Maths M.C.Q (1 Marks)

1. 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.

1. $\pm3$

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,

$=\text{k}^2-9=0$

$[\text{Since }\sin\text{kt}\neq 0]\text{or,}$

$=\text{k}=\pm3.$

3. $\log\text{x}$

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, 

$\text{I.F}=\text{e}^{\int\text{P}\text{dx}}$

$\text{I.F}=\text{e}^{\int\frac{1}{\text{x}\log\text{x}}}\text{dx}$

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

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

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

$\Rightarrow \text{e}^{-\text{u}}=\text{x}-\text{C}$

$\Rightarrow -1=\text{e}^{-\text{u}}(\text{x}-\text{C})$

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

1. 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.

$\therefore\text{Degree}=1$

2. 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.

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

2. $\sec\text{x}$

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,

$\text{I.F}=\text{e}^{\int\tan\text{x}}\text{dx}$

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

$=\sec\text{x}$

3. 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

$\text{y}=\text{x}\frac{\text{dy}}{\text{dx}}+\Big(\frac{\text{dy}}{\text{dx}}\Big)^3$

$\therefore\text{Order}=1$

4. 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.

4.  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. 

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

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

$\log\text{y}=\frac{\text{y}_{1}}{\text{y}}\text{x}$

$\Rightarrow \text{y}\ \log\text{y}=\text{y}_{1}\text{x}$

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

3. 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

2. $\text{y}=\text{x}\text{e}^{-\text{x}}$

Solution:

We have, $\frac{\text{d}\text{y}}{\text{d}\text{x}}+\text{y}=\text{e}^{\text{-x}}$

This is a linear differential equation.

On comparing it with $\frac{\text{d}\text{y}}{\text{d}\text{x}}+\text{P}\text{y}=\text{Q},$we get

$\text{P}=1,\text{Q}=\text{e}^{-\text{x}}$

$\text{I.F.}=\text{e}^{\int\text{Pdx}}=\text{e}^{{\int\text{dx}}}​​$

So, the general solution is,

$\text{y}.\text{e}^{\text{x}}=\int\text{e}^{-\text{x}}\text{e}^{\text{x}}\text{dx}+\text{C}$

$\Rightarrow\text{y}.\text{e}^{\text{x}}=\int\text{dx}+\text{C}$

$\Rightarrow\text{y}.\text{e}^{\text{x}}=\text{x}+\text{C}$

Given that when x = 0 and y = 0

$\Rightarrow0 = 0 + \text{C}$

$\Rightarrow\text{C}=0$

Eq. (i) becomes $\text{y}.\text{e}^{\text{x}}=\text{x}$

$\Rightarrow\text{y}=\text{x}\text{e}^{-\text{x}}$

4. $\tan\text{x}.\tan\text{y}=\text{k}$

Solution:

We have, $\tan\text{y}\sec^2\text{xdx} + \tan\text{x }\sec^2\text{ydy}=0$

$\Rightarrow\tan\text{y}\sec^2\text{xdx} =- \tan\text{x }\sec^2\text{ydy}$

$\Rightarrow\frac{\sec^2\text{x}}{\tan\text{x}}\text{dx}=-\frac{\sec^2\text{y}}{\tan\text{y}}\text{dy}$

$\Rightarrow\int\frac{\sec^2\text{x}}{\tan\text{x}}\text{dx}=-\int\frac{\sec^2\text{y}}{\tan\text{y}}\text{dy}$

$\Rightarrow\log\tan\text{x}=-\log\tan\text{y}+\log\text{k}$

$\Rightarrow\log\tan\text{x}+\log\tan\text{y}=\log\text{k}$

$\Rightarrow\log(\tan\text{x}\tan\text{y})=\log\text{k}$

$\Rightarrow\tan\text{x}\tan\text{y}=\text{k}$

3. y =Cx

Solution:

