Differentiation (Part-38) | S N De

$~1.~$ If $~f'(x)=xe^x~$ and $~f(0)=1,~$ find $~f(x).$

Sol. $~~f'(x)=xe^x \\ \therefore f(x)=\displaystyle\int{xe^x~dx} \\ \text{or,}~~ f(x)=x\displaystyle\int{e^x~dx}\\-\displaystyle\int{\left(\frac{d}{dx}(x)\int{e^x~dx}\right)~dx} \\ \text{or,}~~f(x)=xe^x-\displaystyle\int{e^x~dx}\\ \text{or,}~~f(x)=xe^x-e^x+c \\ \text{or,}~~ f(x)=e^x(x-1)+c\rightarrow(1)\\ \text{where}~~c\rightarrow \text{constant of integration.}$

Now, $~f(0)=1~\text{(given)}~~$ and so by $~(1)~$ we get,

$~~f(0)=e^0(0-1)+c \\ \text{or,}~~ 1=-1+c \\ \text{or,}~~ c=2.$ 

So, putting the value of $~c~$ in $~(1),~$ we get,

$~~f(x)=e^x(x-1)+2~~\text{(ans.)}$ 

$~2.~~f'(x)=\log x~$ and $~f(1)=0~$; show that, $~f(x)=x(\log x-1)+1.$

Sol. $~f'(x)=\log x \\ \therefore f(x)=\displaystyle\int{\log x~dx}\\ \text{or,}~~f(x)=\log x\displaystyle\int{dx}\\~~~-\displaystyle\int{\left(\frac{d}{dx}(\log x)\int{dx}\right)}~dx \\ \text{or,}~~f(x)=x\log x-\displaystyle\int{\frac 1x \cdot x~dx} \\ \text{or,}~~f(x)=x\log x-\displaystyle\int{dx}\\ \text{or,}~~f(x)=x\log x-x+c\rightarrow(1)\\ ~~\text{where}~~c\rightarrow \text{constant of integration.} \\ \text{or,}~~f(1)=\log 1-1+c \\ \text{or,}~~0=0-1+c~~[\because~~f(1)=0] \\ \text{or,}~~c=1$

So, putting the value of $~c~$ in $~(1)~$ we get,

$~~f(x)=x\log x-x+1 \\ \text{or,}~~f(x)=x(\log x-1)+1 $ 

$~3.~$ A curve passes through the point $\,(3,-4)\,$ and the slope of the tangent to the curve at any point $\,(x, y)\,$ is $\,\left(-\frac{x}{y}\right).\,$  Find the equation of the curve. 

Sol. By the given condition,

$~~\frac{dy}{dx}=-\frac xy \\ \text{or,}~~ y~dy=-x~dx \\ \text{or,}~~\displaystyle \int{y~dy}=-\int{x~dx} \\ \text{or,}~~ \frac{y^2}{2}=-\frac{x^2}{2}+c\\~~\text{where}~~c\rightarrow \text{constant of integration.} \\ \text{or,}~~c=\frac{x^2+y^2}{2}\rightarrow(1)$

At $~(3,-4),~$ the value of $~c=\frac{3^2+(-4)^2}{2}=\frac 12(9+16)=\frac{25}{2}.$

So, putting the value of $~c~$ in $~(1)~$, we get

$~~\frac{25}{2}=\frac{x^2+y^2}{2} \\ \text{or,}~~ x^2+y^2=25~~\text{(ans.)}$

$~4.~~$ Determine the equation of the curve passing through the origin, the slope of the tangent of which at any point $~(x, y)~$ is $~\frac{x+1}{y+1}.$

Sol.  By the given condition,

$~~~~~\frac{dy}{dx}=\frac{x+1}{y+1} \\ \text{or,}~~(y+1)~dy=(x+1)~dx \\ \therefore \displaystyle\int{(y+1)~dy}=\int{(x+1)~dx} \\ \text{or,}~~\displaystyle\int{y~dy}+\int{dy}=\int{x~dx}+\int{dx} \\ \text{or,}~~\frac{y^2}{2}+y=\frac{x^2}{2}+x+c \rightarrow(1)\\~~~~\text{where}~~c \rightarrow\text{constant of integration.}$

