MCQ 511 Mark
What is the general solution of the differential equation $x\ dy - y\ dx\ y^2 ?$
- A
$x = cy$
- B
$y^2 = cx$
- C
$x + xy - cy = 0$
- ✓
Answer$xdy - y^3dx = 0$ rearranging
$\frac{\text{dx}}{\text{x}}=\frac{\text{dy}}{\text{y}^3}$
integrating both sides ln
$\text{x}=\frac{\text{y}^{-2}}{-2}+\text{c}$
View full question & answer→MCQ 521 Mark
Choose the correct answer from the given four option.
The number of solutions of $\frac{\text{d}\text{y}}{\text{d}\text{x}}=\frac{\text{y+1}}{\text{x}-1}$ when y(1) = 2:
AnswerGiven that, $\frac{\text{d}\text{y}}{\text{d}\text{x}}=\frac{\text{y+1}}{\text{x}-1}$
$\Rightarrow\frac{\text{d}\text{y}}{\text{y+1}}=\frac{\text{dx}}{\text{x}-1}$
On integrating both sides, we get
$\int\frac{\text{d}\text{y}}{\text{y+1}}=\int\frac{\text{dx}}{\text{x}-1}$
$\log(\text{y}+1)=\log(\text{x}-1)-\log\text{C}$
$\text{C}(\text{y}+1)=(\text{x}-1)$
$\Rightarrow\text{C}=\frac{\text{x}-1}{\text{y}+1}$
When x = 1 and y = 2, then C = 0
So, the required solution is x - 1 = 0
Hence, only one solution exists.
View full question & answer→MCQ 531 Mark
The order and degree of the differential equation$, \frac{\text{d}^2\text{y}}{\text{dx}^2}+\Big(\frac{\text{dy}}{\text{dx}}\Big)^{\frac{1}{4}}+\text{x}^{\frac{1}{5}}=0$ respectively are:
- ✓
$2$ and not defined
- B
$2$ and $2$
- C
$2$ and $3$
- D
$3$ and $3$
AnswerCorrect option: A. $2$ and not defined
View full question & answer→MCQ 541 Mark
Choose the correct answer from the given four options.The solution of the equation $(2\text{y}-1)\text{dx}-(2\text{x}+3)\text{dy}=0$ is:
- A
$\frac{2\text{x}-1}{2\text{y}+3}=\text{k}$
- B
$\frac{2\text{y}+1}{2\text{x}-3}=\text{k}$
- ✓
$\frac{2\text{x}+3}{2\text{y}-1}=\text{k}$
- D
$\frac{2\text{x}-1}{2\text{y}-1}=\text{k}$
AnswerCorrect option: C. $\frac{2\text{x}+3}{2\text{y}-1}=\text{k}$
Given is, $(2\text{y}-1)\text{dx}-(2\text{x}+3)\text{dy}=0$
$\Rightarrow(2\text{y}-1)\text{dx}=(2\text{x}+3)\text{dy}$
$\Rightarrow\frac{\text{dx}}{2\text{x}+3}=\frac{\text{dy}}{2\text{y}-1}$
On integrating both sides, we get
$\frac{1}{2}\log(2\text{x}+3)=\frac{1}{2}\log(2\text{y}-1)+\log\text{C}$
$\Rightarrow\frac{1}{2}\log(2\text{x}+3)-\log(2\text{y}-1)=\log\text{C}$
$\Rightarrow\frac{1}{2}\log\Big(\frac{2\text{x}+3}{2\text{y}-1}\Big)=\text{C}^2$
$\Rightarrow\Big(\frac{2\text{x}+3}{2\text{y}-1}\Big)^{\frac{1}{2}}=\text{C}$
$\Rightarrow\frac{2\text{x}+3}{2\text{y}-1}=\text{C}^2$
$\Rightarrow\frac{2\text{x}+3}{2\text{y}-1}=\text{C}^2$
$\Rightarrow\frac{2\text{x}+3}{2\text{y}-1}=\text{k},\text{Where}\ \text{k}=\text{c}^2$
View full question & answer→MCQ 551 Mark
Choose the correct answer from the given four options.General solution of $\frac{\text{dy}}{\text{dx}}+\text{y}\tan\text{x}=\sec\text{x}$ is:
- ✓
$\text{y}\sec\text{x}=\tan\text{x}+\text{c}$
- B
$\text{y}\tan\text{x}=\sec\text{x}+\text{c}$
- C
$\tan\text{x}=\sec\text{x}+\text{c}$
- D
$\text{x}\sec\text{x}=\tan\text{y}+\text{c}$
AnswerCorrect option: A. $\text{y}\sec\text{x}=\tan\text{x}+\text{c}$
Given differential equation is
$\frac{\text{dy}}{\text{dx}}+\text{y}\tan\text{x}=\sec\text{x}$
This is a linear differential equation
Here, $\text{P}=\tan\text{x},\text{Q}=\sec\text{x},$
$\therefore\text{I.F.}=\text{e}^{\int\tan\text{xdx}}$
$=\text{e}^{\log|\sec\text{x}|}=\sec\text{x}$
Thus, the general solution is
$\text{y}.\sec\text{x}=\int\sec\text{x}.\sec\text{x}+\text{C}$
$\Rightarrow\text{y}.\sec\text{x}=\int\sec^2\text{x}\text{dx}+\text{C}$
$\Rightarrow\text{y}.\sec\text{x}=\tan\text{x}+\text{C}$
View full question & answer→MCQ 561 Mark
What is the solution of the differential equation $\frac{\text{dy}}{\text{dx}}+\frac{\text{y}}{\text{x}}=0?$ Where c is a constant.
