Maths 4 Marks

We have,

$\text{y}=4\text{ax}\ ...(1)$

Differentiating both sides of equation (1) with respect to 3, we get

$2\text{y}\frac{\text{dy}}{\text{dx}}=4\text{a}$

$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{2\text{a}}{\text{y}}\ ...(2)$

now, differentiating both sided of (1) with respect to y, we get

$2\text{y}=4\text{a}\frac{\text{dy}}{\text{dx}}$

$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{\text{y}}{2\text{a}}\ ...(3)$

$\therefore\text{x}\frac{\text{dy}}{\text{dx}}+\text{a}\frac{\text{dy}}{\text{dx}}=\text{x}\Big(\frac{2\text{a}}{\text{y}}\Big)+\text{a}\Big(\frac{\text{y}}{2\text{a}}\Big)$

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

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

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

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

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

Hence, the given function is the solution to the given differential equation.

The equation of the family of curves is

y2 - 2ay + 2+ = a² ...(1)

 where a is a parameter.

This equation contains only one arbitrary constant, so we shall get a differential equation of first order.

Differentiating equation (1) with respect to x, we get

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

$\Rightarrow2\text{y}\frac{\text{dy}}{\text{dx}}+2\text{x}=2\text{a}\frac{\text{dy}}{\text{dx}}$

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

Substituting the value of a in equation (2), we get

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

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

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

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

It is the required differential equation.

$\int\limits_{\text{a}}^{\text{b}}\text{f}\text{(x)}\text{dx}=\int\limits_{\text{a}}^{\text{b}}\text{f}(\text{a}+\text{b}-\text{x})\text{dx}$

Hence,

$\int\limits_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\frac{\cos^2\text{x}}{1+\text{e}^\text{x}}\text{dx}=\int\limits_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\frac{\cos^2\text{(-x)}}{1+\text{e}^\text{-x}}\text{dx}$

$\int\limits_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\frac{\cos^2\text{x}}{1+\text{e}^\text{x}}\text{dx}=\int\limits_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\frac{\cos^2\text{x}}{1+\text{e}^\text{-x}}\text{dx}$

If,

$\text{I}=\int\limits_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\frac{\cos^2\text{x}}{1+\text{e}^\text{x}}\text{dx}$

Then

$\text{I}=\int\limits_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\frac{\cos^2\text{x}}{1+\text{e}^\text{x}}\text{dx}$

So,

$2\text{I}=\int\limits_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\frac{\cos^2\text{x}}{1+\text{e}^\text{x}}+\frac{\cos^2\text{x}}{1+\text{e}^\text{-x}}\text{dx}$

$2\text{I}=\int\limits_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\frac{\cos^2\text{x}}{1+\text{e}^\text{x}}+\frac{\text{e}^\text{x}\cos^2\text{x}}{1+\text{e}^\text{-x}}\text{dx}$

$2\text{I}=\int\limits_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\frac{(1+\text{e}^\text{x})\cos^2\text{x}}{1+\text{e}^\text{x}}$

$2\text{I}=\int\limits_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\cos^2\text{x}\text{dx}$

$2\text{I}=\int\limits_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\frac{1+\cos2\text{x}}{2}\text{dx}$

$\text{I}=\int\limits_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\frac{1}{4}\bigg\{\text{x}+\frac{\sin2\text{x}}{2}\bigg\}^\frac{\pi}{2}_{-\frac{\pi}{2}}$

$\text{I}=\frac{1}{4}\bigg\{\bigg(\frac{\pi}{2}\bigg)-\bigg(-\frac{\pi}{2}\bigg)\bigg\} $

$\text{I}=\frac{\pi}{4}$

 $\int_\limits{a}^{b}\text{f}\text{(x)}\text{dx}=\int_\limits{a}^{b} \text{f}(\text{a}+\text{b}-\text{x}) \text{dx}$

Hence,

$\int_\limits{-\frac{\pi}{4}}^{\frac{\pi}{4}}\frac{\tan^2\text{x}}{{1}+\text{e}^\text{x}}\text{dx}=\int_\limits{-\frac{\pi}{4}}^{\frac{\pi}{4}}\frac{\tan^2\text{(-x)}}{1-\text{e}^\text{-x}}\text{dx}$

