Differentiation (Part-38) | S N De

 

$~8.~~4~\frac{dy}{dx}+8y=5e^{-3x}$

Sol. $~~4~\frac{dy}{dx}+8y=5e^{-3x} \\ \therefore \frac{dy}{dx}+2y=\frac 54 e^{-3x}\rightarrow(1)$

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

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

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

$~~~~~~e^{2x}~\frac{dy}{dx}+2e^{2x}y=\frac 54 e^{-x} \\ \text{or,}~~ \frac{d}{dx}(ye^{2x})=\frac 54 e^{-x} \\ \therefore ye^{2x}=\displaystyle\int{\frac 54 e^{-x}~dx} \\ \text{or,}~~ ye^{2x}=\frac 54 \cdot \frac{e^{-x}}{(-1)} +c \\ \text{or,}~~ y=-\frac 54 e^{-3x}+ce^{-2x}~~\text{(ans)}\\~~~~~[\text{where}~~c \rightarrow \text{constant of integration.}]$

$~9.~~\cos^2x~\frac{dy}{dx}+y=\tan x~~\left(0 \leq x \leq \pi/2\right)$

Sol. $~~\cos^2x~\frac{dy}{dx}+y=\tan x \\ \therefore \frac{dy}{dx}+y\sec^2x=\sec^2\tan x \rightarrow(1)\\~~~[\text{Dividing both sides by}~~\cos^2x]$

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

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

Now, multiplying both sides of $\,(1)\,$ by $~e^{\tan x}~$ we get,

$~~e^{\tan x}~\frac{dy}{dx}+ye^{\tan x}\sec^2x=e^{\tan x}\sec^2x \tan x \\ \text{or,}~~ \frac{d}{dx}(ye^{\tan x})=e^{\tan x}\sec^2x \tan x \\ \therefore ye^{\tan x}= \displaystyle\int{e^{\tan x}\sec^2x \tan x }~dx  \\ \text{or,}~~ ye^{\tan x}=\displaystyle\int{ze^z~dz} \\~~~[ \text{let}~\tan x=z \Rightarrow \sec^2x~dx=dz] \\ \text{or,}~~ ye^{\tan x}=ze^z-e^z+c~[*] \\ \text{or,}~~ ye^{\tan x}=\tan x~e^{\tan x}-e^{\tan x}+c \\ \therefore y=\tan x-1+ce^{-\tan x}~~\text{(ans.)}$

Note[*] : $~~\displaystyle\int{ze^z}~dz\\=z\displaystyle\int{e^z}~dz-\int{\left[\left(\frac{d}{dz}(z)\right)\int{e^z}~dz\right]dz}\\=ze^z-\displaystyle\int{1.e^z~dz}\\=ze^z-e^z$

$~10.~~\frac{dy}{dx}-\frac nx \cdot y=e^x x^n$

Sol. $~~~\frac{dy}{dx}-\frac nx \cdot y=e^x x^n \rightarrow(1)$

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

$=e^{\int(-n/x)~dx}\\=e^{-\int\frac{dx}{x}}\\=e^{-n\log x}\\=e^{\log x^{-n}}\\=x^{-n}\\=\frac{1}{x^n}$

Now, multiplying both sides of $\,(1)\,$ by $~\frac{1}{x^n}~$ we get,

$~~\frac{1}{x^n}~\frac{dy}{dx}-\frac{n}{x^{n+1}}\cdot y=e^x \\ \Rightarrow \frac{d}{dx}\left(y \cdot \frac{1}{x^n}\right)=e^x \\ \therefore  \frac{y}{x^n}=\int{e^x~dx} \\ \text{or,}~~ \frac{y}{x^n}=e^x+c \\ \text{or,}~~ y=x^n(e^x+c)~~\text{(ans.)}$

Note : $~~c \rightarrow  \text{constant of integration.}$ 

$~11.~~\frac{dy}{dx}-y\tan x=e^x\sec x$

Sol. $~~~\frac{dy}{dx}-y\tan x=e^x\sec x\rightarrow(1)$

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

$=e^{\int{(-\tan x)}~dx}\\=e^{-\log (\sec x)}\\=e^{\log (\sec x)^{-1}}\\=e^{\log(\cos x)}\\=\cos x$

