Maths M.C.Q (1 Marks)

4. not defined.

The given differential equation is $\bigg(\frac{\text{d}^2\text{y}}{\text{dx}^2}\bigg)^3 + \bigg(\frac{\text{dy}}{\text{dx}}\bigg)^2+\text{sin} \bigg(\frac{\text{dy}}{\text{dx}}\bigg) + 1 =0 \ $

Since the differential equation is not a polynomial equation in its derivatives.

$\therefore$ its degree is not defined.

1. $\frac{\text{d}^3\text{y}}{\text{dx}^3}+\frac{\text{d}^2\text{y}\text{ dx}}{\text{dx}^2\text{dx}}+\text{y}=\text{x}$

Solution:

Linear equation is an equation between two variables that gives a straight line and order of linear equation is highest order derivative in linear equation.Since, in option A equation is linear equation in which highest order is 3.

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

Solution:

We have,

$\frac{\text{dy}}{\text{dx}}=\text{e}^{\text{x}+\text{y}}$

$\Rightarrow \frac{\text{dy}}{\text{dx}}=\text{e}^{\text{x}}\times\text{e}^{\text{y}}$

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

Integrating both sides, we get

$\int\text{e}^{-\text{y}}\text{dy}=\int\text{e}^{\text{x}}\text{dx}$

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

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

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

3. $\text{xe}^{\int\text{P}_{1}\text{dy}}=\int \left\{\text{Q}_{1}\text{e}^{\int\text{P}_{1}\text{dy}}\right\}\text{dy}+\text{C}$

Solution:

We have,

$\frac{\text{dx}}{\text{dy}}+\text{P}_{1}\text{x}=\text{Q}_{1}$

Comparing with the equation $\frac{\text{dx}}{\text{dy}}+\text{P}\text{x}=\text{Q}$ we get,

$\text{P}=\text{P}_{1}, \text{Q}=\text{Q}_{1}$

The solution of the equation $\frac{\text{dx}}{\text{dy}}+\text{P}\text{x}=\text{Q}$ is given by 

$\text{xe}^{\int\text{P}_{1}\text{dy}}=\int \left\{\text{Q}_{1}\text{e}^{\int\text{P}_{1}\text{dy}}\right\}\text{dy}+\text{C}\ ...(\text{i})$

Putting the value of P and Q in (i),

$\text{xe}^{\int\text{P}_{1}\text{dy}}=\int \left\{\text{Q}_{1}\text{e}^{\int\text{P}_{1}\text{dy}}\right\}\text{dy}+\text{C}$

4. $\tan1-1$

3. $\frac{\text{x}}{\text{y}}-\text{y}^2=\text{c}$

4. $\tan^{-1}\text{y}+\tan^{-1}\text{x}=\tan^{-1}\text{C}$

Solution:

We have,

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

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

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

Integrating both sides, we get

$\int\frac{1}{(1+\text{y}^{2})}\text{dy}=-\int\frac{1}{(1+\text{x}^{2})}\text{dx}$

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

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

2. 2

Solution:

$=\text{y}_3^\frac{2}{3}+2+3\text{y}_2+\text{y}_1=0$

$\Rightarrow\text{y}_3^\frac{2}{3}=-(2+3\text{y}+\text{y}_1)$

$\Rightarrow\text{y}^\frac{2}{3}=-(2+3\text{y}_2+\text{y}_1)^2$

cubing both sides,

Hence degree of given differential equation is 2 (power on y₃)

1. y = xe^a + c

Solution:

Calculation:

Given: $\text{In}\Big(\frac{\text{dy}}{\text{dx}}\Big)-\text{a}=0$

$\Rightarrow\text{In}\Big(\frac{\text{dy}}{\text{dx}}\Big)=\text{a}$

$\Rightarrow\frac{\text{dy}}{\text{dx}}=\text{e}^\text{a}$

$\Rightarrow\int\frac{\text{dy}}{\text{dx}}=\int\text{e}^\text{a}$

On integrating both sides, we get

$\Rightarrow\text{y}=\text{xe}^\text{a}+\text{c}$

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

Solution:

