# Current Affairs JEE Main & Advanced

#### System of Measurement of Angles

There are three system for measuring angles   (1) Sexagesimal or English system : Therefore,   1 right angle = 90 degree$(={{90}^{o}})$   ${{1}^{o}}=60$ minutes $(={{60}^{'}})$   $1'\,\,\,\,\,=60$second $(=60'')$   (2) Centesimal or French system : Therefore,   1 right angle  = 100 grades $(={{100}^{g}})$       1 grade  = 100 minutes $(=100')$   1 minute  = 100 seconds $(=100'')$   (3) Circular system : The measure of an angle subtended at the centre of a circle by an arc of length equal to the radius of the circle.   Consider a circle of radius r having centre at O. Let A be a point on the circle. Now cut off an arc AP whose length is equal to the radius r of the circle. Then by the definition the measure of $\tan ({{60}^{o}}-\theta ).\tan \theta .\tan ({{60}^{o}}+\theta )=\tan 3\theta$ is 1 radian $(={{1}^{c}})$.

#### Relation Between Three Systems of Measurement of an Angle

Let D be the number of degrees, R be the number of radians and G be the number of grades in an angle $\theta ,$ then $\frac{D}{90}=\frac{G}{100}=\frac{2R}{\pi }$   This is the required relation between the three systems of measurement of an angle.   Therefore, one radian $\cos A.\cos 2A.\cos {{2}^{2}}A.\cos {{2}^{3}}A.......\cos {{2}^{n-1}}A=\frac{\sin {{2}^{n}}A}{{{2}^{n}}\sin A},\,\text{if }A=n\pi$radians $={{180}^{o}}$   i.e., 1 radian $=57{}^\circ 1{7}'44.{{8}'}'\approx {{57}^{o}}1{7}'4{{5}'}'$.

#### Relation Between an arc and an Angle

If $s$ is the length of an arc of a circle of radius $r,$ then the angle $22\frac{{{1}^{o}}}{2}$ (in radians) subtended by this arc at the centre of the circle is given by $\theta =\frac{s}{r}$ or $A=2n\pi$.     i.e., Arc = radius $\times$ angle in radians Sectorial area : Let OAB be a sector having central angle ${{\theta }^{C}}$ and radius r. Then area of the sector OAB is given by  $\frac{1}{2}{{r}^{2}}\theta$.

#### Miscellaneous Differential Equation

A special type of second order differential equation :      $\frac{{{d}^{2}}y}{d{{x}^{2}}}=f(x)$                                                                                                                      .....(i)     Equation (i) may be re-written as $\frac{d}{dx}\left( \frac{dy}{dx} \right)=f(x)$     $\Rightarrow$$d\left( \frac{dy}{dx} \right)=f(x)dx$     Integrating, $\frac{dy}{dx}=\int_{{}}^{{}}{f(x)dx+{{c}_{1}}}$ i.e. $\frac{dy}{dx}=F(x)+{{c}_{1}}$                     .....(ii)     Where $F(x)=\int_{{}}^{{}}{f(x)dx}+{{c}_{1}}dx$     From (ii), $dy=f(x)dx+{{c}_{1}}dx$     Integrating, $y=\int_{{}}^{{}}{F(x)dx+{{c}_{1}}x+{{c}_{2}}}$     $\therefore$$y=H(x)+{{c}_{1}}x+{{c}_{2}}$     where $H(x)=\int_{{}}^{{}}{F(x)dx}$  ${{c}_{1}}$ and ${{c}_{2}}$ are arbitrary constants.

#### Application of Differential Equation

Differential equation is applied in various practical fields of life. It is used to define various physical laws and quantities. It is widely used in physics, chemistry, engineering etc.     Some important fields of application are ;     (i) Rate of change       (ii) Geometrical problems etc.

