M.C.Q (1 Marks) - Page 2 - MATHS STD 12 Science Questions - Vidyadip

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M.C.Q (1 Marks)

Question 511 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:

$\frac{2\text{x}-1}{2\text{y}+3}=\text{k}$
$\frac{2\text{y}+1}{2\text{x}-3}=\text{k}$
$\frac{2\text{x}+3}{2\text{y}-1}=\text{k}$
$\frac{2\text{x}-1}{2\text{y}-1}=\text{k}$

Answer

$\frac{2\text{x}+3}{2\text{y}-1}=\text{k}$

Solution:
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
$\frac{1}{2}\log(2\text{x}+3)=\frac{1}{2}\log(2\text{y}-1)+\log\text{C}$
$\Rightarrow\frac{1}{2}\log(2\text{x}+3)-\log(2\text{y}-1)=\log\text{C}$
$\Rightarrow\frac{1}{2}\log\Big(\frac{2\text{x}+3}{2\text{y}-1}\Big)=\text{C}^2$
$\Rightarrow\Big(\frac{2\text{x}+3}{2\text{y}-1}\Big)^{\frac{1}{2}}=\text{C}$
$\Rightarrow\frac{2\text{x}+3}{2\text{y}-1}=\text{C}^2$
$\Rightarrow\frac{2\text{x}+3}{2\text{y}-1}=\text{C}^2$
$\Rightarrow\frac{2\text{x}+3}{2\text{y}-1}=\text{k},\text{Where}\ \text{k}=\text{c}^2$

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Question 521 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?

$\frac{\text{d}^2\text{y}}{\text{dx}^2}+\text{y}=0$
$\frac{\text{d}^2\text{y}}{\text{dx}^2}-\text{y}=0$
$\frac{\text{d}^2\text{y}}{\text{dx}^2}+1=0$
$\frac{\text{d}^2\text{y}}{\text{dx}^2}-1=0$

Answer

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.

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Question 531 Mark

Which of the following differential equations has y = x as one of its particular solution?

$\frac{\text{d}^2\text{y}}{\text{dx}^2}-\text{x}^2\frac{\text{dy}}{\text{dx}}+\text{xy}=\text{x}$
$\frac{\text{d}^2\text{y}}{\text{dx}^2}+\text{x}\frac{\text{dy}}{\text{dx}}+\text{xy}=\text{x}$
$\frac{\text{d}^2\text{y}}{\text{dx}^2}-\text{x}^2\frac{\text{dy}}{\text{dx}}+\text{xy}=0$
$\frac{\text{d}^2\text{y}}{\text{dx}^2}+\text{x}\frac{\text{dy}}{\text{dx}}+\text{xy}=0$

Answer

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.

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Question 541 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:

1
2
3
Not defined

Answer

Not difined

Solution:
The 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.

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Question 551 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:

$\tan^{-1}\big(\frac{\text{x}}{\text{y}}\big)-\log\text{y}+\text{C}$
$\tan^{-1}\big(\frac{\text{y}}{\text{x}}\big)-\log\text{x}+\text{C}$
$\tan^{-1}\big(\frac{\text{x}}{\text{y}}\big)=\log\text{x}+\text{C}$
$\tan^{-1}\big(\frac{\text{y}}{\text{x}}\big)=\log\text{y}+\text{C}$

Answer

$\tan^{-1}\big(\frac{\text{y}}{\text{x}}\big)=\log\text{y}+\text{C}$

Solution:
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
$\int\big(\frac{1}{1+\text{u}^{2}}\big)\text{du}=\int\frac{1}{\text{x}}\text{dx}$
$\Rightarrow \tan^{-1}\text{u}=\log\text{x}+\text{C}$
$\Rightarrow \tan^{-1}\big(\frac{\text{y}}{2}\big)=\log\text{x}+\text{C}$

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Question 561 Mark

Choose the correct answer from the given four option.
Which of the following is a second order differential equation?

$(\text{y}')^2+\text{x}=\text{y}^2$
$\text{y}'\text{y}''+\text{y}=\sin\text{x}$
$\text{y}'''+(\text{y}'')^2+\text{y}=0$
$\text{y}'=\text{y}^2$

Answer

$\text{y}'\text{y}''+\text{y}=\sin\text{x}$

Solution:
The second order differential equation is $\text{y}'\text{y}''+\text{y}=\sin\text{x}.$

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MCQ 571 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
None of the above

Answer

Correct 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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Question 581 Mark

The general solution of the differential equation $\frac{\text{y}\ \text{dx}-\text{x}\ \text{dx}}{\text{y}}=0\ \text{is}$

$\text{xy}=\text{C}$
$\text{x}=\text{Cy}^2$
$\text{y}=\text{Cx}$
$\text{y}=\text{Cx}^2$

Answer

$\ \text{y}=\text{Cx}$

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
$\frac{\text{y dx}-\text{x dy}}{\text{y}}=0$
$\text{or}\ \ \frac{\text{y dx}-\text{x dy}}{\text{x}^2}=0\ \ \text{or}\ \ \text{d}\Big(\frac{\text{x}}{\text{y}}\Big)=0$
$\therefore\ \ \frac{\text{x}}{\text{y}}=\text{constant}.$
$\therefore\ \ \frac{\text{y}}{\text{x}}=\text{C}\ \ \text{or}\ \ \text{y}=\text{Cx}$

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MCQ 591 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
None of these.