Given is, $\frac{\text{d}\text{y}}{\text{d}\text{x}}-\text{y}=1$

$\Rightarrow\frac{\text{d}\text{y}}{\text{d}\text{x}}=\text{y}+1$

$\Rightarrow\frac{\text{d}\text{y}}{1+\text{y}}=\text{dx}$

On integrating both sides, we get

$\log(1+\text{y})=\text{x}+\text{C}\ ......(\text{i})$

When x = 0 and y = 1, then

$\log2=0+\text{C}$

$\Rightarrow\text{C}=\log2$

The required solution is

$\log(1+\text{y})=\text{x}+\log2$

$\Rightarrow\log\Big(\frac{1+\text{y}}{2}\Big)=\text{x}$

$\Rightarrow\frac{1+\text{y}}{2}=\text{e}^\text{x}$

$\Rightarrow1+\text{y}=2\text{e}^\text{x}$

$\Rightarrow\text{y}=2\text{e}^\text{x}-1$

4. $\text{y}\sin\text{x}=\text{x}+\text{C}$

Solution:

$\frac{\text{dy}}{\text{dx}}+\text{y}\cot\text{x}=\text{coses}\ \text{x}$

Comparting with $\frac{\text{dy}}{\text{dx}}+\text{Py}=\text{Q}$ we get

$\text{P}=\cot\text{x}$

$\text{Q}=\text{coses} \ \text{x}$

Now,

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

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

$=\sin\text{x}$

So, the solution is given by

$\Rightarrow \text{y}\sin\text{x}=\int\sin\text{x}\times\text{cosec}\ \text{x}\text{dx}+\text{C}$

$\Rightarrow \text{y}\sin\text{x}=\text{x}+\text{C}$

3. 3

Solution:

The general equation of all conics with center at origin can be written as

ax² + 2hxy + by² + c = 0

Dividing by aa, we get

$\text{x}^2+\Big(\frac{2\text{h}}{\text{a}}\Big)\text{ax}+\Big(\frac{\text{b}}{\text{a}}\Big)\text{y}^2+\Big(\frac{\text{c}}{\text{a}}\Big)=0$

Since, it has three arbitrary constants.So, the differential equation is of order 3.

3. $\text{y e}^\text{x}+\text{x}^2=\text{C}$

The given differential equation is

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

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

$\text{Comparing it with }\frac{\text{dy}}{\text{dx}}+\text{Py}=\text{Q},$ $\text{we get},\ \text{P}=1,\ \text{Q}=-2\text{x e}^{-\text{x}}$

$\therefore\ \ \int\text{P dx}=\int1\ \text{dx}=\text{'x},\ \text{e}^{\int\text{P dx}}=\text{e}^\text{x}$

Solution of given differential equation is

$\text{y e}^{\int\text{P dx}}=\int\text{Q e}^{\int\text{P dx}}+\text{C}$ $\text{or}\ \ \text{y e}^\text{x}=\int(-2\text{x e}^\text{x}).\text{e}^{-\text{x}}\ \text{dx}+\text{C}$

$\text{or}\ \ \text{y e}^\text{x}=-2\int\text{x dx}+\text{C}\ \ \text{or}\ \ \text{y e}^\text{x}=-\text{x}^2+\text{C}$

$\text{or}\ \ \text{y e}^\text{x}+\text{x}^2=\text{C}$

$\therefore$ (C) is correct answer.

2. $\text{a}=-\text{b}$

Solution:

We have,

$​​\frac{\text{dy}}{\text{dx}}=\frac{\text{ax}+\text{g}}{\text{by}+\text{f}}$

$\Rightarrow (\text{by}+\text{f})\text{dy}=(\text{ax}+\text{g})\text{dx}$

Intergrating both sides, we get

$\Rightarrow \int(\text{by}+\text{f})\text{dy}=\int(\text{ax}+\text{g})\text{dx}$

$\Rightarrow \text{b}\frac{\text{y}^{2}}{2}+\text{fy}=\text{a}\frac{\text{x}^{2}}{2}+\text{gx}+\text{C}$

$\Rightarrow \text{b}\frac{\text{y}^{2}}{2}+\text{fy}-\text{a}\frac{\text{x}^{2}}{2}-\text{gx}=\text{C}$

$\Rightarrow \text{b}\text{y}^{2}+2\text{fy}-\text{a}\text{x}^{2}-2\text{gx}-2\text{C}=0$

The above equation resprasents a circle.

Therefore, the coffrcients of x² and y² must be equal.