AnswerGivenexpression is
$\frac{\text{dy}}{\text{dx}}+\frac{\text{y}}{\text{x}}=0$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=-\frac{\text{y}}{\text{x}}$
$\Rightarrow\frac{\text{dy}}{\text{y}}=-\frac{\text{dx}}{\text{x}}=0$
Integrating we get
$\int\frac{\text{dy}}{\text{y}}+\frac{\text{dx}}{\text{x}}=0$
$\Rightarrow\text{In}\text{ y}+\text{In}\text{ x}=\text{c}$
$\Rightarrow\text{x}{\text{y}}=\text{c}$
View full question & answer→MCQ 571 Mark
The general solution of the differential equation $\frac{\text{dy}}{\text{dx}}+\text{y}\ \text{g}(\text{x})=\text{g}(\text{x})\ \text{g}'(\text{x})$ is a given function of x, is:
- A
$\text{g}(\text{x})+\log(1+\text{y}+\text{g}(\text{x}))=\text{C}$
- ✓
$\text{g}(\text{x})+\log(1+\text{y}-\text{g}(\text{x}))=\text{C}$
- C
$\text{g}(\text{x})-\log(1+\text{y}-\text{g}(\text{x}))=\text{C}$
- D
AnswerCorrect option: B. $\text{g}(\text{x})+\log(1+\text{y}-\text{g}(\text{x}))=\text{C}$
We have,
$\frac{\text{dy}}{\text{dx}}+\text{y}\ \text{g}'(\text{x})=\text{g}(\text{x})\ \text{g}'(\text{x})\ ...(\text{i})$
Clearly, it is a linear differential equation of the form
$\frac{\text{dy}}{\text{dx}}+\text{Py}=\text{Q}$
Where $\text{P}=\text{g}'(\text{x}), \text{Q}=\text{g}(\text{x})\ \text{g}'(\text{x})$
$\text{I.F}=\text{e}^{\int\text{P}\text{dx}}$
$=\text{e}^{\int\text{g}'(\text{x})\text{dx}}$
$=\text{e}^{\text{g}(\text{x})}$
Multiplying both sides, we get
$\text{e}^{\text{g}(\text{x})}\Big(\frac{\text{dy}}{\text{dx}}+\text{yg}'({\text{x}})\Big)=\text{e}^{\text{g}(\text{x})}\text{g}(\text{x})\ \text{g}'(\text{x})$
$\text{e}^{\text{g}(\text{x})}\frac{\text{dy}}{\text{dx}}+\text{e}^{\text{g}(\text{x})}\text{yg}'({\text{x}})=\text{e}^{\text{g}(\text{x})}\text{g}(\text{x})\ \text{g}'(\text{x})$
Integrating both sides with respect to x, we get
$\text{y}\ \text{e}^{\text{g}(\text{x})}=\text{e}^{\text{g}(\text{x})}\text{g}(\text{x})\ \text{g}'(\text{x})\ \text{dx}+\text{K}$
$\text{y}\ \text{e}^{\text{g}(\text{x})}=\text{I}+\text{K}$
Where, $\text{I}=\text{e}^{\text{g}(\text{x})}\text{g}(\text{x})\ \text{g}'(\text{x})\ \text{dx}$
Now,
$\text{I}=\text{e}^{\text{g}(\text{x})}\text{g}(\text{x})\ \text{g}'(\text{x})\ \text{dx}$
Putting $\text{g}'(\text{dx})=\text{dt}$
$\text{I}=\int\text{t}\ \text{e}^{\text{t}}\ \text{dt}$
$=\text{t}\int\ \text{e}^{\text{t}}\ \text{dt}-\int\Big[\frac{\text{d}}{\text{dx}}(\text{t})\int\text{e}^{\text{t}}\ \text{dt}\Big]\text{dt}$
$=\text{t}\text{e}^{\text{t}}-\text{e}^{\text{t}}$
$=\text{g}(\text{x})\text{e}^{\text{g}(\text{x})}-\text{e}^{\text{g}(\text{x})}$
$\Rightarrow \text{y}\ \text{e}^{\text{g}(\text{x})}=\text{g}(\text{x})\text{e}^{\text{g}(\text{x})}-\text{e}^{\text{g}(\text{x})}+\text{K}$
$\Rightarrow \text{y}\ \text{e}^{\text{g}(\text{x})}=\text{g}(\text{x})\text{e}^{\text{g}(\text{x})}-\text{e}^{\text{g}(\text{x})}=\text{K}$
Taking log on both sides, we get
$\log\Big[\text{y+1}-\text{g}(\text{x})\Big]=-\text{g}(\text{x})+\log\text{K}$
View full question & answer→MCQ 581 Mark
Choose the correct answer from the given four option. Family $y = Ax + A^3$ of curves is represented by the differential equation of degree:
AnswerGiven is, $y = Ax + A^3$
Differentiating both sides $w.r.t. x$, we get
$\frac{\text{d}\text{y}}{\text{d}\text{x}}=\text{A}$
This equation can be differentiated only once because it has only one arbitrary constant.
$\therefore\text{Degree}=1$
View full question & answer→MCQ 591 Mark
The famliy of curve in which the sub tangent at any point of a curve is double is the abscissae, is
AnswerIt is given that subtangent at any point of a curve is doble of the abscissa.
$\therefore \frac{\text{y}}{\frac{\text{dy}}{\text{dx}}}=2\text{x}$
$\text{y}=2\text{x}\frac{\text{dy}}{\text{dx}}$
$\int\frac{\text{dx}}{\text{x}}=2\int\frac{\text{dy}}{\text{y}}$
$\log\text{x}=2\log\text{y}+\text{a}$
$\log\text{x}=\log\text{y}^{2}+\log\text{C}$
$\log\text{x}=\log\text{Cy}^{2}$
$\text{x}=\text{Cy}^{2}$
View full question & answer→MCQ 601 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 order n is always 0.So here it will be 0
View full question & answer→MCQ 611 Mark
The general solution of a differential equation of the type $\frac{\text{dx}}{\text{dy}}+\text{P}_1\text{x}=\text{Q}_1\ \text{is}$
- A
$\text{y e}^{\int\text{P}_1\text{dy}}=\int\Big(\text{Q}_1\text{e}^{\int\text{P}_1\text{dy}}\Big)\text{dy}+\text{C}$
- B
$\text{y.e}^{\int\text{P}_1\text{dx}}=\int\Big(\text{Q}_1\text{e}^{\int\text{P}_1\text{dx}}\Big)\text{dx}+\text{C}$
- ✓
$\text{x e}^{\int\text{P}_1\text{dy}}=\int\Big(\text{Q}_1\text{e}^{\int\text{P}_1\text{dy}}\Big)\text{dy}+\text{C}$
- D
$\text{x e}^{\int\text{P}_1\text{dx}}=\int\Big(\text{Q}_1\text{e}^{\int\text{P}_1\text{dx}}\Big)\text{dx}+\text{C}$
AnswerCorrect option: C. $\text{x e}^{\int\text{P}_1\text{dy}}=\int\Big(\text{Q}_1\text{e}^{\int\text{P}_1\text{dy}}\Big)\text{dy}+\text{C}$
The given differential equation is $\frac{\text{dx}}{\text{dy}}+\text{P}_1\ \text{x}=\text{Q}_1$
$\text{Multiplying through by }\ \text{e}^{\int\text{P}_1\text{dy}},\ \text{we get},$
$\frac{\text{dx}}{\text{dy}}\text{e}^{\int\text{P}_1\text{dy}}+\text{P}_1\ \text{x e}^{\int\text{P}_1\text{dy}}=\text{Q}_1\ \text{e}^{\int\text{P}_1}\text{dy}$
$\text{or}\ \ \frac{\text{d}}{\text{dy}}(\text{x e}^{\int\text{P}_1\text{dy}})=\text{Q}_1\ \text{e}^{\int\text{P}_1}\text{dy}$
$\begin{bmatrix}\because\frac{\text{d}}{\text{dy}}\ (\text{x e}^{\int\text{P}_1\text{dy}})=\text{x e}^{\int\text{P}_1\text{dy}}\frac{\text{d}}{\text{dy}}(\int\text{P}_1\text{dy})+\text{e}^{\int\text{P}_1\text{dy}}\frac{\text{dx}}{\text{dy}} \\=\text{x e}^{\int\text{P}_1\text{dy}}\ \text{P}_1+\text{e}^{\int\text{P}_1\text{dy}}\ \frac{\text{dx}}{\text{dy}} & \\=\frac{\text{dx}}{\text{dy}}\text{e}^{\int\text{P}_1\text{dy}}+\text{P}_1 \ \text{x e}^{\int\text{P}_1\text{dy}}\end{bmatrix}$
Integrating both sides w.r.t.y, we get,
$\text{xe}^{\int\text{P}_1\text{dy}}=\int\text{Q}_ 1\text{e}^{\int\text{P}_1\text{dy}}\text{dy}+\text{C}\ \text{}$ which is required solution.
$\therefore$ (C) is correct answer.