#### Linear Differential Equation

(1) Linear and non-linear differential equations : A differential equation is a linear differential equation if it is expressible in the form ${{P}_{o}}\frac{{{d}^{n}}y}{d{{x}^{n}}}+{{P}_{1}}\frac{{{d}^{n-1}}y}{d{{x}^{n-1}}}+$ ${{P}_{2}}\frac{{{d}^{n-2}}y}{d{{x}^{n-2}}}+...+$ ${{P}_{n-1}}\frac{dy}{dx}+{{P}_{n}}y=Q$, where ${{P}_{0}},\,{{P}_{1}},\,{{P}_{2}},\,...,\,{{P}_{n-1}},\,{{P}_{n}}$ and $Q$ are either constants or functions of independent variable $x$.     Thus, if a differential equation when expressed in the form of a polynomial involves the derivatives and dependent variable in the first power and there are no product of these, and also the coefficient of the various terms are either constants or functions of the independent variable, then it is said to be linear differential equation. Otherwise, it is a non linear differential equation.     It follows from the above definition that a differential equation will be non-linear differential equation if (i) its degree is more than one (ii) any of the differential coefficient has exponent more than one. (iii) exponent of  the dependent variable is more than one. (iv) products containing dependent variable and its differential coefficients are present.     (2) Linear differential equation of first order : The general form of a linear differential equation of first order is     $\frac{dy}{dx}+Py=Q$                                           .....(i)     Where $P$ and $Q$ are functions of x (or constants)     Multiplying both sides of (i) by ${{e}^{\int{Pdx}}}$, we get     ${{e}^{\int{Pdx}}}\left( \frac{dy}{dx}+Py \right)=Q\,{{e}^{\int{Pdx}}}\Rightarrow \frac{d}{dx}\left\{ y\,{{e}^{\int{Pdx}}} \right\}=Q\,{{e}^{\int{Pdx}}}$     On integrating both sides w. r. t. x, we get ;                  $y\,{{e}^{\int{Pdx}}}=\int{Q\,{{e}^{\int{Pdx}}}+c}$                       .....(ii)     which is the required solution, where c is the constant of integration. ${{e}^{\int{Pdx}}}$ is called the integrating factor. The solution (ii) in short may also be written as $y.(I.F.)=\int{Q.(I.F.)\,dx+c}$     (3) Linear differential equations of the form $\frac{dx}{dy}+Rx=S$.  Sometimes a linear differential equation can be put in the form $\frac{dx}{dy}+Rx=S$ where $R$ and $S$ are functions of $y$ or constants. Note that $y$ is independent variable and $x$ is a dependent variable.     (4) Equations reducible to linear form (Bernoulli's differential equation) : The differential equation of type                                     $\frac{dy}{dx}+Py=Q{{y}^{n}}$                                  .....(i)     Where $P$ and $Q$ are constants or functions of $x$ alone and $n$ is a constant other than zero or unity, can be reduced to the linear form by dividing by ${{y}^{n}}$ and then putting ${{y}^{-n+1}}=v$, as explained below.     Dividing both sides of (i) by ${{y}^{n}}$, we get  ${{y}^{-n}}\frac{dy}{dx}+P{{y}^{-n+1}}=Q$     Putting ${{y}^{-n+1}}=v$ so that $(-n+1){{y}^{-n}}\frac{dy}{dx}=\frac{dv}{dx}$, we get $\frac{1}{-n+1}\frac{dv}{dx}+Pv=Q\Rightarrow \frac{dv}{dx}+(1-n)Pv=(1-n)Q$ which is a linear differential equation.     Note : If $n=1$, then we find that the variables in equation (i) are separable and it can be easily integrated by the method discussed in variable separable from.     (5) Differential equation of the form :     $\frac{dy}{dx}+P\varphi (y)=Q\psi (y)$, where $P$ and $Q$ are functions of $x$ alone or constants.     Dividing by $\psi \,(y),$  we get $\frac{1}{\psi (y)}\frac{dy}{dx}+\frac{\varphi (y)}{\psi (y)}P=Q$     Now put$\frac{\varphi (y)}{\psi (y)}=v$, so that $\frac{d}{dx}\left\{ \frac{\varphi (y)}{\psi (y)} \right\}=\frac{dv}{dx}$     or $\frac{dv}{dx}=k\cdot \frac{1}{\psi (y)}\frac{dy}{dx}$, where $k$ is constant     We get $\frac{dv}{dx}+kP\,v=kQ$   more...