Answer

Correct 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,
$\text{C}_{1}=\frac{\text{e}^{\text{C}_{5}}}{\text{C}_{4}}$
$\text{C}_{4}=\text{C}_{2}$
$\text{C}_{6}=\text{C}_{3}$

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Question 601 Mark

Which of the following is a second-order differential equation?

$(\text{y}')^2+\text{x}=\text{y}^2$
$\text{y}'\text{y}''+\text{y}=\sin\text{x}$
$\text{y}'''+(\text{y}'')^2+\text{y}=0$
$\text{y}'=\text{y}^2$

Answer

$\text{y}'\text{y}''+\text{y}=\sin\text{x}$

Soultion:
The order $\text{y}'\text{y}''+\text{y}=\sin\text{x}$ of is 2. Thus, it is a second-order differential equation.

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Question 611 Mark

Integrating factor of the differential equation $\cos\text{x}\frac{\text{dy}}{\text{dx}}+\text{y}\sin\text{x}=1$ is:

$\cos\text{x}$
$\tan\text{x}$
$\sec\text{x}$
$\sin\text{x}$

Answer

$\sec\text{x}$

Solution:
We 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,
$\text{I.F}=\text{e}^{\int\tan\text{x}\text{dx}}$
$=\text{e}^{\log(\sec\text{x})}$
$=\sec{\text{x}}$

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Question 621 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:

$1,\frac{2}{3}$
$3,1$
$1,2$
$3,3$

Answer

$3,3$

Solution:
$=\Big(1+3\frac{\text{dy}}{\text{dx}}\Big)^\frac{2}{3}=4\frac{\text{d}^3\text{y}}{\text{dx}^3}$
$=\Big(1+3\frac{\text{dy}}{\text{dx}}\Big)^2=16\Big(\frac{\text{d}^3\text{y}}{\text{dx}}\Big)^3$
Highest derivative is third order And power of highest order derivative is 3
Hence order is 3 and degree is 3

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Question 631 Mark

The solution of $\frac{\text{d}^2\text{y}}{\text{dx}^2}=0$ represents:

a straight line
a circle
a parabola
a point

Answer

a straight line

Solution:
$\Rightarrow\frac{\text{d}}{\text{dx}}\text{y}_1=0$
$\Rightarrow\int\text{dy}_1=\int0.\text{dx}+\text{c}_1$
$\Rightarrow\int\text{dy}_1=\text{c}_1$
$\Rightarrow\text{y}_1=\text{c}_1$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\text{c}_1$
$\Rightarrow\int\text{dy}=\int\text{c}_1\text{ dx}+\text{c}_2$
$\Rightarrow\text{y}=\text{c}_1\text{x}+\text{c}_2\text{ straight line}$

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Question 641 Mark

The differential equation of the ellipse $\frac{\text{x}^{2}}{\text{a}^{2}}+\frac{\text{y}^{2}}{\text{b}^{2}}=\text{C}$ is:

$\frac{\text{y}''}{\text{y}'}+\frac{\text{y}'}{\text{y}}-\frac{1}{\text{x}}=0$
$\frac{\text{y}''}{\text{y}'}+\frac{\text{y}'}{\text{y}}+\frac{1}{\text{x}}=0$
$\frac{\text{y}''}{\text{y}'}-\frac{\text{y}'}{\text{y}}-\frac{1}{\text{x}}=0$
None of these.

Answer

$\frac{\text{y}''}{\text{y}'}+\frac{\text{y}'}{\text{y}}-\frac{1}{\text{x}}=0$

Solution:
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,
$\frac{\text{y}''}{\text{y}'}+\frac{\text{y}'}{\text{y}}-\frac{1}{\text{x}}=0$
$\Rightarrow \frac{\text{y}''}{\text{y}'}+\frac{\text{y}'}{\text{y}}-\frac{1}{\text{x}}=0$

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Question 651 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:

$\frac{\text{d}^2\text{y}}{\text{d}\text{x}^2}+2\frac{\text{d}\text{y}}{\text{d}\text{x}}=0$
$\frac{\text{d}^2\text{y}}{\text{d}\text{x}^2}-2\frac{\text{d}\text{y}}{\text{d}\text{x}}+2\text{y}=0$
$\frac{\text{d}^2\text{y}}{\text{d}\text{x}^2}+2\frac{\text{d}\text{y}}{\text{d}\text{x}}+2\text{y}=0$
$\frac{\text{d}^2\text{y}}{\text{d}\text{x}^2}+2\text{y}=0$

