Maths

The given D.E. is

$\begin{array}{l}ydx+\left(x-y^2\right)dy=0\\ =y\frac{dx}{dy}+x-y^2=0\\ =\frac{dx}{dy}+\frac{x}{y}-y=0\end{array}$

$=\frac{dx}{dy}+\frac{1}{y}.x=y$ Which is of form.

$\frac{dx}{dy}+Px=Q$

So, $P=\frac{1}{y}\&Q=y$

$\therefore I.F=e^{{\displaystyle \int Pdy}}=e^{{\displaystyle \int \frac{1}{y}dy}}=e^{logy}=y$

Thus the general solution is of form, $x*\left(I.F\right)={\displaystyle \int Q*}\left(I.F\right)dy+c$

$\begin{array}{l}x.y={\displaystyle \int y.ydy+c}\\ =xy={\displaystyle \int y^{2}dy+c}\\ =xy=\frac{y^3}{3}+c\\ =x=\frac{y^2}{3}+\frac{c}{y}\end{array}$

The given D.E is

$\begin{array}{l}\left(x+y\right)\frac{dy}{dx}=1\\ =x+y=\frac{dx}{dy}\end{array}$

$=\frac{dx}{dy}-x=y$ Which is of form $=\frac{dx}{dy}+Px=Q$

So,  $P=-1\&Q=y$

$\therefore I.F=e^{{\displaystyle \int Pdy}}=e^{{\displaystyle \int -1dy}}=e^y$

Thus the general solution is of the form,  $x*\left(I.F\right)={\displaystyle \int Q}\left(I.F\right)dy+c$

The given D.E is

$=\frac{dy}{dx}+y\frac{\left(1+xcotx\right)}{x}=1$ Which is of form $\frac{dy}{dx}+Py=Q$

So, $P=\frac{\left(1+xcotx\right)}{x}\&Q=1$

${\displaystyle \int Pdx={\displaystyle \int \left(\frac{1}{x}+\frac{xcotx}{x}\right)}}dx=log\left|x\right|+log\left|sinx\right|=log\left|xsinx\right|$

$\therefore I.F=e^{{\displaystyle \int Pdx}}-e^{{\displaystyle \int log\left|xsinx\right|}}xsinx$

Thus the solution is of the form.

$\begin{array}{l}y*xsinx={\displaystyle \int 1.xsinxdx+c}\\ =x{\displaystyle \int sinx-{\displaystyle \int \frac{d}{dx}{\displaystyle \int sinxdxdx+c}}}\\ =-2cosx+{\displaystyle \int cosxdx+c}\\ =y*xsinx=-xcosx+sinx+c\\ =y=\frac{-xcosx}{sinx}+\frac{sinx}{xsinx}+\frac{c}{xsinx}\\ =y=-cotx+\frac{1}{x}+\frac{c}{xsinx}\end{array}$

The given D.E. is

$\begin{array}{l}{\left(1+x\right)}^{2}dy+2xydx=cotxdx\\ ={\left(1+x\right)}^2\frac{dy}{dx}+2xy=cotx\end{array}$

$=\frac{dy}{dx}+\frac{2x}{1+x^2}*y=\frac{cotx}{1+x^2}$ Which is of form $\frac{dy}{dx}+Py=Q$

So , $P=\frac{2x}{1+x^2}\&Q=\frac{cotx}{1+x^2}$

$\therefore I.F=e^{{\displaystyle \int Pdx}}=e^{{\displaystyle \int \frac{2x}{1+x^2}dx}}=e^{log\left|1+x^2\right|}=1+x^2$

Thus the solution is of the form,

$\begin{array}{l}\Rightarrow y\left(1+x^2\right)={\displaystyle \int \frac{cotx}{1+x^2}*}\left(1+x^2\right)dx+c\\ \Rightarrow y\left(1+x^2\right)={\displaystyle \int cotxdx+c}\end{array}$

$=log\left|sinx\right|+c$

$\begin{array}{l}\Rightarrow y=\frac{log\left|sinx\right|}{1+x^2}+\frac{c}{1+x^2}\\ ={\left(1+x^2\right)}^{-1}log\left|sinx\right|+{\left(1+x^2\right)}^{-1}c\end{array}$