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,

$\int\frac{1}{\text{y}}\ \text{dy}=\int\frac{1}{\text{x}}\ \text{dx}$

$\Rightarrow \log\text{y}=\log\text{x}+\text{D}$

$\Rightarrow \log\text{y}-\log\text{x}=\text{C}$

$\Rightarrow \log\big(\frac{\text{y}}{2}\big)=\log\text{C}$

$\Rightarrow \frac{\text{y}}{2}=\text{C}$

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

3. 1

Solution:

Order:- The number of the highest derivative in a differential equation, that is

$=\frac{\text{dy}}{\text{dx}}=1$

2. $\frac{\text{e}^{\text{x}}}{\text{x}}$

Solution:

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, $\text{P}=\frac{-(1-\text{x})}{\text{x}},\text{Q}=\frac{1}{\text{x}}$

$\text{I.F.}=\text{e}^{\int\text{Pdx}}$

$=\text{e}^{-\int\frac{1+\text{x}}{\text{x}}\text{dx}}=\text{e}^{\frac{\text{x}-1}{\text{x}}}$

$=\text{e}^{{\int\Big(1-\frac{1}{\text{x}}\Big)}\text{dx}}=\text{e}^{\int\text{x}-\log\text{x}}$

$=\text{e}^{\text{x}}.\text{e}^{\log\Big(\frac{1}{\text{x}}\Big)}=\text{e}^{\text{x}}.\frac{1}{\text{x}}$

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

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

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

$\Rightarrow\sqrt{\text{y}}=\text{x}$

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

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

1. $\text{y}=\sin\frac{1}{\text{x}}-\cos\frac{1}{\text{x}}$

3. $\frac{\text{dy}}{\text{dx}}-\text{m} ^2\text{y}=0$

Solution:

Given that, $\text{y}=\text{a}\text{e}^{\text{mx}}+\text{b}\text{e}^{- \text{mx}}$

On differentiating both sides w.r.t.x, we get

$\frac{\text{dy}}{\text{dx}}=\text {ma}\text{e}^{\text{mx}}+\text{bm}\text{e}^{-\text {mx}}$

Again, differentiating both sides w.r.t.x, we get

$\frac{\text{d}^2\text{y}}{\text {d}\text{x}^2}=\text{m}^2\text{a}\text{e}^{\text{mx}}+ \text{b}\text{m}^2\text{e}^{-\text{mx}}$

$\Rightarrow\frac{\text{d}^2\text {y}}{\text{d}\text{x}^2}=\text{m}^2(\text{a}\text{e}^{\text {mx}}+\text{b}\text{e}^{-\text{mx}})$

$\Rightarrow\frac{\text{d}^2\text {y}}{\text{d}\text{x}^2}=\text{m}^2\text{y}$

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

3. $(1+\text{x}^2)\text{dy}+(1+\text{y}^2)\text{dx}=0$

Solution:

Given is, $\tan^{-1}+\tan^{-1}\text{y}=\text{C}$

On differentiating above eqaution w.r. t. x, we get

$\frac{1}{1+\text{x}^2}+\frac{1}{1+\text{y}^2}.\frac{\text{d}\text{y}}{\text{d}\text{x}}=0$

$\Rightarrow\frac{1}{1+\text{y}^2}.\frac{\text{d}\text{y}}{\text{d}\text{x}}=-\frac{1}{1+\text{x}^2}$

$\Rightarrow(1+\text{x}^2)\text{dy}+(1+\text{y}^2)\text{dx}=0$

1. 1, 2

Solution:

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

3. log y + cos x - c = 0

Solution:

Concept:

$\int\frac{\text{dx}}{\text{x}}=\log\text{x}+\text{c}$

$\int\sin{\text{x}}{\text{ dx}}=-\cos\text{x}+\text{c}$

Calculation:

Given: $\frac{\text{dx}}{\text{dy}}=\text{y}\sin\text{x}$

$\Rightarrow\frac{\text{dx}}{\text{dy}}=\sin\text{x}\text{ dx}$

Integrating both sides, we get

$\Rightarrow\int\frac{\text{dy}}{\text{y}}=\int\sin\text{x}\text{ dx}$

$\Rightarrow\log\text{y}=-\cos\text{x}+\text{c}$

$\Rightarrow\log\text{y}+\cos\text{x}-\text{c}=0$

4. y² dx + (x² - xy - y²)dy

Solution:

A differential equation is said to be homogenous if all the in the terms in the equation have equal degree and it can be written in the from $\frac{\text{dy}}{\text{dx}}=\frac{\text{f}(\text{x,}\text{y})}{\text{g}(\text{x,}\text{y})}.$

In (a), (b) and (c), the degree of all the terms is not equal.

But in the equation y² dx + (x² - xy - y²)dy = 0, the degree of all the terms is 2.

Thus, (d) constant a homogeneous differential equation.

2. $\frac{\text{d}^2\text{y}}{\text{dx}^2}+\text{a}^2\text{y}=0$

2. $\text{y}=\text{kx}$

Solution:

We have,

$\frac{\text{dy}}{\text{dx}}=\frac{\text{y}}{\text{x}}$

$\Rightarrow\frac{\text{1}}{\text{y}}\text{dy}=\frac{\text{1}}{\text{x}}\text{dx}$

Integrating both sides, we get

$\int\frac{\text{1}}{\text{y}}\text{dy}=\int\frac{\text{1}}{\text{x}}\text{dx}$

$\log\text{y}=\log\text{x}+\log\text{k}$

$\log\text{y}-\log\text{x}=\log\text{k}$

$\log\frac{\text{y}}{\text{x}}=\log\text{k}$

$\Rightarrow\frac{\text{y}}{\text{x}}=\text{k}$

$\Rightarrow\text{y}=\text{k}{\text{x}}$

2. $(2,\frac{3}{2})$

Solution:

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.

3. $\sqrt{1-\text{x}^2}$

4. Not defined

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:

$\frac{\text{d}^2\text{y}}{\text{dx}^2}+3\Big(\frac{\text{dy}}{\text{dx}}\Big)^2=\text{x}^2\log\Big(\frac{\text{d}^2\text{y}}{\text{dx}^2}\Big)$

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.

3. $\sqrt{1-\text{x}^2}$

Solution:

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, $\text{I.F.}=\text{e}^{\frac{1}{2}\int\frac{\text{dt}}{\text{t}}}$

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

$\Rightarrow\sqrt{1-\text{x}^2}$

1. 1, 2

Solution:

The 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.

3. 3

Solution:

Equation of parabola :- (x - h)² = 4a (y - k)

No of constants in the alove equation are 3

$\therefore$ order of diff eq = 3

1. 1

Solution:

The 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.

4. 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.

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

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

$\int\big(\frac{1}{1+\text{u}^{2}}\big)\text{du}=\int\frac{1}{\text{x}}\text{dx}$

$\Rightarrow \tan^{-1}\text{u}=\log\text{x}+\text{C}$

$\Rightarrow \tan^{-1}\big(\frac{\text{y}}{2}\big)=\log\text{x}+\text{C}$

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

Solution:

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

3. c(x + y) = xy where c is the constant of integration

Solution:

$=\text{x}^2\text{dy}+\text{y}^2\text{dx}=0$

$\Rightarrow\frac{\text{dy}}{\text{y}^2}+\frac{\text{dx}}{\text{x}^2}=0$

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}$

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

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

$\frac{​​​\text{y dx}-​​​\text{x dy}}{​​​\text{y}}=0$

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

$\therefore\ \ \frac{\text{x}}{\text{y}}=\text{constant}.$

$\therefore\ \ \frac{\text{y}}{\text{x}}=\text{C}\ \ \text{or}\ \ \text{y}=\text{Cx}$