Answer

$\frac{\text{d}^2\text{y}}{\text{d}\text{x}^2}+2\frac{\text{d}\text{y}}{\text{d}\text{x}}+2\text{y}=0$

Solution:
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
$\frac{\text{d}^2\text{y}}{\text{d}\text{x}^2}=-\frac{\text{d}\text{y}}{\text{d}\text{x}}+\text{e}^{-\text{x}}(-\text{A}\cos\text{x}-\text{B}\sin\text{x})-\text{e}^{-\text{x}}(-\text{A}\sin\text{x}+\text{B}\cos\text{x})$
$\Rightarrow\frac{\text{d}^2\text{y}}{\text{d}\text{x}^2}=-\frac{\text{d}\text{y}}{\text{d}\text{x}}-\text{y}-\Big[\frac{\text{d}\text{y}}{\text{d}\text{x}}+\text{y}\Big]$
$\Rightarrow\frac{\text{d}^2\text{y}}{\text{d}\text{x}^2}=-2\frac{\text{d}\text{y}}{\text{d}\text{x}}-2\text{y}$
$\Rightarrow\frac{\text{d}^2\text{y}}{\text{d}\text{x}^2}+2\frac{\text{d}\text{y}}{\text{d}\text{x}}+2\text{y}=0$

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MCQ 661 Mark

Find the order of differential equations$:2\text{x}^2\frac{\text{d}^2\text{y}}{\text{d}\text{y}^2}-3\frac{\text{dx}}{\text{dx}}+\text{y}=0$

✓
$2$
B
$1$
C
$0$
D
Undefined

Answer

Correct option: A.

$2$

Given, 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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Question 671 Mark

If $\frac{\text{dy}}{\text{dx}}=\sin(\text{x}+\text{y})+\cos(\text{x}+\text{y}),\text{y}(0)=0,$ then $\tan\frac{\text{x}+\text{y}}{2}=$

$\text{e}^\text{x}-1$
$\frac{\text{e}^\text{x}-1}{2}$
$2(\text{e}^\text{x}-1)$
$1-\text{e}^\text{x}$

Answer

$\text{e}^\text{x}-1$

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Question 681 Mark

Which of the following are true?

Particular solution is a solution of a differential equation containing no arbitrary constants.
Particular Solution is a solution to a differential equation that contains arbitrary, unevaluated constants.
General solution is a solution of a differential equation containing no arbitrary constants.
General Solution is a solution to a differential equation that contains arbitrary, unevaluated constants.

Answer

Particular solution is a solution of a differential equation containing no arbitrary constants.

Solution:
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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Question 691 Mark

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:

$\text{z}=\text{y}^{\text{n}-1}$
$\text{z}=\text{y}^{\text{n}}$
$\text{z}=\text{y}^{\text{n}+1}$
$\text{z}=\text{y}^{1-\text{n}}$

Answer

$\text{z}=\text{y}^{1-\text{n}}$

Solution:
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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Question 701 Mark

The differential equation of all ‘Simple Harmonic Motions’ of given period $\frac{2\pi}{\text{n}}$ is:

$\frac{\text{d}^2\text{x}}{\text{dt}^2}+\text{nx}=0$
$\frac{\text{d}^2\text{x}}{\text{dt}^2}+\text{n}^2\text{x}=0$
$\frac{\text{d}^2\text{x}}{\text{dt}^2}-\text{n}^2\text{x}=0$
$\frac{\text{d}^2\text{x}}{\text{dt}^2}+\frac{1}{\text{n}^2}=0$

Answer

$\frac{\text{d}^2\text{x}}{\text{dt}^2}+\text{n}^2\text{x}=0$

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Question 711 Mark

If $\frac{\text{dy}}{\text{dx}}=\text{y}\sin2\text{x},\text{y}(0)=1$ then solution is:

$\text{y}=\text{e}\sin^2\text{x}$
$\text{y}=\sin^2\text{x}$
$\text{y}=\cos^2\text{x}$
$\text{y}=\text{e}\cos^2\text{x}$

Answer

$\text{y}=\text{e}\sin^2\text{x}$

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MCQ 721 Mark

The solution of the differential equation $dy = (1 + y^2) dx$ is:

A
$\text{y}=\tan\text{x}+\text{c}$
✓
$\text{y}=\tan(\text{x}+\text{c})$
C
$\tan^{-1}(\text{y}+\text{c})=\text{x}$
D
$(\tan^{-1}(\text{y}+\text{c})=2\text{x}$

Answer

Correct option: B.