The given differential equation is $\frac{\text{dx}}{\text{dy}}+\text{P}_1\text{x}=\text{Q}_1$
$\text{Multiplying through by}\ \text{e}^{\int\text{P}_1\text{dy}},\ \text{we get},$
$\frac{\text{dx}}{\text{dy}}\text{e}^{\int\text{P}_1\text{dy}}+\text{P}_1\text{x e}^{\int\text{P}_1\text{dy}}=\text{Q}_1\ \text{e}^{\int\text{P}_1}\text{dy}$
$\text{or}\ \ \frac{\text{d}}{\text{dy}}(\text{x e}^{\int\text{P}_1\text{dy}})=\text{Q}_1\ \text{e}^{\int\text{P}_1}\text{dy}$
$\begin{bmatrix}\because\frac{\text{d}}{\text{dy}}\ (\text{x e}^{\int\text{P}_1\text{dy}})=\text{x e}^{\int\text{P}_1\text{dy.}}\frac{\text{d}}{\text{dy}}(\int\text{P}_1\text{dy})+\text{e}^{\int\text{P}_1\text{dy}}\frac{\text{dx}}{\text{dy}} \\=\text{x e}^{\int\text{P}_1\text{dy}}\ \text{P}_1+\text{e}^{\int\text{P}_1\text{dy}}\ \frac{\text{dx}}{\text{dy}} & \\=\frac{\text{dx}}{\text{dy}}\text{e}^{\int\text{P}_1\text{dy}}+\text{P}_1 \ \text{xe}^{\int\text{P}_1\text{dy}}\end{bmatrix}$
Integrating both sides w.r.t.y, we get,
$\text{x e}^{\int\text{P}_1\text{dy}}=\int\text{Q}_1\ \text{e}^{\int\text{P}_1\text{dy}}\text{dy}+\text{C}$ which is required solution.
View full question & answer→MCQ 621 Mark
The solution of the differential equation $\frac{\text{dy}}{\text{dx}}=1+\text{x}+\text{y}^{2}+\text{xy}^{2}, \text{y}=(0)$ is:
AnswerCorrect option: D. $\text{y}=\tan\Big(\text{x}+\frac{\text{x}^{2}}{2}\Big)$
We have,
$\frac{\text{dy}}{\text{dx}}=1+\text{x}+\text{y}^{2}+\text{xy}^{2}$
$\Rightarrow \frac{\text{dy}}{\text{dx}}=(\text{x}+1)\text{y}^{2}(\text{x}+1)$
$\Rightarrow \frac{\text{dy}}{\text{dx}}=(\text{x}+1)(1+\text{y})$
$\Rightarrow \frac{\text{dy}}{(1+\text{y}^{2})}=(\text{x}+1)\text{dx}$
Integrating both sides, we get
$\int \frac{\text{dy}}{(1+\text{y}^{2})}=\int(\text{x}+1)\text{dx}$
$\Rightarrow \tan^{-1}=\frac{\text{x}^{2}}{2}+\text{x}+\text{C}\ ...(\text{i})$
Now, y(0) = 0
$\therefore\ \tan^{-1}(0)=\frac{\text{0}}{2}+\text{0}+\text{C}$
$\Rightarrow \text{C}=0$
Putting the value of C in (i),
$\Rightarrow \tan^{-1}\text{y}=\frac{\text{x}^{2}}{2}+\text{x}$
$\Rightarrow \text{y}=\tan\big(\frac{\text{x}^{2}}{2}+\text{x}\big)$
View full question & answer→MCQ 631 Mark
Which of the following is the integrating factor of $(\text{x}\log\text{x})\frac{\text{dy}}{\text{dx}}+\text{y}=2\log\text{x}?$
AnswerWe have,
$(\text{x}\log\text{x})\frac{\text{dy}}{\text{dx}}+\text{y}=2\log\text{x}$
Dividing both sides by, we get
$ \frac{\text{dy}}{\text{dx}}+\frac{\text{y}}{\text{x}\log\text{x}}=2\frac{\log\text{x}}{\text{x}\log\text{x}}$
$\Rightarrow \frac{\text{dy}}{\text{dx}}+\frac{\text{y}}{\text{x}\log\text{x}}=\frac{2}{\text{x}}$
$\Rightarrow \frac{\text{dy}}{\text{dx}}+\big(\frac{\text{1}}{\text{x}\log\text{x}}\big)\text{y}=\frac{2}{\text{x}}$
Comparing with we get,
$\text{P}=\frac{1}{\text{x}\log\text{x}}, \text{Q}=\frac{2}{\text{x}}$
$\text{I.F}=\text{e}^{\int\frac{1}{\text{x}\log\text{x}}\text{dx}}$
$=\text{e}^{\log({\log\text{x})}}$
$=\log\text{x}$
View full question & answer→MCQ 641 Mark
Choose the correct answer from the given four option.
Integrating factor of the differential equation $\cos\frac{\text{d}\text{y}}{\text{d}\text{x}}+\text{y}\sin\text{x}=1$ is:
- A
$\cos\text{x}$
- B
$\tan\text{x}$
- ✓
$\sec\text{x}$
- D
$\sin\text{x}$
AnswerCorrect option: C. $\sec\text{x}$
we have, $\cos\frac{\text{d}\text{y}}{\text{d}\text{x}}+\text{y}\sin\text{x}=1$
$\Rightarrow\frac{\text{d}\text{y}}{\text{d}\text{x}}+\text{y}\tan\text{x}=\sec\text{x}$
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
$\text{P}=\tan\text{x}$ and $\text{Q}=\sec\text{x}$
$\text{I.F.}=\text{e}^{\int\text{Pdx}}=\text{e}^{\int\tan\text{xdx}}$
$=\text{e}^{\log\sec\text{x}}=\sec\text{x}$
View full question & answer→MCQ 651 Mark
If $\text{y}=\text{e}^{-\text{x}}(\text{A}\cos\text{x}+\text{B}\sin\text{x}),$ then $y$ is a solution of:
- A
$\frac{\text{d}^2\text{y}}{\text{dx}^2}+2\frac{\text{dy}}{\text{dx}}=0$
- B
$\frac{\text{d}^2\text{y}}{\text{dx}^2}-2\frac{\text{dy}}{\text{dx}}+2\text{y}=0$
- ✓
$\frac{\text{d}^2\text{y}}{\text{dx}^2}+2\frac{\text{dy}}{\text{dx}}+2\text{y}=0$
- D
$\frac{\text{d}^2\text{y}}{\text{dx}^2}+2\text{y}=0$
AnswerCorrect option: C. $\frac{\text{d}^2\text{y}}{\text{dx}^2}+2\frac{\text{dy}}{\text{dx}}+2\text{y}=0$
View full question & answer→MCQ 661 Mark
The Integrating Factor of the differential equation $\text{x}\frac{\text{dy}}{\text{dx}}-\text{y}=2\text{x}^2\ \text{is}$
- A
$\text{e}^{-\text{x}}$
- B
$\text{e}^{-\text{y}}$
- ✓
$\frac{1}{\text{x}}$
- D
$\text{x}$
AnswerCorrect option: C. $\frac{1}{\text{x}}$
The given differential equation is:
$\text{x}\frac{\text{dy}}{\text{dx}}-\text{y}=2\text{x}^2$
$\Rightarrow\frac{\text{dy}}{\text{dx}}-\frac{\text{y}}{\text{x}}=2\text{x}$
This is a linear differential equation of the form:
$\frac{\text{dy}}{\text{dx}}+\text{py}=\text{Q}\ (\text{where p}=-\frac{1}{\text{x}}\ \text{and}\ \text{Q}=2\text{x})$
The integrating factor (I.F) is given by the relation,
$\text{e}^{\int\text{pdx}}$
$\therefore\text{I.F}=\text{e}^{\int-\frac{1}{\text{x}}\text{dx}}=\text{e}^{-\log\text{x}}=\text{e}^{\log(\text{x}^-1)}=\text{x}^{-1}=\frac{1}{\text{x}}$
View full question & answer→MCQ 671 Mark
Choose the correct answer from the given four options.Which of the following is the general solution of $\frac{\text{d}^2\text{y}}{\text{d}\text{x}^2}-2\frac{\text{d}\text{y}}{\text{d}\text{x}}+\text{y}=0?$
- ✓
$\text{y}=(\text{Ax}+\text{B})\text{e}^{\text{x}}$
- B
$\text{y}=(\text{Ax}+\text{B})\text{e}^{-\text{x}}$
- C
$\text{y}=\text{Ax}\text{e}^{\text{x}}+\text{B}\text{e}^{\text{x}}$
- D
$\text{y}=\text{A}\cos\text{x}+\text{B}\sin\text{x}$
AnswerCorrect option: A. $\text{y}=(\text{Ax}+\text{B})\text{e}^{\text{x}}$
Given that, $\frac{\text{d}^2\text{y}}{\text{d}\text{x}^2}-2\frac{\text{d}\text{y}}{\text{d}\text{x}}+\text{y}=0$
$\text{D}^2\text{y}-2\text{D}\text{y}+\text{y}=0,$
Where, $\text{D}=\frac{\text{d}}{\text{dx}}$
$\text{D}^2\text{y}-2\text{D}\text{y}+\text{y}=0$
The auxiliary equation is $\text{m}^2-2\text{m}+1=0$
$(\text{m}-1)^2=0$
$\Rightarrow\text{m}=1,1$
Since, the roots are real and equal.
