Maths M.C.Q (1 Marks)

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

Solution:

We have,

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

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

Let $\text{y}=\upsilon\text{x}$

$\Rightarrow \frac{\text{dy}}{\text{dx}}=\upsilon+\text{x}\frac{\text{d}\upsilon}{\text{dx}}$

Putting both value in (i)

$\upsilon+\text{x}\frac{\text{d}\upsilon}{\text{dx}}=\upsilon+\tan\upsilon$

$\Rightarrow \frac{\text{d}\upsilon}{\tan\upsilon}=\frac{\text{dx}}{\text{x}}$

Integrating both sides, we get

$\log\sin\upsilon=\log\text{x}+\log\text{C}$

$\Rightarrow \log\frac{\sin\upsilon}{\text{x}}=\log\text{C}$

$\Rightarrow \frac{\sin\upsilon}{\text{x}}=\text{C}$

$\Rightarrow\sin\upsilon=\text{Cx}$

$\Rightarrow\sin(\frac{\text{y}}{\text{x}})=\text{Cx}$

4. parabolas

Solution:

We have,

$2\text{x}\frac{\text{dy}}{\text{dx}}-\text{y}=3$

$\Rightarrow 2\text{x}\frac{\text{dy}}{\text{dx}}=3+\text{y}$

$\Rightarrow \frac{1}{3+\text{y}}\text{dy}=\frac{1}{2\text{x}}\text{dx}$

Interating both sides, we get

$\Rightarrow \int\frac{1}{3+\text{y}}\text{dy}=\frac{1}{2}\int\frac{1}{\text{x}}\text{dx}$

$\Rightarrow \log|3+\text{y}|=\frac{1}{2}\log|\text{x}|+\log|\text{C}|$

$\Rightarrow\log|\frac{3+\text{y}}{\sqrt{\text{x}}}|=\log\text{C}$

$\Rightarrow\frac{3+\text{y}}{\sqrt{\text{x}}}=\text{C}$

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

Squaring both sides, we get

$(3+\text{y})^{2}=\text{C}{\text{x}}\ ...(\text{i})$

Thus, (i) the equation of parabolas.

3. K < 0

Solution:

We have,

$\Rightarrow \frac{\text{dy}}{\text{dx}}-\text{Ky}=0$

$\Rightarrow \frac{\text{dy}}{\text{dx}}=\text{Ky}$

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

Integrating both sides, we get

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

$\Rightarrow \log|\text{y}|=\text{Kx}+\text{C}\ ...(\text{i})$

Now,

$\text{y}(0)=1$

$\text{C}=0$

Putting C = 0 in (i),

$\log|\text{y}|=\text{Kx}$

$\Rightarrow \text{e}^{\text{Kx}}=\text{y}$

According to the quation,

$\text{e}^{\text{K}\propto}=0$

2. $\text{xyy}_{2}+\text{xy}_{1}^{2}-\text{yy}_{1}=0$

Solution:

We have,

$\text{ax}^{2}+\text{by}^{2}=1\ ...(\text{i})$

Differential both sides of (i) with x, we get

$2\text{ax}+2\text{by}\frac{\text{dy}}{\text{dy}}=0\ ...(\text{ii})$

Differential both sides of (ii) with x, we get

$2\text{ax}+2\text{b}\Big(\frac{\text{dy}}{\text{dy}}\Big)^{2}+2\text{}by\frac{\text{d}^{2}\text{y}}{\text{dx}^{2}}=0$

$\Big[\text{y}\frac{\text{d}^{2}\text{y}}{\text{dx}^{2}}\Big(\frac{\text{dy}}{\text{dx}^{2}}\Big)\Big]=-\frac{2\text{a}}{2\text{b}}$

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

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

$\text{xyy}_{2}+\text{x}(\text{y}_{1}^{2})-\text{yy}_{1}=0$

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.

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

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

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

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

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

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

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

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. x = vy

Solution:

A homogeneous differential of the from $\frac{\text{dx}}{\text{dy}}=\text{h}(\frac{\text{x}}{\text{y}})$ can be solved by sunstituting x = vy.

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,

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

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

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

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

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

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, 

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

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

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

3. ​​​​​$\frac{1}{\text{x}}$

Solution:

We have,

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

$\Rightarrow \frac{\text{dy}}{\text{dx}}-\frac{1}{\text{x}}\text{y}=2\text{x}$

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

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

Now,

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

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

$=\text{e}^{\log|\frac{1}{\text{x}}|}$

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

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

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

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

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

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,

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

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

$=\sec\text{x}$

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

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

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

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

4. $\text{y}=\tan\big(\text{x}+\frac{\text{x}^{2}}{2}\big)$

Solution:

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),

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

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

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,

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

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

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

$=\log\text{x}$

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. $\text{y}=\text{C}_{1}\text{e}^{\text{C}_{2}\text{x}}+\text{C}_{3}$

Solution:

$\text{y}_{1}\text{y}_{3}=\text{y}^{2}_{2}$

$\frac{\text{y}_{3}}{\text{y}_{2}}=\frac{\text{y}_{2}}{\text{y}_{1}}$

$\Rightarrow \frac{\Big(\frac{\text{d}^{3}\text{y}}{\text{dx}^{3}}\Big)}{\Big(\frac{\text{d}^{2}\text{y}}{\text{dx}^{2}}\Big)}=\frac{\Big(\frac{\text{d}^{2}\text{y}}{\text{dx}^{2}}\Big)}{\Big(\frac{\text{d}\text{y}}{\text{dx}}\Big)}$

$\Rightarrow\int \frac{\frac{\text{d}}{\text{dx}}\Big(\frac{\text{d}^{3}\text{y}}{\text{dx}^{3}}\Big)}{\Big(\frac{\text{d}^{2}\text{y}}{\text{dx}^{2}}\Big)}=\int\frac{\frac{\text{d}}{\text{dx}}\Big(\frac{\text{d}\text{y}}{\text{dx}}\Big)}{\Big(\frac{\text{d}\text{y}}{\text{dx}}\Big)}$

$\Rightarrow\log\Big(\frac{\text{d}^{2}\text{y}}{\text{dx}^{2}}\Big)=\log\Big(\frac{\text{dy}}{\text{dx}}\Big)+\log\text{C}_{4}$

$\Rightarrow \frac{\text{d}^{2}\text{y}}{\text{dx}}^{2}=\text{C}_{4}\frac{\text{dy}}{\text{dx}}$

$\Rightarrow\int \frac{\frac{\text{d}}{\text{dx}}\Big(\frac{\text{d}^{3}\text{y}}{\text{dx}^{3}}\Big)}{\Big(\frac{\text{d}^{2}\text{y}}{\text{dx}^{2}}\Big)}=\int\text{C}_{4}\ \text{dx}$

$\Rightarrow \log\Big(\frac{\text{dy}}{\text{dx}}\Big)=\text{C}_{4}\text{x}+\text{C}_{5}$

$\Rightarrow \int\text{dy}=\int\text{e}^{\text{C}_{4}\text{x+}\text{C}_{5}}\ \text{dx}$

$\Rightarrow\text{y}=​​\frac{\text{e}^{\text{C}_{4}\text{x}+\text{C}_{3}}}{\text{C}_{4}}+\text{C}_{6}$

$\Rightarrow\text{y}=​\text{C}_{1}\text{e}^{\text{C}_{2}\text{x}}+\text{C}_{3}$

Where,

$\text{C}_{1}=\frac{\text{e}^{\text{C}_{5}}}{\text{C}_{4}}$

$\text{C}_{4}=\text{C}_{2}$

$\text{C}_{6}=\text{C}_{3}$

3. $\sec\text{x}$

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,

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

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

$=\sec{\text{x}}$

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

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,

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

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

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

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 $\text{z}=\text{y}^{1-\text{n}}.$ 

Solution:

We 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.

1. $\phi(\frac{\text{y}}{\text{x}})=\text{Kx}$

Solution:

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}$

$\Rightarrow \log|\phi(\text{v})|=\log|\text{x}|+\log|\text{K}|$

$\Rightarrow \log|\phi(\frac{\text{y}}{2})|-\log|\text{x}|=\log\text{K}$

$\Rightarrow \log|\phi(\frac{\text{y}}{2})|=\log\text{K}$

$\Rightarrow\phi(\frac{\text{y}}{2})|=\text{Kx}$

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.

3. $\text{y}''=-\omega^{2}\text{y}$

Solution:

We have,

$\text{y}=\text{A}\cos\omega\text{t}+\text{B}\sin\omega\text{t}\ ...(\text{i})$

Differentiating both sides of (i) with respect to x, we get

$\frac{\text{dy}}{\text{dt}}=-\text{A}\omega\sin\omega\text{t}+\text{B}\omega\cos\omega\text{t}\ ...(\text{ii})$

Differentiating both sides of (ii)

$\frac{\text{d}^{2}\text{y}}{\text{dt}^{2}}=-\text{A}\omega^{2}\cos\omega\text{t}+\text{B}\omega^{2}\sin\omega\text{t}$

$\Rightarrow \frac{\text{d}^{2}\text{y}}{\text{dt}^{2}}=-\omega^{2}(\text{A}\cos\omega\text{t}+\text{B}\sin\omega\text{t})$

$\Rightarrow \frac{\text{d}^{2}\text{y}}{\text{dt}^{2}}=-\omega^{2}\text{y}$

$\text{y}''=-\omega^{2}\text{y}$

1. 1

Solution:

The order of a differention depends on the number of constent in it.

Since $\sqrt{1-\text{x}^{4}}+\sqrt{1-\text{y}^{4}}=\text{a}(\text{x}^{2}-\text{y}^{2})$ constant only 1 constant, the order of the differential equation is 1.

