=1$ is:
AnswerCorrect option: C. $\sqrt{1-\text{x}^2}$
Given is, $(1-\text{x}^2)\frac{\text{d}\text{y}}{\text{d}\text{x}}-\text{xy}=1$
$\Rightarrow\frac{\text{d}\text{y}}{\text{d}\text{x}}-\frac{\text{x}}{1-\text{x}^2}\text{y}=\frac{1}{1-\text{x}}^2$
Which is a linear differential equation.
$\therefore\text{I.F.}=\text{e}^{-\int\frac{\text{x}}{1-\text{x}^2}\text{dx}}$
Put $1-\text{x}^2=\text{t}$
$\Rightarrow-2\text{xdx}=\text{dt}$
$\Rightarrow\text{xdx}=-\frac{\text{dt}}{2}$
Now, $\text{I.F.}=\text{e}^{\frac{1}{2}\int\frac{\text{dt}}{\text{t}}}$
$\text{e}^{\frac{1}{2}\log\text{t}}=\text{e}^{\frac{1}{2}\log(1-\text{x}^2)}$
$\Rightarrow\sqrt{1-\text{x}^2}$
View full question & answer→MCQ 1731 Mark
The number of arbitrary constants in the particular solution of differential equation of fourth order is:
AnswerThe number of arbitray constant in the particular solution of a differential equation is always zero.
View full question & answer→MCQ 1741 Mark
Choose the correct answer from the given four option. Integrating factor of $\frac{\text{xd}\text{y}}{\text{d}\text{x}}-\text{y}=\text{x}^4-3\text{x}$ is :
- A
$\text{x}$
- B
$\log\text{x}$
- ✓
$\frac{1}{\text{x}}$
- D
$-\text{x}$
AnswerCorrect option: C. $\frac{1}{\text{x}}$
Given that $\frac{\text{xd}\text{y}}{\text{d}\text{x}}-\text{y}=\text{x}^4-3\text{x}$
Dividing both sides by $x,$ we get
$\Rightarrow\frac{\text{d}\text{y}}{\text{d}\text{x}}-\frac{\text{y}}{\text{x}}=\text{x}^3-3$
Here, $\text{P}=-\frac{1}{\text{x}},\text{Q}=\text{x}^3-3$
$\therefore\text{I.F.}=\text{e}^{\int\text{Pdx}}=\text{e}^{-\int\frac{1}{\text{x}}\text{dx}}$
$\text{e}^{-\log\text{x}}=\frac{1}{\text{x}}$
View full question & answer→MCQ 1751 Mark
The general solution of the differential 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. $\text{e}^{\text{x}}+\text{e}^{-\text{y}}=\text{C}$
The given differential equation is
$\frac{\text{dy}}{\text{dx}}=\text{e}^{\text{x+y}}$ or $ \frac{\text{dy}}{\text{dx}}=\text{e}^{\text{x}}.\text{e}^{\text{y}}$
Separating the variables, we get,
$\frac{1}{\text{e}^{\text{y}}}\text{dy}=\text{e}^\text{x}\ \text{dx}$
Integrating $\int\text{e}^{-\text{y}}\text{dy}=\int\text{e}^\text{x}\ \text{dx}$
$\therefore\ \frac{\text{e}^{-\text{y}}}{-1}=\text{e}^\text{x}+\text{c}'$
$\therefore\ -\text{e}^{-\text{y}}=\text{e}^\text{x}+\text{c}'$ or $ \text{e}^\text{x}+\text{e}^{-\text{y}}=-\text{c}'$
$\therefore\ \text{e}^{\text{x}}+\text{e}^{-\text{y}}=\text{C},$ which is required solution.