$\therefore\text{CF}=(\text{Ax}+\text{B})\text{e}^{\text{x}}$
$\Rightarrow\text{y}=(\text{Ax}+\text{B})\text{e}^{\text{x}}$
$\big[$ Since, if roots of Auxiliary equation are real and equal say (m), then $\text{CF}=(\text{C}_1\text{x}+\text{C}_2)\text{e}^{\text{mx}}\big]$
View full question & answer→MCQ 681 Mark
Degree and order of the differential equation $\frac{\text{d}^2\text{y}}{\text{dx}^2}=\Big(\frac{\text{dy}}{\text{dx}}\Big)^2$ are respectively:
AnswerThe given differential equation is
$\frac{\text{d}^2\text{y}}{\text{dx}^2}=\Big(\frac{\text{dy}}{\text{dx}}\Big)^2$
The degree and order of this equation are 1 and 2 respectively.
View full question & answer→MCQ 691 Mark
The order of the differential equation of the parabola whose axis is parallel to $y-$axis is:
AnswerEquation of parabola $:-(x - h)^2 = 4a (y - k)$
No of constants in the alove equation are $3$
$\therefore$ order of diff $eq = 3$
View full question & answer→MCQ 701 Mark
The differential equation $\frac{\text{d}^2\text{y}}{\text{dx}^2}+\frac{\text{dy}}{\text{dx}}+\text{y}+\text{x}^2=0$ of the following type:
AnswerOrder of highest order derivative is 2 and
power of highest order derivative is 1
Hence order is 2 and degree is 1
View full question & answer→MCQ 711 Mark
The solution of the differention equation $\frac{\text{dy}}{\text{dx}}=\frac{\text{x}^{2}+\text{xy}+\text{y}^{2}}{\text{x}^{2}}$ is:
- A
$\tan^{-1}\big(\frac{\text{x}}{\text{y}}\big)-\log\text{y}+\text{C}$
- ✓
$\tan^{-1}\big(\frac{\text{y}}{\text{x}}\big)-\log\text{x}+\text{C}$
- C
$\tan^{-1}\big(\frac{\text{x}}{\text{y}}\big)=\log\text{x}+\text{C}$
- D
$\tan^{-1}\big(\frac{\text{y}}{\text{x}}\big)=\log\text{y}+\text{C}$
AnswerCorrect option: B. $\tan^{-1}\big(\frac{\text{y}}{\text{x}}\big)-\log\text{x}+\text{C}$
We have,
$\frac{\text{dy}}{\text{dx}}=\frac{\text{x}^{2}+\text{xy}+\text{y}^{2}}{\text{x}^{2}}\ ...(\text{i})$
This is homogenous differential equation.
Let $\text{y}=\text{ux}$
$\Rightarrow \frac{\text{dy}}{\text{dx}}=\text{u}+\text{x}\frac{\text{du}}{\text{dx}}$
Now, putting equation (i),
$\text{u}+\text{x}\frac{\text{du}}{\text{dx}}=\frac{\text{x}^{2}+\text{x}^{2}\text{u}+\text{x}^{2}\text{u}^{2}}{\text{x}^{2}}$
$\Rightarrow \text{u}+\text{x}\frac{\text{du}}{\text{dx}}=1+\text{u}+\text{u}^{2}$
$\Rightarrow \text{x}\frac{\text{du}}{\text{dx}}=1+\text{u}^{2}$
$\Rightarrow \big(\frac{1}{1+\text{u}^{2}}\big)\text{du}=\frac{1}{\text{x}}\text{dx}$
Intergreting both sides, we get
$\int\big(\frac{1}{1+\text{u}^{2}}\big)\text{du}=\int\frac{1}{\text{x}}\text{dx}$
$\Rightarrow \tan^{-1}\text{u}=\log\text{x}+\text{C}$
$\Rightarrow \tan^{-1}\big(\frac{\text{y}}{2}\big)=\log\text{x}+\text{C}$
View full question & answer→MCQ 721 Mark
Which of the following differential equations 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
$\frac{\text{d}^2\text{y}}{\text{dx}^2}+1=0$
- D
$\frac{\text{d}^2\text{y}}{\text{dx}^2}-1=0$
AnswerCorrect option: B. $\frac{\text{d}^2\text{y}}{\text{dx}^2}-\text{y}=0$
The given equation is:
$\text{y} = \text{c}_1\text{e}^\text{x} + \text{c}_2\text{e}^{-\text{x}} \ \ ...(1)$
Differentiating with respect to x, we get:
$\frac{\text{dy}}{\text{dx}} = \text{c}_{1}\text{e}^\text{x}-\text{c}_{2}\text{e}^{-\text{x}}$
Again, differentiating with respect to x, we get:
$\frac{\text{d}^2\text{y}}{\text{dx}^2} = \text{c}_{1}\text{e}^\text{x}+\text{c}_2\text{e}^{-\text{x}}$
$\Rightarrow\frac{\text{d}^2\text{y}}{\text{dx}^2} =\text{y}$
$\Rightarrow\frac{\text{d}^2\text{y}}{\text{dx}^2}-\text{y}=0$
This is the required differential equation of the given equation of curve.
Hence, the correct answer is B.