At $~(0,0),~~c=0~~[\text{By (1)}]$ 

So, putting the value of $~c~$ in $~(1),~$ we get,

$~~\frac{x^2}{2}+x+0=\frac{y^2}{2}+y \\ \text{or,}~~x^2+2x=y^2+2y \\ \text{or,}~~x^2-y^2+2(x-y)=0~~\text{(ans.)}$

$~5(i)~~$ The slope of the tangent to a curve at any point $\,(x, y)\,$ is $\frac{3y + 2x+4}{4x+6y+5}.~$If the curve passes through $\,(0, -1)\,$, find the equation of the curve.

Sol. By the given condition, we have 

$~~\frac{dy}{dx}=\frac{3y+2x+4}{4x+6y+5} \\ \text{or,}~~\frac{dy}{dx}=\frac{3y+2x+4}{2(3y+2x)+5}\rightarrow(1)$

Let $~~~~z=3y+2x \\ \text{or,}~~\frac{dz}{dx}=3\frac{dy}{dx}+2 \\ \text{or,}~~\frac{dy}{dx}=\frac 13\left(\frac{dz}{dx}-2\right) \rightarrow(2)$

Hence, by $~(1),(2)~$ we get,

$~~\frac 13\left(\frac{dz}{dx}-2\right)=\frac{z+4}{2z+5} \\ \text{or,}~~\frac{dz}{dx}-2=\frac{3z+12}{2z+5} \\ \text{or,}~~\frac{dz}{dx}=\frac{3z+12}{2z+5}+2 \\ \text{or,}~~\frac{dz}{dx}=\frac{3z+12+2(2z+5)}{2z+5} \\ \text{or,}~~ \frac{dz}{dx}=\frac{7z+22}{2z+5}\\ \text{or,}~~\frac{2z+5}{7z+22}~dz=dx \\ \text{or,}~~\frac{(2/7)(7z+22)+5-(44/7)}{7z+22}~dz=dx \\ \text{or,}~~\frac 27\displaystyle\int{dz}-\frac 97\int{\frac{dz}{7z+22}}=\int{dx} \\ \text{or,}~~\frac 27\displaystyle\int{dz}-\frac{9}{7\times 7}\int{\frac{7~dz}{7z+22}}=\int{dx} \\ \text{or,}~~\frac 27z-\frac{9}{49}\log|7z+22|\\~~~=x+c \\ \text{or,}~~\frac 27(2x+3y)-\frac{9}{49}\log|7(2x+3y)+22|\\~~~~~=x+c \\ \text{or,}~~\frac{6y}{7}-\frac{9}{49}\log|14x+21y+22|\\~~~=\frac{3x}{7}+c\rightarrow(3) $ 

Now, if the curve passes through $\,(0, -1)\,$, then from $~(3),~$ we get,

$~~ -\frac 67-\frac{9}{49}\log|0-21+22|=0+c \\ \text{or,}~~ c=-\frac 67.$

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

$~~\frac{6y}{7}-\frac{9}{49}\log|14x+21y+22|=\frac{3x}{7}-\frac 67\rightarrow(4) \\ \text{or,}~~ 7\times 6y-9\log|14x+21y+22|\rightarrow(5)\\~~~~~=7(3x-6)\\~~~[\text{Multiplying both sides of (4) by}~~49] \\ \text{or,}~~7\times 2y-3\log|14x+21y+22|\\~~~~=7(x-2)\\~~[\text{Dividing both sides of (5) by}~3] \\ \text{or,}~~7(x-2y-2)\\~~~~+3\log|14x+21y+22|=0~~\text{(ans.)}$

$~5(ii)~~$ Show that, the family of curves for which the slope of the tangent at any point $~(x, y)~$ on it is $~\frac{x^2+y^2}{2xy}~$  given by $~x^2-y^2=cx.$ 