View full question & answer→MCQ 1761 Mark
The integrating factor of the differential equation$(1-\text{y}^{2})\frac{\text{dx}}{\text{dy}}+\text{yx}=\text{ay}(-1<\text{y}<1)$ is:
- A
$\frac{1}{\text{y}^{2}-1}$
- B
$\frac{1}{\sqrt{\text{y}^{2}+1}}$
- C
$\frac{1}{1-\text{y}^{2}}$
- ✓
$\frac{1}{\sqrt{1-\text{y}^{3}}}$
AnswerCorrect option: D. $\frac{1}{\sqrt{1-\text{y}^{3}}}$
We have,
$(1-\text{y}^{2})\frac{\text{dx}}{\text{dy}}+\text{yx}=\text{ay}$
$\frac{\text{dx}}{\text{dy}}+\frac{\text{y}}{1-\text{y}^{2}}\text{x}=\frac{\text{ay}}{1-\text{y}^{2}}$
Comparing with we get,
$\text{P}=\frac{\text{y}}{1-\text{y}^{2}}, \text{Q}=\frac{\text{ay}}{1-\text{y}^{2}}$
Now,
$\text{I.F}=\text{e}^{\int\frac{\text{y}}{1-\text{y}^{2}}\text{dy}}$
$=\text{e}^{-\frac{1}{2}\int\frac{-2\text{y}}{1-\text{y}^{2}}\text{dy}}$
$=\text{e}^{-\frac{1}{2}\log|1-\text{y}^{2}|}$
$=\text{e}^{\log\Big|\frac{1}{\sqrt{1-\text{y}^{2}}}\Big|}$
$=\frac{1}{\sqrt{1-\text{y}^{2}}}$
View full question & answer→MCQ 1771 Mark
The integrating factor of the differential equation $\left(x+2 y^2\right) \frac{d y}{d x}=y(y>0)$ is :
- A
$\frac{1}{x}$
- B
$x$
- C
$y$
- ✓
$\frac{1}{y}$
AnswerCorrect option: D. $\frac{1}{y}$
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→MCQ 1781 Mark
The general solution of the differential equation $x d y+y d x=0$ is:
- ✓
$x y=c$
- B
$x+y=c$
- C
$x^2+y^2=c^2$
- D
$\log y=\log x+c$
AnswerCorrect option: A. $x y=c$
We 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→MCQ 1791 Mark
The differential equation $\frac{d y}{d x}=F(x, y)$ will not be a homogeneous differential equation, if $F(x, y)$ is:
AnswerCorrect option: A. $\cos x-\sin \left(\frac{y}{x}\right)$
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→MCQ 1801 Mark
The order of the following differential equation $\frac{d^3 y}{d x^3}+x\left(\frac{d y}{d x}\right)^5=4 \log \left(\frac{d^4 y}{d x^4}\right)$ is:
View full question & answer→MCQ 1811 Mark
The degree of the differential equation $\left(y^{\prime \prime}\right)^2+\left(y^{\prime}\right)^3$ $=x \sin \left(y^{\prime}\right)$ is :
AnswerSince, the given differential equation is not a polynomial in $\frac{d y}{d x}$.
$\therefore \quad$ Its degree is not defined.
View full question & answer→MCQ 1821 Mark
The degree and order of differential equation $y^{\prime \prime 2}+\log \left(y^{\prime}\right)=x^5$ respectively are:
AnswerWe have, $y^{\prime \prime 2}+\log \left(y^{\prime}\right)=x^5$
As the highest order derivative is $y^{\prime \prime}$. So, order $=2$
But, the given differential equation is not a polynomial.
Therefore, its degree is not defined.
View full question & answer→MCQ 1831 Mark
The general solution of the differential equation $y d x-x d y=0$; (Given $x, y>0)$, is of the form
- A
$x y=c$
- B
$x=c y^2$
- C
$y=c x$
- D
$y=c x^2$
AnswerGiven, $y d x-x d y=0 \Rightarrow y d x=x d y \Rightarrow \frac{d y}{y}=\frac{d x}{x}$;
Integrating both sides, we get $\int \frac{d y}{y}=\int \frac{d x}{x}$
$\log _{ e }|y|=\log _{ e }|x|+\log _{ e }|c|$
Since $x, y, c>0$, we write $\log _{ c } y=\log _{ e } x+\log _{ e } c \Rightarrow y=c x$.
$(1 / 2)$
View full question & answer→MCQ 1841 Mark
Kapila is trying to find the general solution of the following differential equations.
(i) $x e^{\frac{x}{y}} d x-y e^{\frac{3 x}{y}} d y=0$
(ii) $(2 x+1) \frac{d y}{d x}=3-2 y$
(iii) $\frac{d y}{d x}=\sin x-\cos y$
Which of the above become variable separable by substituting $y=b . x$, where $b$ is a variable?