View full question & answer→MCQ 731 Mark
What is the order of the differential equation:
$\Big(\frac{\text{dy}}{\text{dx}}\Big)^2+\frac{\text{dy}}{\text{dx}}-\sin^2\text{y}=0.$
AnswerThe number of the highest derivative in a differential equation is called order.So, for
$=\Big(\frac{\text{dy}}{\text{dx}}\Big)^2+\frac{\text{dy}}{\text{dx}}-\sin^2\text{y}=0,$ the order is 1.
View full question & answer→MCQ 741 Mark
The curve for which the slope of the tangent at any point is equal to the ratio of the abscissa to the ordinate of the point is:
MCQ 751 Mark
The degree of the differential equation $\frac{\text{d}^3\text{y}}{\text{dx}^3}+3\frac{\text{d}^2\text{y}}{\text{dx}^2}=\text{x}^2\log\frac{\text{d}^3\text{y}}{\text{dx}^3}$ is:
AnswerDegree of given differential equation is not defined.
View full question & answer→MCQ 761 Mark
Choose the correct answer from the given four option.
The degree of the differential equation $\Big(\frac{\text{d}^2\text{y}}{\text{d}\text{x}^2}\Big)+\Big(\frac{\text{d}\text{y}}{\text{d}\text{x}}\Big)^2-\sin\Big(\frac{\text{d}\text{y}}{\text{d}\text{x}}\Big)$ is:
AnswerThe degree of above differential equation is not defined because on solving $\sin\Big(\frac{\text{dy}}{\text{dx}}\Big)$ we will get an infinite series in the increasing powers of $\frac{\text{dy}}{\text{dx}}.$ Therefore its degree is not defined.
View full question & answer→MCQ 771 Mark
Choose the correct answer from the given four option.
Which of the following is a second order differential equation?
- A
$(\text{y}')^2+\text{x}=\text{y}^2$
- ✓
$\text{y}'\text{y}''+\text{y}=\sin\text{x}$
- C
$\text{y}'''+(\text{y}'')^2+\text{y}=0$
- D
$\text{y}'=\text{y}^2$
AnswerCorrect option: B. $\text{y}'\text{y}''+\text{y}=\sin\text{x}$
The second order differential equation is $\text{y}'\text{y}''+\text{y}=\sin\text{x}.$
View full question & answer→MCQ 781 Mark
What is the general solution of the differential equation $x^2 dy + y^2 dx = 0?$
- A
$x + y = c$ where $c$ is the constant of integration
- B
$xy = c$ where $c$ is the constant of integration
- ✓
$c(x + y) = xy$ where $c$ is the constant of integration
- D
AnswerCorrect option: C. $c(x + y) = xy$ where $c$ is the constant of integration
$=\text{x}^2\text{dy}+\text{y}^2\text{dx}=0$
$\Rightarrow\frac{\text{dy}}{\text{y}^2}+\frac{\text{dx}}{\text{x}^2}=0$
On integrating, we get
$\Rightarrow -\frac { 1 }{ \text{y} } -\frac { 1 }{ \text{x} } +\frac { 1 }{ \text{c} } =0$
$ \Rightarrow \text{c}(\text{x}+\text{y})=\text{x}\text{y}$
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The general solution of the differential equation $\frac{\text{y}\ \text{dx}-\text{x}\ \text{dx}}{\text{y}}=0\ \text{is}$
- A
$\text{xy}=\text{C}$
- B
$\text{x}=\text{Cy}^2$
- ✓
$\text{y}=\text{Cx}$
- D
$\text{y}=\text{Cx}^2$
AnswerCorrect option: C. $\text{y}=\text{Cx}$
The given differential equation is
$\frac{\text{y dx}-\text{x dy}}{\text{y}}=0$
$\text{or}\ \ \frac{\text{y dx}-\text{x dy}}{\text{x}^2}=0\ \ \text{or}\ \ \text{d}\Big(\frac{\text{x}}{\text{y}}\Big)=0$
$\therefore\ \ \frac{\text{x}}{\text{y}}=\text{constant}.$
$\therefore\ \ \frac{\text{x}}{\text{y}}=\text{C}\ \text{or}\ \text{y}=\text{Cx}$
$\therefore\ \text{(C)}\ \text{is correct answer}.$
The given differential equation is
$\frac{\text{y dx}-\text{x dy}}{\text{y}}=0$
$\text{or}\ \ \frac{\text{y dx}-\text{x dy}}{\text{x}^2}=0\ \ \text{or}\ \ \text{d}\Big(\frac{\text{x}}{\text{y}}\Big)=0$
$\therefore\ \ \frac{\text{x}}{\text{y}}=\text{constant}.$
$\therefore\ \ \frac{\text{y}}{\text{x}}=\text{C}\ \ \text{or}\ \ \text{y}=\text{Cx}$
View full question & answer→MCQ 801 Mark
Choose the correct answer from the given four options.The order and degree of the differential equation $\Big(\frac{\text{d}^3\text{y}}{\text{d}\text{x}^3}\Big)^2-3\frac{\text{d}^2\text{y}}{\text{d}\text{x}^2}+2\Big(\frac{\text{d}\text{y}}{\text{d}\text{x}}\Big)^4=\text{y}^4$ are:
AnswerGiven that $\Big(\frac{\text{d}^3\text{y}}{\text{d}\text{x}^3}\Big)^2-3\frac{\text{d}^2\text{y}}{\text{d}\text{x}^2}+2\Big(\frac{\text{d}\text{y}}{\text{d}\text{x}}\Big)^4=\text{y}^4$
$\therefore\text{Order}=3\text{ and degree}=2$
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The solution of $\frac{\text{d}^2\text{y}}{\text{dx}^2}=0$ represents:
Answer$\Rightarrow\frac{\text{d}}{\text{dx}}\text{y}_1=0$
$\Rightarrow\int\text{dy}_1=\int0.\text{dx}+\text{c}_1$
$\Rightarrow\int\text{dy}_1=\text{c}_1$
$\Rightarrow\text{y}_1=\text{c}_1$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\text{c}_1$
$\Rightarrow\int\text{dy}=\int\text{c}_1\text{ dx}+\text{c}_2$
$\Rightarrow\text{y}=\text{c}_1\text{x}+\text{c}_2\text{ straight line}$
View full question & answer→MCQ 821 Mark
The solution of the differential equartion $y_1y_3 = y_2$ is:
- A
$\text{x}=\text{C}_{1}\text{e}^{\text{C}_{2}\text{y}}+\text{C}_{3}$
- ✓
$\text{y}=\text{C}_{1}\text{e}^{\text{C}_{2}\text{x}}+\text{C}_{3}$
- C
$2\text{x}=\text{C}_{1}\text{e}^{\text{C}_{2}\text{y}}+\text{C}_{3}$
- D
AnswerCorrect option: B. $\text{y}=\text{C}_{1}\text{e}^{\text{C}_{2}\text{x}}+\text{C}_{3}$