$\int_\limits{-\frac{\pi}{4}}^{\frac{\pi}{4}}\frac{\tan^2\text{x}}{{1}+\text{e}^\text{x}}\text{dx}=\int_\limits{-\frac{\pi}{4}}^{\frac{\pi}{4}}\frac{\tan^2\text{x}}{1-\text{e}^\text{-x}}\text{dx}$

If,

$\text{I}=\int_\limits{-\frac{\pi}{4}}^{\frac{\pi}{4}}\frac{\tan^2\text{x}}{{1}+\text{e}^\text{x}}\text{dx}$

Then

$\text{I}=\int_\limits{-\frac{\pi}{4}}^{\frac{\pi}{4}}\frac{\tan^2\text{x}}{{1}+\text{e}^\text{-x}}\text{dx}$ 

So,

$2\text{I}=\int_\limits{-\frac{\pi}{4}}^{\frac{\pi}{4}}\frac{\tan^2\text{x}}{{1}+\text{e}^\text{x}}+\frac{\tan^2\text{x}}{1+\text{e}^\text{-x}}\text{dx}$

$2\text{I}=\int_\limits{-\frac{\pi}{4}}^{\frac{\pi}{4}}\frac{\tan^2\text{x}}{{1}+\text{e}^\text{x}}+\frac{\tan^2\text{x}}{1+\text{e}^\text{-x}}\text{dx}$

$2\text{I}=\int_\limits{-\frac{\pi}{4}}^{\frac{\pi}{4}}\frac{\tan^2\text{x}}{{1}+\text{e}^\text{x}}+\frac{\text{e}^x\tan^2\text{x}}{1+\text{e}^\text{x}}\text{dx}$

$2\text{I}=\int_\limits{-\frac{\pi}{4}}^{\frac{\pi}{4}}\frac{\tan^2\text{x}}{1+\text{e}^\text{x}}+\frac{\text{e}^\text{x}\tan^2\text{x}}{1+\text{e}^2}\text{dx}$

$2\text{I}=\int_\limits{\frac{\pi}{4}}^{\frac{\pi}{4}}\frac{\tan^2\text{x}+\text{e}^\text{x}\tan^2\text{x}}{1+\text{e}^\text{x}}\text{dx}$

$2\text{I}=\int_\limits{\frac{\pi}{4}}^{\frac{\pi}{4}}\frac{(1+\text{e}^\text{x})\tan^2\text{x}}{1+\text{e}^\text{x}}\text{dx}$

$2\text{I}=\int_\limits{\frac{\pi}{4}}^{\frac{\pi}{4}}\frac{\tan^2\text{x}+\text{e}^\text{x}\tan^2\text{x}}{1+\text{e}^\text{x}}\text{dx}$

$2\text{I}=\int_\limits{\frac{\pi}{4}}^{\frac{\pi}{4}}\frac{(1+\text{e}^\text{x})\tan^2\text{x}}{1+\text{e}^\text{x}}\text{dx}$

$2\text{I}=\int_\limits{-\frac{\pi}{4}}^{\frac{\pi}{4}}\tan^2\text{x}\text{dx}$.

$\text{I}=\frac{1}{2}\int_\limits{-\frac{\pi}{4}}^{\frac{\pi}{4}}\tan^2\text{x}\text{dx}$

We know 

If f(x)is even

$\int_\limits{-a}^{a} \text{f}\text{(x)}\text{dx}=2\int_\limits{0}^{a}\text{f}\text{(x)}\text{dx}$

If f(x)is odd

$\int_\limits{-a}^{a} \text{f}\text{(x)}\text{dx}=0$

Here

$\text{f}\text{(x)}=\tan^2\text{x}$

f(x)is even,hence

$\text{I}=\int\limits_{0}^{\frac{\pi}{4}}\tan^2\text{x}\text{dx}$

$\text{I}=\int\limits_{0}^{\frac{\pi}{4}}\sec^2\text{x}-1\text{dx}$.