$-\text{a}=\text{b}$

$\Rightarrow \text{a}=-\text{b}$ 

1. $\text{e}^{\text{x}}\cos\text{y}=\text{k}$

Solution:

Given is, $\text{e}^{\text{x}}\cos\text{ydx}-\text{e}^\text{x}\sin\text{ydy}=0$

$\Rightarrow\text{e}^{\text{x}}\cos\text{ydx}=\text{e}^\text{x}\sin\text{ydy}$

$\Rightarrow\frac{\text{d}\text{y}}{\text{d}\text{x}}=\tan\text{ydy}$

$\Rightarrow\text{dx}=\tan\text{ydy}$

On integrating both sides, we get

$\text{x}=\log\sec\text{y}+\text{C}$

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

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

$\Rightarrow\frac{1}{\cos\text{y}}=\frac{\text{e}^\text{x}}{\text{e}^\text{c}}$

$\Rightarrow\text{e}^{\text{x}}\cos\text{y}=\text{k}$ $[\text{where},\text{K}=\text{e}^\text{c}]$

4. $\text{None of these}$

2. 2 only

Solution:

$\frac{\text{d}^2\text{y}}{\text{d}\text{x}^2}+\cos\Big(\frac{\text{dx}}{\text{dy}}\Big)=0:$ 

The order of differential equation is 2.

Degree of a differential equation is power of highest order differential equation

$\therefore$ The degree of this differential equation is 1.

1. $(\text{x}^{2}-\text{y}^{2})\ \frac{\text{dy}}{\text{dx}}=2\text{xy}$

Solution:

Given equation is, $\text{x}^2+\text{y}^2-2\text{ay}=0$

$\Rightarrow\frac{\text{x}^2+\text{y}^2}{\text{y}}=2\text{a}$

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

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

$\Rightarrow2\text{xy}+2{\text{y}^2\frac{\text{dy}}{\text{dx}}}-(\text{x}^2+\text{y}^2)\frac{\text{dy}}{\text{dx}}=0$

$\Rightarrow(2\text{y}^2-\text{x}^2-\text{y}^2)\frac{\text{dy}}{\text{dx}}=-2\text{xy}$

$\Rightarrow(\text{y}^2-\text{x}^2)\frac{\text{dy}}{\text{dx}}=-2\text{xy}$

$\Rightarrow(\text{x}^2-\text{y}^2)\frac{\text{dy}}{\text{dx}}=2\text{xy}$

2. $\sec\text{x}$

Solution:

Given that, $\frac{\text{d}\text{y}}{\text{d}\text{x}}+\text{y}\tan\text{x}-\sec\text{x}=0$

$\Rightarrow\frac{\text{d}\text{y}}{\text{d}\text{x}}+\text{y}\tan\text{x}=\sec\text{x}$

$\Big($It is a linear differential equation of form $\frac{\text{d}\text{y}}{\text{d}\text{x}}+\text{P}\text{y}=\text{Q}\Big)$

Here, $\text{P}=\tan\text{x},\text{Q}=\sec\text{x}$

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

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

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

$=\sec\text{x}$

1. $\text{x}(\text{y}+\cos\text{x})=\sin\text{x}+\text{C}$

Solution:

We have,

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

$\Rightarrow \frac{\text{dy}}{\text{dx}}+\frac{1}{\text{x}}\text{y}=\sin\text{x}\ ...(\text{i})$

Comparing with we get,

$\text{P}=\frac{1}{\text{x}}$

$\text{Q}=\sin\text{x}$

Now,

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

$=\text{e}^{\log|\text{x}|}$

$=\text{x}$

Therefore, intergration of (i) is given by,

$\text{y}\times\text{I.F}=\int\text{x}^{2}\times\text{I.F.}\ \text{dx}+\text{C}$

$\Rightarrow\text{yx}=\int\text{x}\ \sin\text{x}\ \text{dx}+\text{C}$

$\Rightarrow\text{yx}=\text{x}\int\sin\text{x}\ \text{dx}-\int\Big[\frac{\text{d}}{\text{dx}}(\text{x})\int\sin\text{x}\ \text{dx}\Big]\text{dx}+\text{C}$

$\Rightarrow\text{yx}=-\text{x}\cos\text{x}+\int\cos\text{x}\ \text{dx}+\text{C}$

$\Rightarrow\text{yx}+\text{x}\cos\text{x}=\sin\text{x}+\text{C}$

$\Rightarrow\text{x}(\text{y}+\cos\text{x})=\sin\text{x}+\text{C}$

3. $\text{y}^{3}+2\text{x}-3\text{x}^{2}\text{y}=0$

Solution:

We have,

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

$\Rightarrow (\text{xy}+\text{y}^{4})\text{dx}=(\text{xy}^{3}-\text{x}^{2})\text{dy}=0$

$\Rightarrow\text{xy}\ \text{dx}+\text{y}^{4}\text{dx}-\text{xy}^{3}\text{dy}+\text{x}^{2}\text{dy}=0$