$\text{y}_{1}\text{y}_{3}=\text{y}^{2}_{2}$
$\frac{\text{y}_{3}}{\text{y}_{2}}=\frac{\text{y}_{2}}{\text{y}_{1}}$
$\Rightarrow \frac{\Big(\frac{\text{d}^{3}\text{y}}{\text{dx}^{3}}\Big)}{\Big(\frac{\text{d}^{2}\text{y}}{\text{dx}^{2}}\Big)}=\frac{\Big(\frac{\text{d}^{2}\text{y}}{\text{dx}^{2}}\Big)}{\Big(\frac{\text{d}\text{y}}{\text{dx}}\Big)}$
$\Rightarrow\int \frac{\frac{\text{d}}{\text{dx}}\Big(\frac{\text{d}^{3}\text{y}}{\text{dx}^{3}}\Big)}{\Big(\frac{\text{d}^{2}\text{y}}{\text{dx}^{2}}\Big)}=\int\frac{\frac{\text{d}}{\text{dx}}\Big(\frac{\text{d}\text{y}}{\text{dx}}\Big)}{\Big(\frac{\text{d}\text{y}}{\text{dx}}\Big)}$
$\Rightarrow\log\Big(\frac{\text{d}^{2}\text{y}}{\text{dx}^{2}}\Big)=\log\Big(\frac{\text{dy}}{\text{dx}}\Big)+\log\text{C}_{4}$
$\Rightarrow \frac{\text{d}^{2}\text{y}}{\text{dx}}^{2}=\text{C}_{4}\frac{\text{dy}}{\text{dx}}$
$\Rightarrow\int \frac{\frac{\text{d}}{\text{dx}}\Big(\frac{\text{d}^{3}\text{y}}{\text{dx}^{3}}\Big)}{\Big(\frac{\text{d}^{2}\text{y}}{\text{dx}^{2}}\Big)}=\int\text{C}_{4}\ \text{dx}$
$\Rightarrow \log\Big(\frac{\text{dy}}{\text{dx}}\Big)=\text{C}_{4}\text{x}+\text{C}_{5}$
$\Rightarrow \int\text{dy}=\int\text{e}^{\text{C}_{4}\text{x+}\text{C}_{5}}\ \text{dx}$
$\Rightarrow\text{y}=\frac{\text{e}^{\text{C}_{4}\text{x}+\text{C}_{3}}}{\text{C}_{4}}+\text{C}_{6}$
$\Rightarrow\text{y}=\text{C}_{1}\text{e}^{\text{C}_{2}\text{x}}+\text{C}_{3}$
Where,
$\text{C}_{1}=\frac{\text{e}^{\text{C}_{5}}}{\text{C}_{4}}$
$\text{C}_{4}=\text{C}_{2}$
$\text{C}_{6}=\text{C}_{3}$
View full question & answer→MCQ 831 Mark
Which of the following is a second-order differential equation?
- A
$(\text{y}')^2+\text{x}=\text{y}^2$
- ✓
$\text{y}'\text{y}''+\text{y}=\sin\text{x}$
- C
$\text{y}'''+(\text{y}'')^2+\text{y}=0$
- D
$\text{y}'=\text{y}^2$
AnswerCorrect option: B. $\text{y}'\text{y}''+\text{y}=\sin\text{x}$
The order $\text{y}'\text{y}''+\text{y}=\sin\text{x}$ of is 2. Thus, it is a second-order differential equation.
View full question & answer→MCQ 841 Mark
Integrating factor of the differential equation $\cos\text{x}\frac{\text{dy}}{\text{dx}}+\text{y}\sin\text{x}=1$ is:
AnswerWe have,
$\cos\text{x}\frac{\text{dy}}{\text{dx}}+\text{y}\sin\text{x}=1$
Dividing both sides by, we get
$\frac{\text{dy}}{\text{dx}}+\frac{\sin\text{x}}{\cos\text{x}}\text{y}=\frac{1}{\cos\text{x}}$
$\Rightarrow \frac{\text{dy}}{\text{dx}}+(\tan\text{x})\text{y}=\frac{1}{\cos\text{x}}$
Comparing with we get,
$\text{P}=\tan\text{x}, \text{Q}=\frac{2}{\cos\text{x}}$
Now,
$\text{I.F}=\text{e}^{\int\tan\text{x}\text{dx}}$
$=\text{e}^{\log(\sec\text{x})}$
$=\sec{\text{x}}$
View full question & answer→MCQ 851 Mark
The order and degree of the differential equation $\Big(1+3\frac{\text{dy}}{\text{dx}}\Big)^\frac{2}{3}=4\frac{\text{d}^3\text{y}}{\text{dx}^3}$ are:
- A
$1,\frac{2}{3}$
- B
$3,1$
- C
$1,2$
- ✓
$3,3$
Answer$=\Big(1+3\frac{\text{dy}}{\text{dx}}\Big)^\frac{2}{3}=4\frac{\text{d}^3\text{y}}{\text{dx}^3}$
$=\Big(1+3\frac{\text{dy}}{\text{dx}}\Big)^2=16\Big(\frac{\text{d}^3\text{y}}{\text{dx}}\Big)^3$
Highest derivative is third order And power of highest order derivative is $3$
Hence order is $3$ and degree is $3$
View full question & answer→MCQ 861 Mark
The differential equation of the ellipse $\frac{\text{x}^{2}}{\text{a}^{2}}+\frac{\text{y}^{2}}{\text{b}^{2}}=\text{C}$ is:
AnswerCorrect option: A. $\frac{\text{y}''}{\text{y}'}+\frac{\text{y}'}{\text{y}}-\frac{1}{\text{x}}=0$
We have,
$\frac{\text{x}^{2}}{\text{a}^{2}}+\frac{\text{y}^{2}}{\text{b}^{2}}=\text{C}\ ...(\text{i})$
Differentiating with respect to x, we get
$\frac{2\text{x}^{2}}{\text{a}^{2}}+\frac{2\text{y}^{2}}{\text{b}^{2}}\text{y}'=0$
$\frac{\text{x}^{2}}{\text{a}^{2}}+\frac{\text{y}^{2}}{\text{b}^{2}}\text{y}'=0\ ...(\text{ii})$
Again differentiating with respect to x, we get
$\Rightarrow \frac{1}{\text{a}^{2}}+\frac{1}{\text{b}^{2}}(\text{y'}^{2})+\frac{\text{xy}}{\text{b}^{2}}\text{y}''=0\ ...(\text{iii})$
Multiplying throughout by x, we get
$\Rightarrow \frac{\text{x}}{\text{a}^{2}}+\frac{\text{x}}{\text{b}^{2}}(\text{y'}^{2})+\frac{\text{xy}}{\text{b}^{2}}\text{y}''=0\ ...(\text{iv})$
Subtracting (ii) from (iv),
$\frac{1}{\text{b}^{2}}\big[\text{x}(\text{y}')^{2}+\text{xyy}''-\text{yy}''\big]=0$
$\Rightarrow \text{x}(\text{y}')^{2}+\text{xyy}''-\text{yy}''=0$
Diving both sides by,
$\frac{\text{y}''}{\text{y}'}+\frac{\text{y}'}{\text{y}}-\frac{1}{\text{x}}=0$
$\Rightarrow \frac{\text{y}''}{\text{y}'}+\frac{\text{y}'}{\text{y}}-\frac{1}{\text{x}}=0$
View full question & answer→MCQ 871 Mark
Choose the correct answer from the given four option.If $\text{y}=\text{e}^{-\text{x}}(\text{A}\cos\text{x}+\text{B}\sin\text{x}),$ then y is a solution of:
- A
$\frac{\text{d}^2\text{y}}{\text{d}\text{x}^2}+2\frac{\text{d}\text{y}}{\text{d}\text{x}}=0$
- B
$\frac{\text{d}^2\text{y}}{\text{d}\text{x}^2}-2\frac{\text{d}\text{y}}{\text{d}\text{x}}+2\text{y}=0$
- ✓
$\frac{\text{d}^2\text{y}}{\text{d}\text{x}^2}+2\frac{\text{d}\text{y}}{\text{d}\text{x}}+2\text{y}=0$
- D
$\frac{\text{d}^2\text{y}}{\text{d}\text{x}^2}+2\text{y}=0$