$\text{y}=\tan(\text{x}+\text{c})$

Concept:
$\int\limits\frac{\text{dx}}{1+\text{x}^2}\tan^{-1}\text{x}=\text{c}$
Calculation:
Given: $dy = (1 + y^2) dx$
$\Rightarrow\frac{\text{dy}}{1+\text{y}^2}\text{dx}$
Integrating both sides, we get
$\Rightarrow\int\frac{\text{dy}}{1+\text{y}^2}=\int\text{dx}$
$\Rightarrow\tan^{-1}\text{y}=\text{x}+\text{c}$
$\Rightarrow\text{y}=\tan(\text{x}+\text{c})$

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Question 731 Mark

The solution of differential equation $(\text{e}^\text{y}+1)\cos\text{dx}+\text{e}^\text{y}\sin\text{x}\text{dy}=0$ is:

$\text{e}^\text{y}+1\sin\text{x}=\text{c}$
$\text{e}^\text{y}\sin=\text{c}$
$(\text{e}^\text{y}+1)\cos\text{x}=\text{c}$
$\text{None of these}$

Answer

$\text{e}^\text{y}+1\sin\text{x}=\text{c}$

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Question 741 Mark

The solution of the differential equation $\frac{\text{dy}}{\text{dx}}=\frac{\text{x}}{1+\text{x}^2}$ is:

$\text{y}=\frac{1}{2}\log|2+\text{x}^2|+\text{c}$
$\text{y}=\frac{1}{2}\log(1+\text{x})+\text{c}$
$\text{y}=\log\Big(\sqrt{1+\text{x}^2}\Big)+\text{c}$
$\text{None of these}$

Answer

$\text{y}=\log\Big(\sqrt{1+\text{x}^2}\Big)+\text{c}$

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Question 751 Mark

Which of the following differentials equation has y = x as one of its particular solution?

$\frac{\text{d}^{2}\text{y}}{\text{dx}^{2}}-\text{x}^{2}\frac{\text{dy}}{\text{dx}}+\text{xy}=\text{x}$
$\frac{\text{d}^{2}\text{y}}{\text{dx}^{2}}+\text{x}^{2}\frac{\text{dy}}{\text{dx}}+\text{xy}=\text{x}$
$\frac{\text{d}^{2}\text{y}}{\text{dx}^{2}}-\text{x}^{2}\frac{\text{dy}}{\text{dx}}+\text{xy}=\text{0}$
$\frac{\text{d}^{2}\text{y}}{\text{dx}^{2}}+\text{x}^{2}\frac{\text{dy}}{\text{dx}}+\text{xy}=\text{0}$

Answer

$\frac{\text{d}^{2}\text{y}}{\text{dx}^{2}}-\text{x}^{2}\frac{\text{dy}}{\text{dx}}+\text{xy}=0$

Solution:
We have,
$\text{y}=\text{x}\ ...(\text{i})$
Differentiating both sides of (i) with respect to x, we get
$\frac{\text{dy}}{\text{dx}}=1\ ...(\text{ii})$
Differentiating both sides of (ii) with respect to x, we get
$\Rightarrow \frac{\text{d}^{2}\text{y}}{\text{dx}^{2}}=0$
$\Rightarrow \frac{\text{d}^{2}\text{y}}{\text{dx}^{2}}+\text{x}^{2}=\text{x}^{2}$
$\Rightarrow \frac{\text{d}^{2}\text{y}}{\text{dx}^{2}}+\text{x}\times\text{x}=\text{x}^{2}\times1$
$\Rightarrow \frac{\text{d}^{2}\text{y}}{\text{dx}^{2}}+\text{xy}=\text{x}^{2}\times1$
$\Rightarrow \frac{\text{d}^{2}\text{y}}{\text{dx}^{2}}+\text{xy}=\text{x}^{2}\frac{\text{dy}}{\text{dx}}$
$\Rightarrow \frac{\text{d}^{2}\text{y}}{\text{dx}^{2}}-\text{x}^{2}\frac{\text{dy}}{\text{dx}}+\text{xy}=0$

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Question 761 Mark

The degree of the differntial equation $\Big(\frac{\text{d}^{2}\text{y}}{\text{dx}^{2}}\Big)^{2}=\Big(\frac{\text{dy}}{\text{dx}}\Big)=\text{y}^{3}$ is:

$\frac{1}{2}$
$2$
$3$
$4$

Answer

$2$

Solution:
We have,
$\Big(\frac{\text{d}^{2}\text{y}}{\text{dx}^{2}}\Big)^{2}=\Big(\frac{\text{dy}}{\text{dx}}\Big)=\text{y}^{3}$
The highest order derivative is $\frac{\text{d}^{2}\text{y}}{\text{dx}^{2}}$ and its power is 2.
Hence, the degree is 2.