Sol. Sol. By the given condition, we have 

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

let $~~~~y=vx \\ \text{or,}~~\frac{dy}{dx}=v+x\frac{dv}{dx}\rightarrow(2)$

Then from $~(1),(2)~$ we get,

$~~v+x\frac{dv}{dx}=\frac{x^2+(vx)^2}{2x(vx)} \\ \text{or,}~~ v+x\frac{dv}{dx}=\frac{x^2(1+v^2)}{2x^2v}\\ \text{or,}~~ v+x\frac{dv}{dx}=\frac{1+v^2}{2v} \\ \text{or,}~~ x\frac{dv}{dx}=\frac{1+v^2}{2v}-v \\ \text{or,}~~x\frac{dv}{dx}=\frac{1+v^2-2v^2}{2v} \\ \text{or,}~~ x\frac{dv}{dx}=\frac{1-v^2}{2v} \\ \text{or,}~~ -\displaystyle\int{\left(\frac{-2v}{1-v^2}\right)}~dv=\int{\frac{dx}{x}} \\ \text{or,}~~ -\log|1-v^2|=\log|x|+\log c_1 \\ \text{or,}~~-\log\left|1-\frac{y^2}{x^2}\right|=\log(c_1|x|) \\ \text{or,}~~-\log\left|\frac{x^2-y^2}{x^2}\right|=\log(c_1|x|)  \\ \text{or,}~~\log\left|\frac{x^2}{x^2-y^2}\right|=\log(c_1|x|) \\ \therefore \frac{x^2}{x^2-y^2}=c_1x \\ \text{or,}~~\frac{x}{x^2-y^2}=c_1 \\ \text{or,}~~ x^2-y^2=\frac{x}{c_1} \\ \text{or,}~~ x^2-y^2=cx~~(\text{showed})\\~~~~~[~\text{where}~~c=1/c_1.]$ 

$~6.~~$ A curve is passing through the point $~(4, 3)~$ and at any point the gradient of the tangent to the curve is reciprocal of the ordinate of the point. Obtain the equation of the curve.

Sol. By the given condition,

$~~~~\frac{dy}{dx}=\frac 1y \\ \text{or,}~~y~dy=dx \\ \text{or,}~~\displaystyle\int{y~dy}=\int{dx} \\ \text{or,}~~\frac{y^2}{2}=x+c \rightarrow(1)$

Since the curve is passing through the point $~(4, 3)~$, we get from $~(1),$

$~~\frac{3^2}{2}=4+c \\ \text{or,}~~c=\frac 92-4 \\ \text{or,}~~ c=\frac 12$

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

So, putting the value of $~c~$ in $~(1)~$, we get,

$~~\frac{y^2}{2}=x+\frac 12 \\ \text{or,}~~y^2=2x+1~~\text{(ans.)}$

$~7.~~$  Show that, the equation of the curve passing through $~(1, 0)~$ and satisfying $~(1+ y^2)~dx-xy~dy=0~$ is $~x^2-y^2=1$.

Sol. By the given condition,

$~~(1+ y^2)~dx-xy~dy=0 \\ \text{or,}~~ (1+y^2)~dx=xy~dy\\ \text{or,}~~\displaystyle\int{\frac{dx}{x}}=\frac{1}{2}\int{\frac{2y}{1+y^2}}~dy \\ \text{or,}~~\log|x|=\frac 12\log|1+y^2|+\log c \\ \text{or,}~~ 2\log|x|=\log|1+y^2|+2\log c \\ \text{or,}~~ \log|x|^2=\log|1+y^2|+\log c^2\rightarrow(1)$

Now at $~(1,0),~$ we get from $\,(1),\,$

$~~\log 1^2=\log|1+0|+\log c^2 \\ \text{or,}~~0=0+\log c^2 \\ \text{or,}~~\log c^2=0\rightarrow(2)$

Hence from $\,(1),(2)~$ we get,

$~~\log|x|^2=\log|1+y^2| \\ \text{or,}~~ x^2=1+y^2 \\ \therefore~~ x^2-y^2=1~~(\text{showed})$