#### Solution by Inspection

If we can write the differential equation in the form $f({{f}_{1}}(x,\,y))d({{f}_{1}}(x,\,y))+\varphi ({{f}_{2}}(x,\,y))d({{f}_{2}}(x,\,y))+......=0$, then each term can be easily integrated separately. For this the following results must be memorized.     (i) $d(x+y)=dx+dy$     (ii) $d(xy)=xdy+ydx$     (iii) $d\left( \frac{x}{y} \right)=\frac{ydx-xdy}{{{y}^{2}}}$                      (iv) $d\left( \frac{y}{x} \right)=\frac{xdy-ydx}{{{x}^{2}}}$     (v) $d\,\left( \frac{{{x}^{2}}}{y} \right)=\frac{2xydx-{{x}^{2}}dy}{{{y}^{2}}}$       (vi) $d\left( \frac{{{y}^{2}}}{x} \right)=\frac{2xydy-{{y}^{2}}dx}{{{x}^{2}}}$     (vii) $d\left( \frac{{{x}^{2}}}{{{y}^{2}}} \right)=\frac{2x{{y}^{2}}dx-2{{x}^{2}}ydy}{{{y}^{4}}}$     (viii) $d\,\left( \frac{{{y}^{2}}}{{{x}^{2}}} \right)=\frac{2{{x}^{2}}ydy-2x{{y}^{2}}dx}{{{x}^{4}}}$     (ix) $d\,\left( {{\tan }^{-1}}\frac{x}{y} \right)=\frac{ydx-xdy}{{{x}^{2}}+{{y}^{2}}}$     (x) $d\left( {{\tan }^{-1}}\frac{y}{x} \right)=\frac{xdy-ydx}{{{x}^{2}}+{{y}^{2}}}$     (xi) $d[\ln (xy)]=\frac{xdy+ydx}{xy}$        (xii) $d\left( \ln \left( \frac{x}{y} \right) \right)=\frac{ydx-xdy}{xy}$     (xiii) $d\,\left[ \frac{1}{2}\ln ({{x}^{2}}+{{y}^{2}}) \right]=\frac{xdx+ydy}{{{x}^{2}}+{{y}^{2}}}$     (xiv) $d\left[ \ln \left( \frac{y}{x} \right) \right]=\frac{xdy-ydx}{xy}$     (xv) $d\text{ }\left( -\frac{1}{xy} \right)=\frac{xdy+ydx}{{{x}^{2}}{{y}^{2}}}$         (xvi) $d\text{ }\left( \frac{{{e}^{x}}}{y} \right)=\frac{y{{e}^{x}}dx-{{e}^{x}}dy}{{{y}^{2}}}$                    (xvii) $d\left( \frac{{{e}^{y}}}{x} \right)=\frac{x{{e}^{y}}dy-{{e}^{y}}dx}{{{x}^{2}}}$     (xviii) $d({{x}^{m}}{{y}^{n}})={{x}^{m-1}}{{y}^{n-1}}(mydx+nxdy)$     (xix) $d\text{ }\left( \sqrt{{{x}^{2}}+{{y}^{2}}} \right)=\frac{xdx+y\,dy}{\sqrt{{{x}^{2}}+{{y}^{2}}}}$     (xx) $d\text{ }\left( \frac{1}{2}\log \frac{x+y}{x-y} \right)=\frac{x\,dy-y\,dx}{{{x}^{2}}-{{y}^{2}}}$     (xxi) $\frac{d{{[f(x,\,y)]}^{1-n}}}{1-n}=\frac{{f}'(x,\,y)}{{{(f(x,\,y))}^{n}}}$

#### Exact Differential Equation

(1) Exact differential equation : If $M$ and $N$ are functions of $x$ and $y,$ the equation $Mdx+Ndy=0$  is called exact when there exists a function $f(x,\,\,y)$ of $x$ and $y$ such that     $d[f(x,\,\,y)=Mdx+Ndy$ i.e., $\frac{\partial f}{\partial x}dx+\frac{\partial f}{\partial y}dy=Mdx+Ndy$     where $\frac{\partial f}{\partial x}=$Partial derivative of $f(x,\,y)$ with respect to $x$ (keeping y constant) . $\frac{\partial f}{\partial y}=$Partial derivative of $f(x,\,\,y)$ with respect to $y$ (treating $x$as constant)     The necessary and sufficient condition for the differential condition $Mdx+Ndy=0$ to be exact is $\frac{\partial M}{\partial y}=\frac{\partial N}{\partial x}$.     An exact differential equation can always be derived from its general solution directly by differentiation without any subsequent multiplication, elimination etc.     (2) Integrating factor : If an equation of the form $Mdx+Ndy=0$ is not exact, it can always be made exact by multiplying by some function of $x$ and $y$. Such a multiplier is called an integrating factor.     (3) Working rule for solving an exact differential equation :     Step (i) : Compare the given equation with $Mdx+Ndy=0$ and find out $M$ and $N$. Then find out $\frac{\partial M}{\partial y}$ and $\frac{\partial N}{\partial x}$. If $\frac{\partial M}{\partial y}=\frac{\partial N}{\partial x}$, the given equation is exact.     Step (ii) : Integrate $M$ with respect to $x$ treating $y$ as a constant.     Step (iii) : Integrate $N$ with respect to $y$ treating $x$ as constant and omit those terms which have been already obtained by integrating $M$.     Step (iv) : On adding the terms obtained in steps (ii) and (iii) and equating to an arbitrary constant, we get the required solution.     In other words, solution of an exact differential equation is     $\underset{\begin{smallmatrix} \text{Regarding }y\text{ } \\ \text{as constant} \end{smallmatrix}}{\mathop{\int_{{}}^{{}}{Mdx}}}\,+\underset{\begin{smallmatrix} \text{Only those terms } \\ \text{not containing }x \end{smallmatrix}}{\mathop{\int_{{}}^{{}}{Ndy}}}\,=c$