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Question 771 Mark

The solution of the differential equartion $\frac{\text{dy}}{\text{dx}}-\frac{\text{y}(\text{x}+1)}{\text{x}}=0$ is given by:

$\text{y}=\text{xe}^{\text{x}+\text{C}}$
$\text{x}=\text{ye}^{\text{x}}$
$\text{y}=\text{x}+\text{c}$
$\text{xy}=\text{e}^{\text{x}}+\text{C}$

Answer

$\text{y}=\text{xe}^{\text{x}+\text{C}}$

Solution:
We have,
$\frac{\text{dy}}{\text{dx}}-\frac{\text{y}(\text{x}+1)}{\text{x}}=0$
$\Rightarrow \frac{\text{dy}}{\text{dx}}-\frac{\text{y}(\text{x}+1)}{\text{x}}$
$\Rightarrow \frac{\text{dy}}{\text{dx}}=\frac{(\text{x}+1)}{\text{x}}\text{dx}$
Integrating both sides, we get
$ \int\frac{\text{dy}}{\text{y}}=\int\frac{(\text{x}+1)}{\text{x}}\text{dx}$
$ \Rightarrow \int\frac{\text{dy}}{\text{y}}=\int\text{dx}+\int\frac{1}{\text{x}}\text{dx}$
$ \Rightarrow \log{\text{y}}=\text{x}+\log\text{x}+\text{C}$
$\Rightarrow \log{\text{y}}-\log\text{x}=\text{x}+\text{C}$
$ \Rightarrow \log\frac{{\text{y}}}{\text{x}}=\text{x}+\text{C}$
$\Rightarrow \frac{\text{y}}{\text{x}}=\text{e}^{\text{x}+\text{C}}$
$\Rightarrow{\text{y}}=\text{xe}^{\text{x}+\text{C}}$

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Question 781 Mark

What is the order of differential equation y’’ + 5y’ + 6 = 0?

0
1
2
3

Answer

2

Solution
Given, differential equation y’’ + 5y’ + 6 = 0.
The highest order derivative present in the differential equation is y’’. Hence, the order is 2.

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MCQ 791 Mark

The general solution of differention eqution of the $e^x dy + (ye^x + 2x)dx = 0$ is:

A
$xe^y + x^2= C$
B
$xe^y+ y^2= C$
✓
$ye^x + y^2= C$
D
$ye^y+ x^2 = C$

Answer

Correct option: C.

$ye^x + y^2= C$

We have,
$e^x dy + (ye^x + 2x) dx = 0$
Diving both sides by we get,
$\frac{\text{dy}}{\text{dx}}+(\text{y}+\frac{2\text{x}}{\text{e}^{x}})=0$
$\Rightarrow \frac{\text{dy}}{\text{dx}}+\text{y}=-\frac{2\text{x}}{\text{e}^{\text{x}}}$
Comping with $\frac{\text{dy}}{\text{dx}}=\text{Q}$ we get,
$\text{P}=1, \text{Q}=-\frac{2\text{x}}{\text{e}^{\text{x}}}$
Now,
$\text{I.F}=\text{e}^{\int\text{dx}}$
$=\text{e}^{\text{x}}$
Solution is given by,
$\text{y}\times\text{I.F}=\int(\text{Q}\times\text{I.F}) \text{dx}+\text{C}$
$\Rightarrow \text{ye}^{\text{x}}=-\int\text{e}^{\text{x}}\times \frac{2\text{x}}{\text{e}^{\text{x}}}\text{dx}+\text{C}$
$\Rightarrow \text{ye}^{\text{x}}=-2\int\text{x}\ \text{dx}+\text{C}$
$\Rightarrow \text{ye}^{\text{x}}=-\text{x}^{2}+\text{C}$
$\Rightarrow \text{ye}^{\text{x}}+\text{x}^{2}=\text{C}$

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Question 801 Mark

The order of differential equation $\frac{\text{d}^2\text{y}}{\text{dx}^2}-\Big(\frac{\text{dy}}{\text{dx}^2}\Big)=1$ is:

one
two
four
zero

Answer

two

Solution:
The order of differential equation is the order of thehighest derivative in the equation
$\therefore$ the above given equation is of second order

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MCQ 811 Mark

Which of the following is a homogeneous differential equation?

A
$(4x + 6y + 5) dy - (3y + 2x + 4) dx = 0$
B
$(xy) dx - (x^3 + y^3) dy = 0$
C
$(x^3 + 2y^2) dx + 2xy dy = 0$
✓
$y^2 dx + (x^2 - xy - y^2) dy = 0$

Answer

Correct option: D.

$y^2 dx + (x^2 - xy - y^2) dy = 0$

Out of the given four options, option $(D)$ is the only option in which all coefficients of $dx$ and $dy$ are of same degree i.e., $2.$ It may be noted that $xy$ is a term of second degree.
Hence differential equation in option $(D)$ is Homogeneous differential equation.

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Question 821 Mark

The number of arbitrary constants in the general solution of a differential equation of fourth order are:

0
2
3
4

Answer

4

Solution:
The number of arbitrary constants in a solution of a differential equation of order nn is equal to its order.
So, here it is 4.

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Question 831 Mark

The degree of the differential equation $2\text{x}^{2}\frac{\text{d}^{2}\text{y}}{\text{dx}^{2}}+3\frac{\text{dy}}{\text{dx}}+\text{y}=0$ is:

2
1
0
Not defined.

Answer

2

Solution:
We have,
$2\text{x}^{2}\frac{\text{d}^{2}\text{y}}{\text{dx}^{2}}+3\frac{\text{dy}}{\text{dx}}+\text{y}=0$
Here, the highest order is $\frac{\text{d}^{2}\text{y}}{\text{d}^{2}\text{x}}.$
Hence, the order is 2.