AnswerCorrect option: C. $\frac{\text{d}^2\text{y}}{\text{d}\text{x}^2}+2\frac{\text{d}\text{y}}{\text{d}\text{x}}+2\text{y}=0$
Given that, $\text{y}=\text{e}^{-\text{x}}(\text{A}\cos\text{x}+\text{B}\sin\text{x})$
Differentiating both sides w.r.t.x, we get
$\frac{\text{dy}}{\text{dx}}=\text{e}^{-\text{x}}(\text{A}\cos\text{x}+\text{B}\sin\text{x})+\text{e}^{-\text{x}}(-\text{A}\sin\text{x}+\text{B}\cos\text{x})$
$\frac{\text{dy}}{\text{dx}}=-\text{y}+\text{e}^{-\text{x}}(-\text{A}\sin\text{x}+\text{B}\cos\text{x})$
Again, Differentiating both sides W.r.t.x, we get
$\frac{\text{d}^2\text{y}}{\text{d}\text{x}^2}=-\frac{\text{d}\text{y}}{\text{d}\text{x}}+\text{e}^{-\text{x}}(-\text{A}\cos\text{x}-\text{B}\sin\text{x})-\text{e}^{-\text{x}}(-\text{A}\sin\text{x}+\text{B}\cos\text{x})$
$\Rightarrow\frac{\text{d}^2\text{y}}{\text{d}\text{x}^2}=-\frac{\text{d}\text{y}}{\text{d}\text{x}}-\text{y}-\Big[\frac{\text{d}\text{y}}{\text{d}\text{x}}+\text{y}\Big]$
$\Rightarrow\frac{\text{d}^2\text{y}}{\text{d}\text{x}^2}=-2\frac{\text{d}\text{y}}{\text{d}\text{x}}-2\text{y}$
$\Rightarrow\frac{\text{d}^2\text{y}}{\text{d}\text{x}^2}+2\frac{\text{d}\text{y}}{\text{d}\text{x}}+2\text{y}=0$
View full question & answer→MCQ 881 Mark
Find the order of differential equations$:\ 2\text{x}^2\frac{\text{d}^2\text{y}}{\text{d}\text{y}^2}-3\frac{\text{dx}}{\text{dx}}+\text{y}=0$
AnswerGiven, the differential equation is:
$2\text{x}^2\frac{\text{d}^2\text{y}}{\text{d}\text{y}^2}-3\frac{\text{dx}}{\text{dx}}+\text{y}=0$
Or we can write:
$2x^2 y\ ’’ – 3y\ ’ + y = 0$
Order is the highest derivative in the differential equation. Therefore, the order is $2.$
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The order of the differential whose general solution is given by ${\text{y}}=\text{C}_1\cos(2\text{x}+\text{C}_{2})+(\text{C}_{3}+\text{C}_{4})\text{a}^{\text{x}+\text{C}_{5}}+\text{C}_{6}\sin(\text{x}-\text{C}_{7}).$
AnswerThe given equation can be reduced to :
${\text{y}}=\text{C}_1\cos(2\text{x}+\text{C}_{2})+(\text{C}_{3}+\text{C}_{4})\text{a}^{\text{x}+\text{C}_{5}}+\text{C}_{6}\sin(\text{x}-\text{C}_{7})$
Where $C = C_3 + C_4$ be a constant
There are $5$ constant $(C_1, C_2, C_3, C_6, C_7)$ in the given differential equation.
Hence, the order of the dfifferential equation is $5.$
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If $\frac{\text{dy}}{\text{dx}}=\sin(\text{x}+\text{y})+\cos(\text{x}+\text{y}),\text{y}(0)=0,$ then $\tan\frac{\text{x}+\text{y}}{2}=$
AnswerCorrect option: A. $\text{e}^\text{x}-1$
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The order of the differential equation
$2\text{x}^2\frac{\text{d}^2\text{y}}{\text{dx}^2}-3\frac{\text{dy}}{\text{dx}}+\text{y}=0 \ \text{is}$
AnswerThe given differential equation is $2\text{x}^2\frac{\text{d}^2\text{y}}{\text{dx}^2}-3\frac{\text{dy}}{\text{dx}}+\text{y}=0$
The highest order derivative present in the differential equation is $\frac{\text{d}^2\text{y}}{\text{dx}^2}.$
$\therefore$ its order is 2
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Which of the following are true?
- ✓
Particular solution is a solution of a differential equation containing no arbitrary constants.
- B
Particular Solution is a solution to a differential equation that contains arbitrary, unevaluated constants.
- C
General solution is a solution of a differential equation containing no arbitrary constants.
- D
General Solution is a solution to a differential equation that contains arbitrary, unevaluated constants.
AnswerCorrect option: A. Particular solution is a solution of a differential equation containing no arbitrary constants.
The general solution to a differential equation contains arbitrary constants which can be any value and the number of arbitrary constants is the order of the differential equation.
We can determine the values of arbitrary constants with an initial value condition like a point through which the curve passes etc.
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The differention equation $\frac{\text{dy}}{\text{dx}}+\text{P}\text{y}=\text{Qy}^{\text{n}},\text{n}>2$ can be reduced to linear from by substituting:
- A
$\text{z}=\text{y}^{\text{n}-1}$
- B
$\text{z}=\text{y}^{\text{n}}$
- C
$\text{z}=\text{y}^{\text{n}+1}$
- ✓
$\text{z}=\text{y}^{1-\text{n}}$
AnswerCorrect option: D. $\text{z}=\text{y}^{1-\text{n}}$
We have,
$\frac{\text{dy}}{\text{dx}}+\text{P}\text{y}=\text{Qy}^{\text{n}}$
$\Rightarrow \text{y}^{-\text{n}}\frac{\text{dy}}{\text{dx}}+\text{Py}^{1-\text{n}}=\text{Q}\ ...(\text{i})$
Put $\text{z}=\text{y}^{1-\text{n}}$
Integrating both sides with respect to x, we get
$\frac{\text{dz}}{\text{dx}}=(1-\text{n})\text{y}^{-\text{n}}\frac{\text{dy}}{\text{dx}}$
$\Rightarrow \text{y}^{-\text{n}}\frac{\text{dy}}{\text{dx}}=\frac{1}{(1-\text{n})}\frac{\text{dz}}{\text{dx}}$
Now, (i),
$\frac{1}{(1-\text{n})}\frac{\text{dz}}{\text{dx}}+\text{Pz}=\text{Q}$
$\Rightarrow \frac{\text{dz}}{\text{dx}}+\text{P}(1-\text{n})=\text{Q}(1-\text{n})$
Which is linear from of differential equation.