- ✓
- B
- C
all - (i), (ii) and (iii)
- D
View full question & answer→MCQ 1851 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→MCQ 1861 Mark
In which of the following differential equations is the degree equal to its order?
- A
$x^3\left(\frac{d y}{d x}\right)-\frac{d^3 y}{d x^3}=0$
- B
$\left(\frac{d^3 y}{d x^3}\right)^3+\sin \left(\frac{d y}{d x}\right)=0$
- ✓
$x^2\left(\frac{d y}{d x}\right)^4+\sin y-\left(\frac{d^2 y}{d x^2}\right)^2=0$
- D
$\left(\frac{d y}{d x}\right)^3+x\left(\frac{d^2 y}{d x^2}\right)-y^3\left(\frac{d^3 y}{d x^3}\right)+y=0$
AnswerCorrect option: C. $x^2\left(\frac{d y}{d x}\right)^4+\sin y-\left(\frac{d^2 y}{d x^2}\right)^2=0$
$x^2\left(\frac{d y}{d x}\right)^4+\sin y-\left(\frac{d^2 y}{d x^2}\right)^2=0$
View full question & answer→MCQ 1871 Mark
The general solution of the differential equation $x d y-\left(1+x^2\right) d x=d x$ is :
AnswerGiven differential equation is
$x d y-\left(1+x^2\right) d x=d x$
$\Rightarrow x d y=d x+\left(1+x^2\right) d x$
$=\left(2+x^2\right) d x$
$\Rightarrow \quad d y=\left(\frac{2}{x}+x\right) d x$
Integrating both sides, we get
$\int d y=\int\left(\frac{2}{x}+x\right) d x$
$\Rightarrow y=2 \log x+\frac{x^2}{2}+C$
View full question & answer→MCQ 1881 Mark
The general solution of the differential equation $x d y-\left(1+x^2\right) d x=d x$ is :
AnswerCorrect option: D. $y=2 \log x+\frac{x^2}{2}+C$
Given 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→MCQ 1891 Mark
The integrating factor of the differential equation $\left(3 x^2+y\right) \frac{d x}{d y}=x$ is
- ✓
$\frac{1}{x}$
- B
$\frac{1}{x^2}$
- C
$\frac{2}{x}$
- D
$-\frac{1}{x}$
AnswerCorrect option: A. $\frac{1}{x}$
Given, $\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→MCQ 1901 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, $y(1)=2$
$\Rightarrow C=\frac{1-1}{2+1}=0$
$\therefore$ Required solution is $x-1=0$
Hence, only one solution exist.
View full question & answer→MCQ 1911 Mark
The order and the degree of the differential equation $\left(1+3 \frac{d y}{d x}\right)^2=4 \frac{d^3 y}{d x^3}$ respectively are
AnswerWe have, $\left(1+3 \frac{d y}{d x}\right)^2=4 \frac{d^3 y}{d x^3}$
Here, order $=3$ as highest order derivative is $\frac{d^3 y}{d x^3}$.
And degree $=1$, as power of highest order derivative i.e., $\frac{d^3 y}{d x^3}$ is 1.
View full question & answer→MCQ 1921 Mark
The sum of the order and the degree of the differential equation $\frac{d}{d x}\left(\left(\frac{d y}{d x}\right)^3\right)$ is
Answer[There is error in question, the given differential equation should be $\frac{d}{d x}\left(\frac{d y}{d x}\right)^3=0$.]
The given differential equation is,
$
\begin{aligned}
& \frac{d}{d x}\left(\left(\frac{d y}{d x}\right)^3\right)=0 \Rightarrow 3\left(\frac{d y}{d x}\right)^2\left(\frac{d^2 y}{d x^2}\right)=0 \\
\therefore \quad & \text { Order }=2 \text { and degree }=1 . \text { So, required sum }=2+1=3
\end{aligned}
$
View full question & answer→MCQ 1931 Mark
The sum of the order and the degree of the differential equation $\frac{d^2 y}{d x^2}+\left(\frac{d y}{d x}\right)^3=\sin y$ is:
AnswerGiven differential equation is
$
\frac{d^2 y}{d x^2}+\left(\frac{d y}{d x}\right)^3=\sin y
$
Since, the highest order derivative is 2 and its power is 1 .