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Question 841 Mark

$\text{y}=\sin$ kt satisfies the differential equation y′′+9y = 0. Then k:

$\pm3$
$0$
$\pm2$
$\pm4$

Answer

$\pm3$

Solution:
Given, $\text{y}=\sin$ kt satisfies the differential equation y′′ + 9y = 0.
Then, we have,$-\text{k}^2\sin\text{kt}+9\sin\text{kt}=0$ or,
$=\text{k}^2-9=0$
$[\text{Since }\sin\text{kt}\neq 0]\text{or,}$
$=\text{k}=\pm3.$

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Question 851 Mark

The integrating factor of the differential equation $(\text{x}\log\text{x})\frac{\text{dy}}{\text{dx}}+\text{y}=2\ \log\text{x}$ is given by:

$\log(\log\text{x})$
$\text{e}^{\text{x}}$
$\log\text{x}$
$\text{x}$

Answer

$\log\text{x}$

Solution:
We have,
$(\text{x}\log\text{x})\frac{\text{dy}}{\text{dx}}+\text{y}=2\ \log\text{x}$
Dividing both sides by,
$\frac{\text{dy}}{\text{dx}}+\frac{\text{y}}{\text{x}+\log\text{x}}=\frac{2}{\text{x}}$
$\frac{\text{dy}}{\text{dx}}+\Big(\frac{\text{1}}{\text{x}+\log\text{x}}\Big)\text{y}=\frac{2}{\text{x}}$
Comparing with $\frac{\text{dy}}{\text{dx}}+\text{Py}=\text{Q}$
$\text{P}=\frac{1}{\text{x}\log\text{x}}$
$\text{Q}=\frac{2}{\text{x}}$
Now,
$\text{I.F}=\text{e}^{\int\text{P}\text{dx}}$
$\text{I.F}=\text{e}^{\int\frac{1}{\text{x}\log\text{x}}}\text{dx}$
$=\text{e}^{\log(\log\text{x})}$
$=\log\text{x}$

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Question 861 Mark

What are the order and degree, respectively, of the differential equation:
$\Big(\frac{\text{d}^3\text{y}}{\text{dx}^3}\Big)^2=\text{y}^4+\Big(\frac{\text{dy}}{\text{dx}}\Big)^5?$

4, 5
2, 3
3, 2
5, 4

Answer

3, 2

Solution:
Order is the highest derivative of the dependent variable with respect to the independent variable and degree is the highest power to which the highest order derivative in the differential equation is raised.
so, Order = 3 and Degree=2.

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Question 871 Mark

The solution of the differention $\frac{\text{dy}}{\text{dx}}+1=\text{e}^{\text{x}+\text{y}}$ is:

$(\text{x}+\text{y})\text{e}^{\text{x}+\text{y}}=0$
$(\text{x}+\text{C})\text{e}^{\text{x}+\text{y}}=0$
$(\text{x}-\text{C})\text{e}^{\text{x}+\text{y}}=1$
$(\text{x}-\text{C})\text{e}^{\text{x}+\text{y}}+1=0$

Answer

$(\text{x}-\text{C})\text{e}^{\text{x}+\text{y}}+1=0$

Solution:
We have,
$\frac{\text{dy}}{\text{dx}}+1=\text{e}^{\text{x}+\text{y}}$
Let $\text{x}+\text{y}=\text{u}$
$\Rightarrow 1+\frac{\text{dy}}{\text{dx}}=\frac{\text{du}}{\text{dx}}$
$\Rightarrow \frac{\text{dy}}{\text{dx}}+1=\frac{\text{du}}{\text{dx}}$
$\Rightarrow \frac{\text{dy}}{\text{dx}}=\text{e}^{\text{u}}$
$\Rightarrow \text{e}^{-\text{u}}\text{du}=\text{dx}$
Intergrating both sides, we get
$\Rightarrow \text{e}^{-\text{u}}=\text{x}-\text{C}$
$\Rightarrow -1=\text{e}^{-\text{u}}(\text{x}-\text{C})$
$\Rightarrow (\text{x}-\text{C})\text{e}^{\text{x}+\text{y}}+1=0$

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Question 881 Mark

Choose the correct answer from the given four option.
The degree of the differential equation $\frac{\text{d}^2\text{y}}{\text{d}\text{x}^2}+\Big(\frac{\text{d}\text{y}}{\text{d}\text{x}}\Big)^3+6\text{y}^5=0$ is:

1.
2.
3.
5.

Answer

1.

Solution:
$\frac{\text{d}^2\text{y}}{\text{d}\text{x}^2}+\Big(\frac{\text{d}\text{y}}{\text{d}\text{x}}\Big)^3+6\text{y}^5=0$
We know that, the degree of a differential equation is highest exponent of order dervivatibve.
$\therefore\text{Degree}=1$

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Question 891 Mark

What is the degree of the differential equation:
$\text{y}=\text{x}\frac{\text{dy}}{\text{dx}}+\Big(\frac{\text{dy}}{\text{dx}}\Big)^{-2}?$

1
3
-2
Degree does not exist.

