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:
AnswerWe 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
- 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 R.H.S. integral we get
$2\text{x}\text{dx}=\text{dt}$
$\Rightarrow\int\text{e}^{\text{y}}\text{dy}=\int\text{e}^{\text{t}}\text{dt}$
$\Rightarrow\text{e}^{\text{y}}=\text{e}^{\text{t}}+\text{C}$
$\Rightarrow\text{e}^{\text{y}}=\text{e}^{\text{x}^2}+\text{C}$
View full question & answer→MCQ 1551 Mark
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, $\tan\text{y}\sec^2\text{xdx} + \tan\text{x }\sec^2\text{ydy}=0$
$\Rightarrow\tan\text{y}\sec^2\text{xdx} =- \tan\text{x }\sec^2\text{ydy}$
$\Rightarrow\frac{\sec^2\text{x}}{\tan\text{x}}\text{dx}=-\frac{\sec^2\text{y}}{\tan\text{y}}\text{dy}$
$\Rightarrow\int\frac{\sec^2\text{x}}{\tan\text{x}}\text{dx}=-\int\frac{\sec^2\text{y}}{\tan\text{y}}\text{dy}$
$\Rightarrow\log\tan\text{x}=-\log\tan\text{y}+\log\text{k}$
$\Rightarrow\log\tan\text{x}+\log\tan\text{y}=\log\text{k}$
$\Rightarrow\log(\tan\text{x}\tan\text{y})=\log\text{k}$
$\Rightarrow\tan\text{x}\tan\text{y}=\text{k}$
View full question & answer→MCQ 1561 Mark
The general solution of differention eqution $\frac{\text{y}\ \text{dx}-\text{x}\ \text{dy}}{\text{y}}=0$ is:
- A
$xy = C$
- B
$x = Cy^2$
- ✓
$y = Cx$
- D
$y = Cx^2$
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:
AnswerWe 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
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 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:
- A
- ✓
- C
- D
$\text{y}=\text{k}\log\text{x}$
AnswerWe 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 $\frac{\text{dy}}{\text{dx}}=\text{e}^{\text{x}+\text{y}}\text{is}$
- ✓
$\text{e}^{\text{x}}+\text{e}^{-\text{y}}=\text{C}$
- 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
$\frac{\text{dy}}{\text{dx}}=\text{e}^{\text{x+y}}\ \ \text{or}\ \ \frac{\text{dy}}{\text{dx}}=\text{e}^{\text{x}}.\text{e}^{\text{y}}$
Separating the variables, we get,
$\frac{1}{\text{e}^{\text{y}}}\text{dy}=\text{e}^\text{x}\ \text{dx}$
$\text{Integrating},\ \int\text{e}^{-\text{y}}\text{dy}=\int\text{e}^\text{x}\ \text{dx}$
$\therefore\ \frac{\text{e}^{-\text{y}}}{-1}=\text{e}^\text{x}+\text{c}'$
$\therefore\ -\text{e}^{-\text{y}}=\text{e}^\text{x}+\text{c}'\ \ \text{or}\ \ \text{e}^\text{x}+\text{e}^{-\text{y}}=-\text{c}'$
$\therefore\ \text{e}^{\text{x}}+\text{e}^{-\text{y}}=\text{C},$ 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,
$\text{I.F}=\text{e}^{\int\frac{\text{y}}{1-\text{y}^{2}}\text{dy}}$
$=\text{e}^{-\frac{1}{2}\int\frac{-2\text{y}}{1-\text{y}^{2}}\text{dy}}$
$=\text{e}^{-\frac{1}{2}\log|1-\text{y}^{2}|}$
$=\text{e}^{\log\Big|\frac{1}{\sqrt{1-\text{y}^{2}}}\Big|}$
$=\frac{1}{\sqrt{1-\text{y}^{2}}}$
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