So, order $=2$
and degree $=1$
$\therefore \quad$ Required sum $=2+1=3$
View full question & answer→MCQ 1941 Mark
The difference of the order and the degree of the differential equation $\left(\frac{d^2 y}{d x^2}\right)^2+\left(\frac{d y}{d x}\right)^3+x^4=0$ is:
AnswerSince, $\left(\frac{d^2 y}{d x^2}\right)^2+\left(\frac{d y}{d x}\right)^3+x^4=0$
Order $=2$ and Degree $=2$
$\therefore \quad$ Required difference $=2-2=0$
View full question & answer→MCQ 1951 Mark
If $m$ and $n$, respectively, are the order and the degree of the differential equation $\frac{d}{d x}\left[\left(\frac{d y}{d x}\right)\right]^4=0$, then $m+n=$
AnswerThe given differential equation is $\frac{d}{d x}\left[\left(\frac{d y}{d x}\right)\right]^4=0$$
\Rightarrow 4\left(\frac{d y}{d x}\right)^3 \frac{d^2 y}{d x^2}=0
$Here, $m=2$ and $n=1$
Hence, $m+n=3$
View full question & answer→MCQ 1961 Mark
The integrating factor of the differential equation $\left(x+3 y^2\right) \frac{d y}{d x}=y$ is
- A
$y$
- B
$-y$
- C
$\frac{1}{ v }$
- D
$-\frac{1}{y}$
AnswerWe have, $\left(x+3 y^2\right) \frac{d y}{d x}=y$
$\Rightarrow \frac{x+3 y^2}{y}=\frac{d x}{d y} \Rightarrow \frac{d x}{d y}-\frac{x}{y}=3 y
$This is a linear differential equation.
$\therefore \quad \text { I.F. }=e^{-\int \frac{d y}{y}}=e^{-\log y}=e^{\log y^{-1}}=\frac{1}{y}$
View full question & answer→MCQ 1971 Mark
The number of arbitrary constants in the particular solution of a differential equation of second order is (are)
AnswerIn the particular solution of a differential equation of any order, there is no arbitrary constant because in the particular solution of any differential equation, we remove all the arbitrary constant by substituting some particular values.
View full question & answer→MCQ 1981 Mark
Integrating factor of the differential equation $\left(1-x^2\right) \frac{d y}{d x}-x y=1$ is
AnswerCorrect option: C. $\sqrt{1-x^2}$
(c): $\left(1-x^2\right) \frac{d y}{d x}-x y=1 \Rightarrow \frac{d y}{d x}-\frac{x}{1-x^2} \cdot y=\frac{1}{1-x^2}$
$\therefore \quad$ I.F. $=e^{-\int \frac{x}{1-x^2} d x}=e^{\frac{1}{2} \int \frac{-2 x}{1-x^2} d x}$
$=e^{\frac{1}{2} \log \left(1-x^2\right)}=e^{\log \left(1-x^2\right)^{\frac{1}{2}}}=\sqrt{1-x^2}$
View full question & answer→MCQ 1991 Mark
Order and degree of the differential equation $\left(1+\left(\frac{d y}{d x}\right)^3\right)^{\frac{7}{3}}=7 \frac{d^2 y}{d x^2}$ are respectively
Answer(b) : We have $\left(1+\left(\frac{d y}{d x}\right)^3\right)^{\frac{7}{3}}=7 \frac{d^2 y}{d x^2}$
$
\Rightarrow\left(1+\left(\frac{d y}{d x}\right)^3\right)^7=\left(7 \frac{d^2 y}{d x^2}\right)^3
$
$\therefore \quad$ Order is 2 and degree is 3 .
View full question & answer→MCQ 2001 Mark
If $\log _e\left(1+\frac{d^2 y}{d x^2}\right)=x$, then find the sum of order and degree of given differential equation.
Answer(c) : Given differential equation is $\log _e\left(1+\frac{d^2 y}{d x^2}\right)=x$
$\Rightarrow 1+\frac{d^2 y}{d x^2}=e^x$
Here, highest order derivative is $\frac{d^2 y}{d x^2}$, whose power is 1
$\therefore \quad$ Its order is 2 and degree is 1 .
$\therefore \quad$ Required sum $=2+1=3$.
View full question & answer→