$\text{I}=\big\{\tan\text{x}-\text{x}\big\}\frac{\frac{\pi}{4}}{0 }$

$\text{I}=1-\frac{\pi}{4}$

The equation of the family of curves is

$\frac{\text{x}^2}{\text{a}^2}-\frac{\text{y}^2}{\text{b}^2}=1$

where a is a parameter.

As this equation has only one arbitrary constant, we shall get a differential equation of first order.

Differentiating (1) with respect to x, we get

$\frac{2\text{x}}{\text{a}^2}-\frac{2\text{y}}{\text{b}^2}\frac{\text{dy}}{\text{dx}}=0\ ...(2)$

Differentiating (2) with respect to x, we get

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

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

$\Rightarrow\frac{\text{b}^2}{\text{a}^2}=\Big[\text{y}\frac{\text{d}^2\text{y}}{\text{dx}^2}+\Big(\frac{\text{dy}}{\text{dx}}\Big)^2\Big]\ ...(3)$

Now, from (2), we get

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

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

From (3), (4), we get

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

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

It is the required differential equation.

This given equation is homogeneous because each coefficients of dx and dy is of degree 1.

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

 $\text{Putting}\frac{\text{y}}{\text{x}}=\text{v}\ \ \Rightarrow\ \ \text{y}=\text{vx}\ \ $ $\Rightarrow\ \ \frac{\text{dy}}{\text{dx}}=\text{v}.1+\text{x}\frac{\text{dv}}{\text{dx}}=\text{v}+\text{x}\frac{\text{dv}}{\text{dx}}\ \ ....\text{(ii)}$

$\text{Putting value of y and}\ \frac{\text{dy}}{\text{dx}}\ \text{in eq. (ii)}$

$\text{v}+\text{x}\frac{\text{dv}}{\text{dx}}=\frac{1+\text{v}}{1-\text{v}}\ \ \Rightarrow\ \ \text{x}\frac{\text{dv}}{\text{dx}}=\frac{1+\text{v}}{1-\text{v}}-\text{v}$ $\ \ \Rightarrow\ \ \text{x}\frac{\text{dv}}{\text{dx}}=\frac{1+\text{v}-\text{v}+\text{v}^2}{1-\text{v}}$

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

$\Rightarrow\ \ \frac{(1-\text{v})}{1+\text{v}^2}\ \text{dv}=\frac{\text{dx}}{\text{x}}\ \ [\text{Separating variables]}$

$\text{Integrating both sides,}\ \ \int\frac{(1-\text{v})}{1+\text{v}^2}\ \text{dv}=\int\frac{1}{\text{x}}\ \text{dx}$.

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

$\Rightarrow \ \ \tan^{-1}\ \text{v}-\frac{1}{2}\log(1+\text{v}^2)=\log\text{x}+\text{c}$

$\text{Putting v}=\frac{\text{y}}{\text{x}},$ $\ \ \tan^{-1}\ \frac{\text{y}}{\text{x}}-\frac{1}{2}\log\Big(1+\frac{\text{y}^2}{\text{x}^2}\Big)=\log\text{x}+\text{c}$

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

$\Rightarrow\ \ \tan^{-1}\ \frac{\text{y}}{\text{x}}-\bigg[\frac{1}{2}\log(\text{x}^2+\text{y}^2)-\log\text{x}^2\bigg]=\log\text{x}+\text{c}$

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

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

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

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

$\text{v + x}\frac{\text{dv}}{\text{dx}}=\frac{\text{vx}}{2\text{x}-\text{x}\log\text{v}}$

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

$=\frac{\text{x}^2\log\big(\frac{\text{x}}{\text{y}}\big)-\text{y}^2}{\text{xy}\log\big(\frac{\text{x}}{\text{y}}\big)}$

It is the required differential equation.

$\text{v + x}\frac{\text{dv}}{\text{dx}}=\frac{\text{x}^2+\text{v}^2\text{x}^2}{2\text{xvx}}$

The solution of the given differential equation is given by the relation,