139. Kindly go through the solution

The given D.E. is

$xlogx\frac{dy}{dx}+y=\frac{2}{x}logx$

$=\frac{dy}{dx}+\frac{y}{xlogx}=\frac{2}{x^2}$ Which is of form $\frac{dy}{dx}+Py=Q$

So, $P=\frac{1}{xlogx}\&Q=\frac{2}{x^2}$

$\therefore I.F=e^{{\displaystyle \int Pdx}}=e^{{\displaystyle \int \frac{1}{x}logxdx}}=e^{{\displaystyle \int \frac{\frac{1}{x}}{logx}}dx}=e^{log\left|logx\right|}=logx$

Thus, the general solution is of the form

$\begin{array}{l}y*logx={\displaystyle \int \frac{2}{x^2}}*logxdx+c\\ y.logx=2\left[logx{\displaystyle \int \frac{1}{x^2}dx-{\displaystyle \int \frac{d}{dx}logx{\displaystyle \int \frac{1}{x^2}dx.dx}}}\right]+c\\ =2\left[logx*\left(\frac{x^{-1}}{-1}\right)-{\displaystyle \int \frac{1}{x}\left(\frac{x^{-1}}{-1}\right)dx}\right]+c\\ =2\left[-\frac{logx}{x}+{\displaystyle \int x^{-2}dx}\right]+c\\ =2\left[-\frac{logx}{x}-\left(\frac{x^{-1}}{-1}\right)\right]+c\end{array}$

$=ylogx=\frac{-2}{x}\left[logx+1+c\right]$

The given D.E. is

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

$\Rightarrow \frac{dy}{dx}+\frac{2}{x}.y=xlogx$ which is of form $\frac{dy}{dx}+Py=Q$

So, $P=\frac{2}{x}\&Q=xlogx$

$\therefore I.F=e^{{\displaystyle \int Pdx}}=e^{{\displaystyle \int \frac{2}{x}dx}}=e^{2logx}=e^{log{x}^2}=x^2$

Thus, the general solution is of the form.

$y*x^2={\displaystyle \int xlogx.x^{2}dx+c}$

$={\displaystyle \int logx}{x}^{3}dx+c$

$\begin{array}{l}=logx{\displaystyle \int x^{3}dx-{\displaystyle \int \frac{d}{dx}logx{\displaystyle \int x^{3}dxdx+c}}}\\ =\frac{logx.x^4}{4}-{\displaystyle \int \frac{1}{4}*\frac{x^4}{4}dx+c}\end{array}$

$\begin{array}{l}=y{x}^2=\frac{x^4}{4}logx-\frac{x^4}{16}+c\\ =y=\frac{x^2}{4}logx-\frac{x^2}{16}+\frac{c}{x^2}\end{array}$

137. Yes, Let us take $f(x)=|x-1|+|x-2|.$

So, x = 1, x= 2 divides the real line into three disjoint intervals $(-\infty ,1],[1,2]$ and $[2,\infty ).$

For $x\in (-\infty ,1].$

$f(x)=-(x-1)+[-(x-2)]=-x+1-x+2=3-2x.$

For $x\in [1,2].$

$f(x)=(x-1)-(x-2)=1.$

For $x\in [2,\infty )$

$f(x)=x-1+x-2=2x-3.$

Hence, these polynomial fun are all continous and desirable. for all real values of x or, except x = 1 and x = 2.

ie, $\forall x\in R-\{1,2\}.$

For differentiavity at x = 1,

LHD = $=\underset{x\to 1^{-}}{lim}\frac{f(x)-f(1)}{x-1}=\underset{x\to 1^{-}}{lim}\frac{3-2x-1}{x-1}=\underset{x\to 1^{-}}{lim}\frac{2-2x}{x-1}.$

$=\underset{x\to 1^{-}}{lim}\frac{-2(x-1)}{x-1}$

$=\underset{x\to 1^{-}}{lim}-2$

= -2

RHD = $=\underset{x\to 1^{+}}{lim}\frac{f(x)-f(1)}{x-1}=\underset{x\to 1^{+}}{lim}\frac{1-1}{x-1}=\underset{x\to 1^{+}}{lim}\frac{0}{x-1}=0.$

as L.HD ≠ R.HD

f is not differentiable at x =1.