#### Homogeneous Differential Equation

(1) Homogeneous differential equation : A function $f(x,\,\,y)$is called a homogeneous function of degree $n$ if $f(\lambda x,\lambda y\,)=$ ${{\lambda }^{n}}\,f(x,y)\,$.     For example, $f(x,y)={{x}^{2}}-{{y}^{2}}+3xy$ is a homogeneous function of degree 2. A homogenous function $f(x,\,y)$ of degree $n$ can always be written as $f(x,\,y)={{x}^{n}}f\left( \frac{y}{x} \right)$ or$f(x,y)={{y}^{n}}f\left( \frac{x}{y} \right)$. If a first-order first-degree differential equation is expressible in the form $\frac{dy}{dx}=\frac{f(x,\,y)}{g(x,y)}$ where $f(x,\,\,y)$ and $g(x,\,\,y)$ are homogeneous functions of the same degree, then it is called a homogeneous differential equation. Such type of equations can be reduced to variable separable form by the substitution $y=vx$. The given differential equation can be written as $\frac{dy}{dx}=\frac{{{x}^{n}}f(y/x)}{{{x}^{n}}g(y/x)}$$=\frac{f(y/x)}{g(y/x)}$$=F\left( \frac{y}{x} \right)$.  If $y=vx$, then $\frac{dy}{dx}=v+x\frac{dv}{dx}$. Substituting the value of $\frac{dy}{dx}=F\left( \frac{y}{x} \right)$, we get $v+x\frac{dv}{dx}=F(v)$$v+x\frac{dv}{dx}=F(v)\Rightarrow \frac{dv}{F(v)-v}=\frac{dx}{x}$.     On integration, $\int{\frac{1}{F(v)-v}dv=\int{\frac{dx}{x}+c}}$ where $c$ is an arbitrary constant of integration. After integration, $v$ will be replaced by $\frac{y}{x}$ in complete solution.     (2) Equation reducible to homogeneous form : A first order, first degree differential equation of the form      $\frac{dy}{dx}=\frac{{{a}_{1}}x+{{b}_{1}}y+{{c}_{1}}}{{{a}_{2}}x+{{b}_{2}}y+{{c}_{2}}}$, where $\frac{{{a}_{1}}}{{{a}_{2}}}\ne \frac{{{b}_{1}}}{{{b}_{2}}}$                    .....(i)     This is non-homogeneous.     It can be reduced to homogeneous form by certain substitutions. Put $x=X+h,\,y=Y+k$.     Where $h$ and $k$ are constants, which are to be determined.

#### Variable Separable Type Differential Equation

(1) Equations in variable separable form : If the differential equation of the form ${{f}_{1}}(x)dx={{f}_{2}}(y)dy$        .....(i)     where ${{f}_{1}}$ and ${{f}_{2}}$ being functions of $x$ and $y$ only. Then we say that the variables are separable in the differential equation.     Thus, integrating both sides of (i), we get its solution as $\int_{{}}^{{}}{{{f}_{1}}(x)dx}=\int_{{}}^{{}}{{{f}_{2}}(y)dy}+c$, where $c$ is an arbitrary constant.     There is no need of introducing arbitrary constants to both sides as they can be combined together to give just one.     (2) Equations reducible to variable separable form     (i) Differential equations of the form $\frac{dy}{dx}=f(ax+by+c)$ can be reduced to variable separable form by the substitution  $ax+by+c=Z$.     $\therefore$ $a+b\frac{dy}{dx}=\frac{dZ}{dx}$;  $\therefore$ $\left( \frac{dZ}{dx}-a \right)\frac{1}{b}=f(Z)$$\Rightarrow$$\frac{dZ}{dx}=a+bf(Z)$.     This is variable separable form.     (ii) Differential equation of the form     $\frac{dy}{dx}=\frac{ax+by+c}{Ax+By+C}$, where $\frac{a}{A}=\frac{b}{B}=K$ (say)     $\therefore$ $\frac{dy}{dx}=\frac{K(Ax+By)+c}{Ax+By+C}$     Put $Ax+By=Z\Rightarrow$ $A+B\frac{dy}{dx}=\frac{dZ}{dx}$     $\therefore$$\left[ \frac{dZ}{dx}-A \right]\frac{1}{B}=\frac{KZ+c}{Z+C}$$\Rightarrow$ $\frac{dZ}{dx}=A+B\frac{KZ+c}{Z+C}$     This is variable separable form and can be solved.

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