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),

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

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

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

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

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

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

1. x = Cy²

Solution:

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

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

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

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

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

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

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

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

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

Solution:

We have,

$\text{y}=\text{C}_{1}\text{e}^{\text{x}}+\text{C}_{2}\text{e}^{-\text{x}}\ ...(\text{i})$

Differentiating both sides of (i) with we get,

$\frac{\text{dy}}{\text{dx}}=\text{C}_{1}\text{e}^{\text{x}}+\text{C}_{2}\text{e}^{-\text{x}}\ ...(\text{ii})$

Differentiating both sides of (ii) with we get,

$\frac{\text{d}^{2}\text{y}}{\text{dx}^{2}}=\text{C}_{1}\text{e}^{\text{x}}+\text{C}_{2}\text{e}^{-\text{x}}$

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

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

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

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

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

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

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

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

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

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.

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

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. 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,

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

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

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

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

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

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

4. Not defined

Solution:

We have,

$\big(\frac{\text{d}^{2}\text{y}}{\text{dx}^{2}}\big)^{3}+\big(\frac{\text{dy}}{\text{dx}}\big)^{2}+\sin\big(\frac{\text{dy}}{\text{dx}}\big)+1=0$

The highest order derivative in this equation is $\frac{\text{d}^{2}\text{y}}{\text{d}^{2}\text{x}}.$

But the equation cannot be as a polynomial in differential coeffcient.

Hence, the degree is not defined.

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

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

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

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

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

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

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

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

Solution:

We have,

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

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

$\Rightarrow\frac{1}{2}\times\frac{1}{\text{y}}\text{dy}=\frac{-1}{\text{x}}\text{dx}$

Integrating both sides, we get

$\Rightarrow\frac{1}{2}\int\frac{1}{\text{y}}\text{dy}=-\int\frac{1}{\text{x}}\text{dx}$

$\Rightarrow\frac{1}{2}\log\text{y}=-\log{\text{x}}+\log\text{C}$

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

$\Rightarrow \log(\sqrt{\text{yx}})=\log\text{C}$

$\Rightarrow\sqrt{\text{yx}}=\log\text{C}\ ...(\text{i})$

As (i) y(1) = 1, we get

$1=\text{C}$

Putting the valur of C in (i)

$\Rightarrow\sqrt{\text{yx}}=1$

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

4. $\frac{1}{\sqrt{1-\text{y}^{2}}}$

Solution:

We have,

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

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

Comparing with we get,

$\text{P}=\frac{\text{y}}{1-\text{y}^{2}}, \text{Q}=\frac{\text{ay}}{1-\text{y}^{2}}$

Now,

$\text{I.F}=\text{e}^{\int\frac{\text{y}}{1-\text{y}^{2}}\text{dy}}$

$=\text{e}^{-\frac{1}{2}\int\frac{-2\text{y}}{1-\text{y}^{2}}\text{dy}}$

$=\text{e}^{-\frac{1}{2}\log|1-\text{y}^{2}|}$

$=\text{e}^{\log\Big|\frac{1}{\sqrt{1-\text{y}^{2}}}\Big|}$

$=\frac{1}{\sqrt{1-\text{y}^{2}}}$

1. $\sec{\text{x}}+\tan\text{x}$

Solution:

We have,

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

Comparing with We get,

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

Now,

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

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

$=\sec\text{x}+\tan\text{x}$

3. 5

Solution:

The 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 = C₃ + C₄ be a constant

There are 5 constant (C₁, C₂, C₃, C₆, C₇) in the given differential equation.

Hence, the order of the dfifferential equation is 5.

3. $\text{p}>\text{q}$

Solution:

We have,

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

The highest order is $\frac{\text{d}^{2}\text{y}}{\text{dz}^{2}}$ and it's degree is 1.

So, the order is 2 and the degree is 1.

$\text{p}=2, \text{q}=1$

Clearly, $\text{p}>\text{q}$ 

2. $\text{m}=3,\text{n}=2$

Solution:

We have,

$(\text{y}_{2})^{5}+\frac{4(\text{y}_{2})^{3}}{\text{y}^{3}}+\text{y}_{3}=\text{x}^{2}-1$

$\text{y}_{3}(\text{y}_{2})^{5}+{4(\text{y}_{2})^{3}}+(\text{y}_{3})^{2}=\text{y}_{3}(\text{x}^{2}-1)$

The highest order is y₃ and its highest in this equation is 2.

Hence, m = 3, n = 2.

4. $\text{y}=\frac{1-\text{x}}{1+\text{x}}$

Solution:

We have,

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

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

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

Intergrating both sides, we get

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

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

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

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

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

Disclaimer : The initial value given, So the find will be C = 1, So

 $\Rightarrow \text{x}+\text{y}=1-\text{xy}$

$\Rightarrow \text{y}+\text{xy}=1-\text{x}$

$\Rightarrow \text{y}(1+\text{x})=1-\text{x}$

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

1. x² - 1 = C(1 + y²)

Solution:

We have,

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

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

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

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

Integrating both sides, we get

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

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

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

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

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

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

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

Solution:

We have,

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

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

Comparing with we get, 

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

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

Now,

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

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

$=\text{e}^{\log|\frac{1}{\text{x}}|}$

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

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

$\Rightarrow \text{y}\frac{1}{\text{x}}=\int\text{x}^{2}\times\frac{1}{\text{x}}\text{dx}+\text{C}$

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

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

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