$\Rightarrow \text{x}(\text{y}\text{dx}+\text{x}\text{dy})+\text{y}^{3}(\text{y}\text{dx}-\text{x}\text{dy})=0$

$\Rightarrow \text{xd}(\text{xy})+\text{x}^{2}\text{y}^{3}\frac{(\text{y}\text{dx}-\text{x}\text{dy})}{\text{x}^{2}}=0$

$\Rightarrow \frac{\text{d}(\text{xy})}{\text{x}^{2}\text{y}^{2}}-\frac{\text{y}}{\text{x}}\text{d}\Big(\frac{\text{y}}{\text{x}}\Big)=0$

$\Rightarrow \frac{\text{d}(\text{xy})}{\text{x}^{2}\text{y}^{2}}-\frac{\text{y}}{\text{x}}\text{d}\Big(\frac{\text{y}}{\text{x}}\Big)$

Integrating both sides we get,

$\Rightarrow \int\frac{\text{d}(\text{xy})}{\text{x}^{2}\text{y}^{2}}=\int\frac{\text{y}}{\text{x}}\text{d}\Big(\frac{\text{y}}{\text{x}}\Big)$

$\Rightarrow-\frac{1}{\text{xy}}=\frac{\Big(\frac{\text{y}}{\text{x}}\Big)^{2}}{2}-\text{C}$

$\Rightarrow-\frac{1}{\text{xy}}-\frac{1}{2}\Big(\frac{\text{y}^{2}}{\text{x}^{2}}\Big)+\text{C}=0$

$\Rightarrow \text{y}^{3}+2\text{x}+2\text{Cx}^{2}\text{y}=0$

It is given that the curve passes through (1, 1).

Hence,

$\text{y}^{3}+2\text{x}+2\text{Cx}^{2}\text{y}=0$

$(1)^{3}+2(1)+2\text{C}(1)(1)=0$

$1+2+2\text{C}=0$

$2\text{C}=-3$

$\text{C}=-\frac{3}{2}$

The required curve is,

 $\text{y}^{3}+2\text{x}-3\text{x}^{2}\text{y}=0$ 

4. x³ + y³ = 12x + C

Solution:

We have,

$\text{x}^{2}+\text{y}^{2}\frac{\text{dy}}{\text{dx}}=4$

$\Rightarrow\text{y}^{2}\frac{\text{dy}}{\text{dx}}=4-\text{x}^{2}$

$\Rightarrow\text{y}^{2}\frac{\text{dy}}{\text{dx}}=(4-\text{x}^{2})\text{dx}$

Integrating both sides, we get

$\int\text{y}^{2}\frac{\text{dy}}{\text{dx}}=\int(4-\text{x}^{2})\text{dx}$

$\Rightarrow \frac{\text{y}^{3}}{3}=4\text{x}-\frac{\text{x}^{3}}{3}+\text{D} $

$\Rightarrow \text{y}^{3}=12\text{x}-\text{x}^{3}+3\text{D}$

$\Rightarrow \text{x}^{3}+\text{y}^{3}=12\text{x}+\text{C}$

3. $\text{u}=(\log\text{z})^{-1}$

Solution:

We have,

$\frac{\text{dz}}{\text{dx}}+\frac{\text{z}}{\text{x}}\log\text{z}=\frac{\text{z}}{\text{x}^{2}}(\log\text{z})^{2}\ ...(\text{i})$

Let $\text{u}=(\log\text{z})^{-1}$

$\frac{\text{du}}{\text{dx}}=-\frac{1}{(\log)^{2}}\times\frac{1}{\text{z}}\times\frac{\text{dz}}{\text{dx}}$

$\frac{\text{du}}{\text{dx}}=-\text{z}(\log\text{z})^{2}\frac{\text{du}}{\text{dx}}$

Substituting the value of the equation (i),

$-\text{z}(\log\text{z})^{2}\frac{\text{du}}{\text{dx}}+\frac{\text{z}}{\text{x}}\log\text{z}=\frac{\text{z}}{\text{x}^{2}}(\log\text{z}^{2})$

$\frac{\text{du}}{\text{dx}}-\frac{1}{\text{x}}\frac{1}{\log\text{z}}=-\frac{1}{\text{x}^{2}}$

$\frac{\text{du}}{\text{dx}}-\frac{1}{\text{x}}({\log\text{z}})^{-1}=-\frac{1}{\text{x}^{2}}$

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

It can be written as,

$\frac{\text{du}}{\text{dx}}+\text{P}(\text{x})\text{u}=\text{Q}(\text{x})$

Where, $\text{p}(\text{x})=-\frac{1}{\text{x}}$

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

3. $\text{e}^{\text{y}}=\text{e}^{\text{x}^{2}}+\text{C}$

Solution:

We have $\frac{\text{dy}}{\text{dx}}=2\text{x}\ \text{e}^{\text{x}^{2}-\text{y}}$

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

$\Rightarrow\int\text{e}^{\text{y}}\text{dy}=2\int\text{x}\text{e}^{\text{x}^{2}}\text{dx}$

Put $\text{x}^2=\text{t}$ in R.H.S. integral we get

$2\text{x}\text{dx}=\text{dt}$

$\Rightarrow\int\text{e}^{\text{y}}\text{dy}=\int\text{e}^{\text{t}}\text{dt}$

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

$\Rightarrow\text{e}^{\text{y}}=\text{e}^{\text{x}^2}+\text{C}$

2. $\text{y}=\frac{\sqrt{1+\text{x}^2}}{\text{x}}+\frac{\text{c}}{\text{x}}$

2. One.

Solution:

Given 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.

1. xy = c

Solution:

Given expression 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

$\int\frac{\text{dy}}{\text{y}}+\frac{\text{dx}}{\text{x}}=0$

$\Rightarrow\text{In}\text{ y}+\text{In}\text{ x}=\text{c}$

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

2. $\text{g}(\text{x})+\log(1+\text{y}-\text{g}(\text{x}))=\text{C}$

Solution:

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

$\log\Big[\text{y+1}-\text{g}(\text{x})\Big]=-\text{g}(\text{x})+\log\text{K}$

1. 1.

Solution:

Given is, y = Ax + A³

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.

$\therefore\text{Degree}=1$

1. x = Cy²

Solution:

It is given that subtangent at any point of a curve is doble of the abscissa.

$\therefore \frac{\text{y}}{\frac{\text{dy}}{\text{dx}}}=2\text{x}$

$\text{y}=2\text{x}\frac{\text{dy}}{\text{dx}}$

$\int\frac{\text{dx}}{\text{x}}=2\int\frac{\text{dy}}{\text{y}}$

$\log\text{x}=2\log\text{y}+\text{a}$

$\log\text{x}=\log\text{y}^{2}+\log\text{C}$

$\log\text{x}=\log\text{Cy}^{2}$

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

3. $\text{x e}^{\int\text{P}_1\text{dy}}=\int\Big(\text{Q}_1\text{e}^{\int\text{P}_1\text{dy}}\Big)\text{dy}+\text{C}$

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.

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, 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,

$\text{P}=\frac{1}{\text{x}\log\text{x}}, \text{Q}=\frac{2}{\text{x}}$

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

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

$=\log\text{x}$

3. $\sec\text{x}$

Solution:

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 $\text{Q}=\sec\text{x}$

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

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

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

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

Solution:

Concept:

$\int\limits\frac{\text{dx}}{1+\text{x}^2}\tan^{-1}\text{x}=\text{c}$

Calculation:

Given: dy = (1 + y²) dx

$\Rightarrow\frac{\text{dy}}{1+\text{y}^2}\text{dx}$

Integrating both sides, we get

$\Rightarrow\int\frac{\text{dy}}{1+\text{y}^2}=\int\text{dx}$

$\Rightarrow\tan^{-1}\text{y}=\text{x}+\text{c}$

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

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

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

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

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

$\Rightarrow \frac{\text{d}^{2}\text{y}}{\text{dx}^{2}}=0$

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

$\Rightarrow \frac{\text{d}^{2}\text{y}}{\text{dx}^{2}}+\text{x}\times\text{x}=\text{x}^{2}\times1$

$\Rightarrow \frac{\text{d}^{2}\text{y}}{\text{dx}^{2}}+\text{xy}=\text{x}^{2}\times1$

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

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

2. $2$

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.

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

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}$

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

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

3. ye^x + x² = C

Solution:

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,

$\text{y}\times\text{I.F}=\int(\text{Q}\times\text{I.F}) \text{dx}+\text{C}$

$\Rightarrow \text{ye}^{\text{x}}=-\int\text{e}^{\text{x}}\times \frac{2\text{x}}{\text{e}^{\text{x}}}\text{dx}+\text{C}$

$\Rightarrow \text{ye}^{\text{x}}=-2\int\text{x}\ \text{dx}+\text{C}$

$\Rightarrow \text{ye}^{\text{x}}=-\text{x}^{2}+\text{C}$

$\Rightarrow \text{ye}^{\text{x}}+\text{x}^{2}=\text{C}$

2. 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