Answer

3

Solution:
Concept:
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:
Given:
$\text{y}=\text{x}\frac{\text{dy}}{\text{dx}}+\Big(\frac{\text{dy}}{\text{dx}}\Big)^{-2}\text{y} $
$=\text{x}\frac{\text{dy}}{\text{dx}}+\frac{1}{\Big(\frac{\text{dy}}{\text{dx}}\Big)2}\text{y}\Big(\frac{\text{dy}}{\text{dx}}\Big)^2$
$=\text{x}\frac{\text{dy}}{\text{dx}}^3+1$
For the given differential equation the highest order derivative is 1.
Now, the power of the highest order derivative is 3.
We know that the degree of a differential equation is the power of the highest derivative.
Hence, the degree of the differential equation is 3.

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MCQ 901 Mark

If $(x^2 + y^2) dy = xy \ dx, y(1) = 1,$ and $y(x_0) = e,$ then $x_0 =$

A
$\sqrt{2(\text{e}^2-1)}$
B
$\sqrt{2(\text{e}^2+1)}$
✓
$\sqrt{3}\text.{e}$
D
$\sqrt{\frac{\text{e}^2+1}{2}}$

Answer

Correct option: C.

$\sqrt{3}\text.{e}$

$\sqrt{3}\text.{e}$

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Question 911 Mark

Integrating factor of the differntial equation $\cos\text{x}\frac{\text{dy}}{\text{dx}}+\text{y}\sin\text{x}=1$is:

$\sin\text{x}$
$\sec\text{x}$
$\tan\text{x}$
$\cos\text{x}$

Answer

$\sec\text{x}$

Solution:
We 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 $ \frac{\text{dy}}{\text{dx}}+\text{Py}=\text{Q}$
$\text{P}=\tan\text{x}$
$\text{Q}=\frac{1}{\cos\text{x}}$
Now,
$\text{I.F}=\text{e}^{\int\tan\text{x}}\text{dx}$
$=\text{e}^{\log(\sec\text{x})}$
$=\sec\text{x}$

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Question 921 Mark

Choose the correct answer from the given four options.
Family $\text{y}=\text{A}\text{x}+\text{A}^3$ of curves will correspond to a differential equation of order:

3
2
1
Not defined

Answer

1

Solution:
Given family of curves is $\text{y}=\text{A}\text{x}+\text{A}^3\ .....(\text{i})$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\text{A}$
Replacting A by $\frac{\text{dy}}{\text{dx}}$ in Eq. (i), we get
$\text{y}=\text{x}\frac{\text{dy}}{\text{dx}}+\Big(\frac{\text{dy}}{\text{dx}}\Big)^3$
$\therefore\text{Order}=1$

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Question 931 Mark

Determine the degree and order of the given differential equation respectively:
$\text{y}=\text{x}\Big(\frac{\text{dy}}{\text{dx}}\Big)^2\Big(\frac{\text{dx}}{\text{dy}}\Big)^{-2}?$

1, 2
2, 1
1, 4
4, 1

Answer

4, 1

Solution:
Concept:
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.
Note:
Degree is defined if function is a polynomial, if differential contains logarithmic, exponential and trigonometric function of the highest derivative, then degree is not defined.
Degree and order is always a positive integer.
Calculation:
Given: $\text{y}=\text{x}\Big(\frac{\text{dy}}{\text{dx}}\Big)^2\Big(\frac{\text{dx}}{\text{dy}}\Big)^{-2}$
To find: Order & Degree
$\text{y}=\text{x}\Big(\frac{\text{dy}}{\text{dx}}\Big)^2\frac{1}{\Big(\frac{\text{dy}}{\text{dx}}\Big)^2}$
$\text{y}=\Big(\frac{\text{dy}}{\text{dx}}\Big)^2=\text{x}\Big(\frac{\text{dx}}{\text{dy}}\Big)^4+1$
Hence, the degree is 4 & order is 1.

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Question 941 Mark

Find the degree of the differential equation:
$\Big(1+\frac{\text{dx}}{\text{dy}}\Big)^3=\Big(\frac{\text{dy}}{\text{dx}}\Big)^2$

0
1
2
3

Answer

3

Solution:
Given, the differential equation is:
$\Big(1+\frac{\text{dx}}{\text{dy}}\Big)^3=\Big(\frac{\text{dy}}{\text{dx}}\Big)^2$
We can expand it and get:
$1+3\frac{\text{dx}}{\text{dy}}^3+3\Big(\frac{\text{dy}}{\text{dx}}\Big)^2+\Big(\frac{\text{dy}}{\text{dx}}\Big)^3=\Big(\frac{\text{dy}}{\text{dx}}\Big)^2$
The exponent of highest derivative is the degree. Therefore, the degree is 3.