Therefore the given differential equation can be to linear by the $\text{z}=\text{y}^{1-\text{n}}.$
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The differential equation of all ‘Simple Harmonic Motions’ of given period $\frac{2\pi}{\text{n}}$ is:
- A
$\frac{\text{d}^2\text{x}}{\text{dt}^2}+\text{nx}=0$
- ✓
$\frac{\text{d}^2\text{x}}{\text{dt}^2}+\text{n}^2\text{x}=0$
- C
$\frac{\text{d}^2\text{x}}{\text{dt}^2}-\text{n}^2\text{x}=0$
- D
$\frac{\text{d}^2\text{x}}{\text{dt}^2}+\frac{1}{\text{n}^2}=0$
AnswerCorrect option: B. $\frac{\text{d}^2\text{x}}{\text{dt}^2}+\text{n}^2\text{x}=0$
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If $\frac{\text{dy}}{\text{dx}}=\text{y}\sin2\text{x},\text{y}(0)=1$ then solution is:
- ✓
$\text{y}=\text{e}\sin^2\text{x}$
- B
$\text{y}=\sin^2\text{x}$
- C
$\text{y}=\cos^2\text{x}$
- D
$\text{y}=\text{e}\cos^2\text{x}$
AnswerCorrect option: A. $\text{y}=\text{e}\sin^2\text{x}$
View full question & answer→MCQ 961 Mark
The degree of the differntial equation $\left\{5+\Big(\frac{\text{dy}}{\text{dx}}\Big)^{2}\right\}^{\frac{5}{3}}=\text{x}^{5}\Big(\frac{\text{d}^{2}\text{y}}{\text{dx}^{2}}\Big)$ is:
AnswerWe have,
$\left\{5+\Big(\frac{\text{dy}}{\text{dx}}\Big)^{2}\right\}^{\frac{5}{3}}=\text{x}^{5}\Big(\frac{\text{d}^{2}\text{y}}{\text{dx}^{2}}\Big)$
Taking cube power on both sides, we get
$\left\{5+\Big(\frac{\text{dy}}{\text{dx}}\Big)^{2}\right\}^{5}=\text{x}^{15}\Big(\frac{\text{d}^{2}\text{y}}{\text{dx}^{2}}\Big)^{3}$
The highest order derivative is $\frac{\text{d}^{2}\text{y}}{\text{dx}^{2}}$ and its power is 3.
Hence, the degree is 3.
Disclaimer: The correct potion is not given in the quation.
View full question & answer→MCQ 971 Mark
The solution of the differential equation $\frac{\text{dy}}{\text{dx}}=\frac{\text{y}}{\text{x}}+\frac{\phi(\frac{\text{y}}{\text{x}})}{\phi'(\frac{\text{y}}{\text{x}})}$ is:
- ✓
$\phi(\frac{\text{y}}{\text{x}})=\text{Kx}$
- B
$\text{x}\phi(\frac{\text{y}}{\text{x}})=\text{K}$
- C
$\phi(\frac{\text{y}}{\text{x}})=\text{Ky}$
- D
$\text{y}\phi(\frac{\text{y}}{\text{x}})=\text{K}$
AnswerCorrect option: A. $\phi(\frac{\text{y}}{\text{x}})=\text{Kx}$
We have,
$\frac{\text{dy}}{\text{dx}}=\frac{\text{y}}{\text{x}}+\frac{\phi(\frac{\text{y}}{\text{x}})}{\phi'(\frac{\text{y}}{\text{x}})}$
Let $\text{y}=\text{ux}$
$\Rightarrow \frac{\text{dy}}{\text{dx}}=\text{u}+\text{x}\frac{\text{du}}{\text{dx}}$
$\therefore \text{u}+\text{x}\frac{\text{du}}{\text{dx}}=\text{u}+\frac{\phi(\text{u})}{\phi'(\text{u})}$
$\Rightarrow \text{x}\frac{\text{du}}{\text{dx}}=\frac{\phi(\text{u})}{\phi'(\text{u})}$
$\Rightarrow \frac{\phi(\text{u})}{\phi'(\text{u})}\text{du}=\frac{1}{\text{x}}\text{dx}$
Integrating both sides, we get
$ \int\frac{\phi(\text{u})}{\phi'(\text{u})}\text{du}=\int\frac{1}{\text{x}}\text{dx}$
$\Rightarrow \log|\phi(\text{v})|=\log|\text{x}|+\log|\text{K}|$
$\Rightarrow \log|\phi(\frac{\text{y}}{2})|-\log|\text{x}|=\log\text{K}$
$\Rightarrow \log|\phi(\frac{\text{y}}{2})|=\log\text{K}$
$\Rightarrow\phi(\frac{\text{y}}{2})|=\text{Kx}$
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Choose the correct answer from the given four options.The order and degree of the differential equation $\Big[1+\Big(\frac{\text{dy}}{\text{dx}}\Big)^2\Big]=\frac{\text{d}^2\text{y}}{\text{d}\text{x}^2}$ are:
AnswerGiven that, $\Big[1+\Big(\frac{\text{dy}}{\text{dx}}\Big)^2\Big]=\frac{\text{d}^2\text{y}}{\text{d}\text{x}^2}$
$\therefore\text{Order}=2\text{ and degree}=1$
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The general solution of differential equation is $(y + c)^2= cx$ where ccis an arbitrary constant. The order and degree of the differential equation are respectively:
- ✓
$1, 2$
- B
$2, 2$
- C
$1, 1$
- D
$2, 1$
AnswerCorrect option: A. $1, 2$
Given equation is
$=(\text{y}+\text{c})^2=\text{cx}$
$\Rightarrow\text{y}=\sqrt{\text{cx}}-\text{c}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{\sqrt{\text{c}}}{\text{x}}$
$\Rightarrow\Big(\frac{\text{dy}}{\text{dx}}\Big)^2=\frac{\text{c}}{\text{x}}$
The order of differential equation is the order of the highest derivative in the equation is $1$
The degree of differential equation is the power of the highest order derivative in the equation is $2$
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Choose the correct answer from the given four options.The differential equation for which $\text{y}=\text{a}\cos\text{x}+\text{b}\sin\text{x}$ is a solution, is:
- ✓
$\frac{\text{d}^2\text{y}}{\text{d}\text{x}^2}+\text{y}=0$
- B
$\frac{\text{d}^2\text{y}}{\text{d}\text{x}^2}-\text{y}=0$
- C
$\frac{\text{d}^2\text{y}}{\text{d}\text{x}^2}+(\text{a}+\text{b})\text{y}=0$
- D
$\frac{\text{d}^2\text{y}}{\text{d}\text{x}^2}+(\text{a}-\text{b})\text{y}=0$
AnswerCorrect option: A. $\frac{\text{d}^2\text{y}}{\text{d}\text{x}^2}+\text{y}=0$
Given equation is, $\text{y}=\text{a}\cos\text{x}+\text{b}\sin\text{x}$
On differentiating both sides w.r.t.x. we get
$\frac{\text{dy}}{\text{dx}}=-\text{a}\sin\text{x}+\text{b}\cos\text{dx}$
Again, differentiating w.r.t.x. we get
$\frac{\text{d}^2\text{y}}{\text{d}\text{x}^2}=-\text{a}\sin\text{x}+\text{b}\cos\text{dx}$
$\Rightarrow\frac{\text{d}^2\text{y}}{\text{d}\text{x}^2}=-\text{y}$
$\Rightarrow\frac{\text{d}^2\text{y}}{\text{d}\text{x}^2}+\text{y}=0$
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