For continuity at x = 1.

L.HL= $=\underset{x\to 1^{-}}{lim}f(x)=\underset{x\to 1^{-}}{lim}=1.$

RHL = $\underset{x\to 1^{+}}{lim}f(x)=\underset{x\to 1^{+}}{lim}1=1$ \ LHL = RHS

f is continuous at x = 1

For continuity & differentiability at x = 2

$=\underset{x\to 2^{-}}{lim}f(x)=\underset{x\to 2^{-}}{lim}1=1.$

$=\underset{x\to 2^{+}}{lim}f(x)=\underset{x\to 2^{+}}{lim}(2x-3)=4-3=1.$

? LHL = RHL

f is continuous at x = 2

$\begin{array}{l}=\underset{x\to 2^{-}}{lim}\frac{f(x)-f(2)}{x-2}=\underset{x\to 2^{-}}{lim}\frac{1-1}{x-2}=\underset{x\to 2^{-}}{lim}=\frac{0}{x-2}\\ \end{array}$

$=\underset{x\to 2^{+}}{lim}\frac{f(x)-f(2)}{x-2}=\underset{x\to 2^{+}}{lim}\frac{2x-3-1}{x-2}$

$=\underset{x\to 2^{+}}{lim}\frac{2(x-2)}{x-2}$

$=\underset{x\to 2^{+}}{lim}2$

= 2

? LHD ≠ RHD

f is not differentiable at x = 2.

The given D.E.is

$\begin{array}{l}co{s}^{2}x\frac{dy}{dx}+y=tanx\\ =\frac{dy}{dx}+\frac{1}{co{s}^{2}x}y=\frac{tanx}{co{s}^{2}x}\end{array}$

$=\frac{dy}{dx}+se{c}^{2}xy=se{c}^{2}xtanx$ Which is of form $\frac{dy}{dx}+Py=Q$

So, $P=se{c}^{2}x\&Q=se{c}^{2}xtanx$

$\therefore I.F=e^{{\displaystyle \int Pdx}}=e^{{\displaystyle \int se{c}^{2}dx}}=e^{tanx}$

Thus, the general solution is of the form.

$y.e^{tanx}={\displaystyle \int se{c}^{2}xtanx.}{e}^{tanx}dx+c$

Let, $tanx=t=se{c}^{2}xdx=dt$

$\begin{array}{l}=y{e}^t={\displaystyle \int t.}{e}^{t}dt+c\\ =t{\displaystyle \int e^{t}dt-{\displaystyle \int \frac{d}{dt}t{\displaystyle \int e^{t}dt.dt+c}}}\\ =t{e}^t-{\displaystyle \int e^{t}dt+c}\\ =t{e}^t-e^t+c\\ =e^t\left(t-1\right)+c\end{array}$

$\begin{array}{l}\Rightarrow y{e}^{tanx}=e^{tanx}\left(tanx-1\right)+c\\ y=\left(tanx-1\right)+c{e}^{-tanx}\end{array}$

136. Given, $sin(A+B)=sinAcosB+cosAsinB$

Differentiating w r t. 'x' we get,

$\frac{d}{dx}sin(A+B)=\frac{d}{dx}(sinAcosB+cosAsinB)$

$\to cos(A+B)-\frac{d}{dx}(A+B)=sinA\frac{d}{dx}cosB+cosB\frac{d}{dx}sinA+cosA\frac{d}{dx}sinB+sinB\frac{d}{dx}cosA$

$\to cos(A+B)\left(\frac{dA}{dx}+\frac{dB}{dx}\right)=-sinAsinB\frac{dB}{dx}+cosBcosA\frac{dA}{dx}+cosAcosB\frac{dB}{dx}-sinAsinB\frac{dA}{dx}$

$\to cos(A+B)\left(\frac{dA}{dx}+\frac{dB}{dt}\right)=cosAcosB\left(\frac{dA}{dx}+\frac{dB}{dx}\right)-sinAsinB\left(\frac{dA}{dx}+\frac{dB}{dx}\right)$