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Question 951 Mark

The differential equation which respresents the famliy of curves $\text{y}=\text{e}^{\text{Cx}}$ is:

$\text{y}_{1}=\text{C}^{2}\text{y}$
$\text{xy}_{1}-\log\text{y}=0$
$\text{x}\log\text{y}=\text{yy}_{1}$
$\text{y}\log\text{y}=\text{xy}_{1}$

Answer

$\text{y}\log\text{y}=\text{xy}_{1}$

Solution:
We have,
$\text{y}=\text{e}^{\text{Cx}}$
Taking in both sides, we get
$\Rightarrow \log\text{y}=\text{Cx}\ ...(\text{1})$
Differentiating both sides of (i) with respect to x, we get
$\frac{1}{\text{y}_{1}}=\text{C}$
Substituting the value of C in in (i). we get
$\log\text{y}=\frac{\text{y}_{1}}{\text{y}}\text{x}$
$\Rightarrow \text{y}\ \log\text{y}=\text{y}_{1}\text{x}$

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Question 961 Mark

The Solution of $\cos(\text{x}+\text{y})\text{ dy}=\text{dx}$ is:

$\text{y}=\tan\Big(\frac{\text{x}+\text{y}}{2}\Big)+\text{c}$
$\text{y}=\cos^{-1}\Big(\frac{\text{y}}{\text{x}}\Big)+\text{c}$
$\text{y}=\text{x}\sec\Big(\frac{\text{y}}{\text{x}}\Big)+\text{c}$
$\text{None of these}$

Answer

$\text{y}=\tan\Big(\frac{\text{x}+\text{y}}{2}\Big)+\text{c}$

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Question 971 Mark

Order of $\Big( \frac{\text{dy}}{\text{dx}}\Big)^{3}+ \Big( \frac{\text{dy}}{\text{dx}}\Big)^{2}+\text{y}^{4}=0$ is:

4
3
1
2

Answer

1

Solution:
$=\Big( \frac{\text{dy}}{\text{dx}}\Big)^{3}+ \Big( \frac{\text{dy}}{\text{dx}}\Big)^{2}+\text{y}^{4}=0$
$=\text{i.e} {\text{ y}_{1}}^{3.}+{\text{y}_{1}}^{2}+\text{y}^{4}=0$
= Order = 1
= Degree = 3

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Question 981 Mark

If $\frac{\text{dy}}{\text{dx}}=\text{e}^{-2\text{y}}$ and y = 0, when x = 5, then the value of x when y = 3 is:

$\text{e}^5$
$\text{e}^5+1$
$\frac{\text{e}^5+9}{2}$
$\log_\text{e}6$

Answer

$\frac{\text{e}^5+9}{2}$

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Question 991 Mark

The equation of the curve whose slope is given by $\frac{\text{dy}}{\text{dx}}=\frac{2\text{y}}{\text{x}};\text{x}>0,\text{y}>0$ and which passes through the point (1, 1) is:

$\text{x}^{2}=\text{y}$
$\text{y}^{2}=\text{x}$
$\text{x}^{2}=2\text{y}$
$\text{y}^{2}=2\text{x}$

Answer

$\text{x}^{2}=\text{y}$

Solution:
We have,
$\frac{\text{dy}}{\text{dx}}=\frac{2\text{y}}{\text{x}}$
$\Rightarrow\frac{1}{2}\times\frac{1}{\text{y}}\text{dy}=\frac{1}{\text{x}}\text{dx}$
Interating both sides, we get
$\Rightarrow\frac{1}{2}\int\frac{1}{\text{y}}\text{dy}=\int\frac{1}{\text{x}}\text{dx}$
$\Rightarrow\frac{1}{2}\ \log{\text{y}}=\log{\text{x}}+\log\text{C}$
$\Rightarrow\log{\text{y}}^{\frac{1}{2}}-\log{\text{x}}=\log\text{C}$
$\Rightarrow\log\big(\frac{\sqrt{\text{y}}}{2}\big)=\log\text{C}$
$\Rightarrow\frac{\sqrt{\text{y}}}{2}=\text{C}$
$\Rightarrow\sqrt{\text{y}}=\text{Cx}\ ...(\text{i})$
As (i) passes through (1, 1), we get
$1=\text{C}$
Putting the value of C in (1), we get
$\Rightarrow\sqrt{\text{y}}=\text{x}$
$\Rightarrow{\text{y}}=\text{x}^{2}$

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Question 1001 Mark

The solution of $\frac{\text{dy}}{\text{dx}}=1+\text{x}+\text{y}+\text{xy}$ is:

$\text{x}-\text{y}=\text{k}(1+\text{xy})$
$\log(1+\text{y})=\text{x}+\frac{\text{x}^2}{2}+\text{k}$
$\log(1+\text{y})=\text{x}+\frac{\text{y}^2}{2}=\text{k}$
$\text{None of these}$

Answer

$\log(1+\text{y})=\text{x}+\frac{\text{x}^2}{2}+\text{k}$

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