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

$~~~\cos x\frac{dy}{dx}-y \tan x \cdot \cos x= e^x \sec x \cdot \cos x \\ \text{or,}~~ \cos x \frac{dy}{dx}-y\sin x=e^x \\ \text{or,}~~ \frac{d}{dx}(y\cos x)=e^x \\ \therefore y\cos x=\int{e^x~dx} \\ \text{or,}~~ y \cos x=e^x+c~~\text{(ans.)}~~[*]$

Note[*] : $~~c \rightarrow \text{constant of integration.}$

$~12.~~\frac{dy}{dx}+\frac nx y=\frac{a}{x^n}$

Sol. $~~~\frac{dy}{dx}+\frac nx y=\frac{a}{x^n}\rightarrow(1)$

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

$=e^{\int{\frac nx ~dx}}\\=e^{n\log x}\\=e^{\log x^n}\\=x^n$

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

$~~~x^n\frac{dy}{dx}+nx^{n-1}=a \\ \text{or,}~~\frac{d}{dx}(yx^n)=a \\ \therefore yx^n=\int{a~dx} \\ \text{or,}~~ x^ny=ax+c~~\text{(ans.)}$

Note[*] : $~~c \rightarrow \text{constant of integration.}$

$~13.~~\frac{dy}{dx}-3y\cot x=\sin 2x,\\~~~\text{given}~~y=2~~\text{when}~~x=\pi/2.$

Sol. $~~\frac{dy}{dx}-3y\cot x=\sin 2x\rightarrow(1)$

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

$=e^{\int{(-3\cot x)~dx}}\\=e^{-3\log_e \sin x}\\=e^{\log_e (\sin x)^{-3}}\\=(\sin x)^{-3}\\=\frac{1}{\sin^3 x}$

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

$~~~\frac{1}{\sin^3x}\cdot \frac{dy}{dx}-\frac{3\cos x}{\sin^4x}y=\frac{\sin2x}{\sin^3x} \\ \Rightarrow \frac{d}{dx} \left(y \cdot \frac{1}{\sin^3x}\right)=\frac{\sin 2x}{\sin^3x} \\ \therefore y \cdot \frac{1}{\sin^3x}=\displaystyle\int{\frac{2\sin x\cos x}{\sin^3x}} \\ \text{or,}~~  \frac{y}{\sin^3x}=2\displaystyle\int{\frac{\cos x}{\sin^2x}}~dx \\ \text{or,}~~ \frac{y}{\sin^3x}=2\displaystyle\int{\frac{d(\sin x)}{\sin^2x}} \\ \text{or,}~~ \frac{y}{\sin^3x}=2 \cdot \left(\frac{\sin^{-2+1}x}{(-2+1)}\right)+c \\ \text{or,}~~\frac{y}{\sin^3x}=2\cdot \frac{1}{-\sin x}+c\rightarrow(2) $

Now, $~\text{it has been given}~~y=2~~\text{when}~~x=\pi/2.$

So, from $\,(2)\,$ we get,

$~~\frac{2}{\sin^3(\pi/2)}=2\cdot \frac{1}{-\sin(\pi/2)} +c \\ \Rightarrow 2=-2+c \\ \therefore c=2+2=4.$

Now, putting the value of $\,c\,$ in $\,(2)\,$ we get,

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

$~14.~~\frac{dy}{dx}+2y=6e^x~~~[\text{CBSE-2007}]$

Sol. $~~\frac{dy}{dx}+2y=6e^x\rightarrow(1)$

I.F. of $\,(1)$ 

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

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

$~~e^{2x}~\frac{dy}{dx}+2e^{2x}y=6e^{3x} \\ \Rightarrow \frac{d}{dx}(ye^{2x})=6e^{3x} \\ \therefore ye^{2x}=\displaystyle\int{6e^{3x}~dx}\\ \Rightarrow ye^{2x}=6 \cdot \frac{e^{3x}}{3}+c \\ \Rightarrow y=2e^{x}+ce^{-2x}~~\text{(ans.)}$

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