Differentiation (Part-38) | S N De

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

Sol.

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

The integrating factor (I.F.) of $\,(1)$

$=e^{\displaystyle\int{\frac{2x}{1+x^2}~dx}}$

$=e^{\log_e(1+x^2)}$

$=1+x^2$

Now, multiplying both sides of $\,(1)\,$ by $\,(1+x^2)\,$ we get,

$~~~(1+x^2)\frac{dy}{dx}+2xy=\frac{1}{1+x^2}$

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

$\therefore y(1+x^2)=\displaystyle\int{\frac{1}{1+x^2}~dx}$

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

16. $2\cos x~\frac{dy}{dx}+4y\sin x=\sin2x$

Sol.

$~~~2\cos x~\frac{dy}{dx}+4y\sin x=\sin2x$

$\therefore \frac{dy}{dx}+2y\tan x=\frac{\sin2x}{2\cos x}\rightarrow(1)$

I.F. of $\,(1)$

$=e^{\int{2\tan x~dx}} = e^{2\log_e{\sec x}} = e^{\log_e \sec^2x} = \sec^2x$

Now, multiplying both sides of $\,(1)\,$ by $~\sec^2x~$ we get,

$~~~~\sec^2x ~\frac{dy}{dx}+2y \tan x\sec^2x=\frac{\sin2x}{2\cos x}\cdot \sec^2x$

$\Rightarrow \frac{d}{dx}(y\sec^2x)=\frac{\sin2x}{2\cos x}\cdot \sec^2x$

$\therefore y\sec^2x=\displaystyle\int{\frac{\sin2x}{2\cos x}\sec^2x~dx}$

$\Rightarrow y\sec^2x=\displaystyle\int{\frac{-d(\cos x)}{\cos^2x}~dx}$

$\Rightarrow y\sec^2x=-\frac{\cos^{-2+1}}{(-2+1)}+c$

$\therefore y\sec^2x=\frac{1}{\cos x}+c$

$\text{or,}~~ y\sec^2x=\sec x+c ~~\text{(ans.)}$

17. $\frac{dy}{dx}+\frac{4x}{x^2+1}y=\frac{1}{(x^2+1)^3}$

Sol.

$~~~\frac{dy}{dx}+\frac{4x}{x^2+1}y=\frac{1}{(x^2+1)^3} \rightarrow(1)$

I.F. of $\,(1)$

$=e^{\int{\frac{4x}{x^2+1}}~dx} = e^{2\log_e(x^2+1)} = e^{\log_e(x^2+1)^2} = (x^2+1)^2$

Now, multiplying both sides of $\,(1)\,$ by $~(x^2+1)^2~$ we get,

$~~~(x^2+1)^2~\frac{dy}{dx}+4x(x^2+1)y= \frac{1}{(x^2+1)}$

$\Rightarrow \frac{d}{dx}[y(x^2+1)^2]=\frac{1}{(x^2+1)}$

$\therefore y (x^2+1)^2=\displaystyle\int{\frac{1}{(x^2+1)}~dx}$

$\text{or,}~y(x^2+1)^2=\tan^{-1}x+c~~\text{(ans.)}$

18. $\frac{dy}{dx}+\frac{1}{x \log x}~y=\frac 2x$

Sol.

$~~~\frac{dy}{dx}+\frac{1}{x \log x}~y=\frac 2x\rightarrow(1)$

I.F. of $\,(1)$

$=e^{\displaystyle \int{\frac{dx}{x\log x}}} = e^{\displaystyle \int{\frac{d(\log x)}{\log x}}} = e^{\log(\log x)} = \log x$

Now, multiplying both sides of $\,(1)\,$ by $~\log x~$ we get,

$~~\log x ~\frac{dy}{dx}+\frac 1x~y=\frac 2x \log x$

$\text{or,}~~ \frac{d}{dx} (y \log x)=\frac 2x \log x$

$\therefore y \log x=\displaystyle \int{\frac 2x \log x~dx}$

$\text{or,}~~ y \log x=2 \displaystyle \int{\log x~d(\log x)}$

$\text{or,}~~ y \log x=2 \cdot \frac{(\log x)^2}{2}+c$

$\therefore y \log x=(\log x)^2+c~~\text{(ans.)}$

19(i). $(x^2-1)\frac{dy}{dx}+2xy=\frac{2}{x^2-1}$

Sol.

$~~~~(x^2-1)~\frac{dy}{dx}+2xy=\frac{2}{x^2-1}$

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

I.F. of $\,(1)$

$=e^{\displaystyle\int{\frac{2x}{x^2-1}~dx}} = e^{\log_e(x^2-1)} = x^2-1$

Now, multiplying both sides of $\,(1)\,$ by $~(x^2-1)~$ we get,

$~~(x^2-1)~\frac{dy}{dx}+2xy=\frac{2}{(x^2-1)}$

$\text{or,}~~ \frac{d}{dx}[y(x^2-1)]=\frac{2}{(x^2-1)}$

$\therefore y(x^2-1)=\displaystyle\int{\frac{2}{(x^2-1)}~dx}$

$\text{or,}~~ y(x^2-1)=2 \cdot \frac 12 \log\left(\frac{x-1}{x+1}\right) +c$

$\text{or,}~~ y(x^2-1)=\log \left(\frac{x-1}{x+1}\right)+c~~\text{(ans.)}$

19(ii). $(1+x^2)~\frac{dy}{dx}+2xy=4x^2 ;\\~~~\text{given}~y=0,~~\text{when}~~x=0.$

Sol.

$~~(1+x^2)~\frac{dy}{dx}+2xy=4x^2$

$\Rightarrow \frac{dy}{dx}+\frac{2x}{1+x^2}y=\frac{4x^2}{1+x^2} \rightarrow(1)$

I.F. of $\,(1)$

$=e^{\displaystyle \int{\frac{2x}{1+x^2}}~dx}=e^{\log_e(1+x^2)}=1+x^2$

Now, multiplying both sides of $\,(1)\,$ by $~1+x^2~$ we get,

$~~~(1+x^2)~\frac{dy}{dx}+2xy=4x^2$

$\text{or,}~~\frac{d}{dx} [y(1+x^2)]=4x^2$

$\therefore y(1+x^2)=\displaystyle \int{4x^2}~dx$

$\text{or,}~~ y(1+x^2)=\frac 43 x^3+c \rightarrow(3)$

Now, $~~y(0)=0 \Rightarrow c=0~~[\text{By (3)}]$

So, putting the value of $~c~$ in $\,(3)\,$ we get,

$~~y(1+x^2)=\frac 43x^3$

$\Rightarrow 4x^3=3y(1+x^2)~~\text{(ans.)}$

20. $\cos t~ \frac{dx}{dt}+x\sin t=1$

Sol.

$~~\cos t~ \frac{dx}{dt}+x\sin t=1 \Rightarrow \frac{dx}{dt}+x\tan t=\sec t\rightarrow(1)$

I.F. of $\,(1)$

$=e^{\displaystyle\int{\tan t}~dt}=e^{\log_e \sec t}=\sec t$

Now, multiplying both sides of $\,(1)\,$ by $~\sec t~$ we get,

$~~~\sec t~\frac{dx}{dt}+x \sec t\tan t=\sec^2t$

$\text{or,}~~ \frac{d}{dt}(x \sec t)=\sec^2t$

$\therefore x\sec t=\displaystyle\int{\sec^2t~dt}$

$\text{or,}~~ x\sec t=\tan t+c$

$\text{or,}~~ x=\sin t+c\cos t~~\text{(ans.)}$

21. $\frac{dy}{dx}+y\sec x=\tan x~~~\left(0 \leq x \leq \frac{\pi}{2}\right)$

Sol.

$~~\frac{dy}{dx}+y\sec x=\tan x \rightarrow(1)$

I.F. of $\,(1)$

$=e^{\displaystyle\int{\sec x}~dx} = e^{\log_e(\sec x+\tan x)} = \sec x+\tan x$

Now, multiplying both sides of $\,(1)\,$ by $~(\sec x+\tan x)~$ we get,

$~~ (\sec x+\tan x)~\frac{dy}{dx}+y\sec x(\sec x+\tan x) = \tan x (\sec x+\tan x)$

$\Rightarrow \frac{d}{dx}[y(\sec x+\tan x)]=\tan x (\sec x+\tan x)$

$\therefore y(\sec x+\tan x)=\displaystyle\int{(\sec x+\tan x)\tan x~dx}$

$\text{or,}~~y (\sec x+\tan x)=\displaystyle\int{\sec x\tan x~dx} + \displaystyle\int{(\sec^2x-1)~dx}$

$~~~~~[\because ~\tan^2x=\sec^2x-1]$

$\text{or,}~~y(\sec x+\tan x)=\sec x+\tan x - x+c~~\text{(ans.)}$