Differential Equations

Let a point P(h, k) on the curve y = y(x), so equation of tangent to the curve at point P is

$y - k = {\left( {{{dy} \over {dx}}} \right)_{h,k}}(x - h)$ ....(i)

Now, the tangent (i) intersect the Y-axis at Y_p, so coordinates Y_p is $\left( {0,k - h{{dy} \over {dx}}} \right)$,

where ${{{dy} \over {dx}}}$ = ${\left( {{{dy} \over {dx}}} \right)}$_(h,k)

So, PY_p = 1 (given)

$ \Rightarrow \sqrt {{h^2} + {h^2}{{\left( {{{dy} \over {dx}}} \right)}^2}} = 1$

$ \Rightarrow {{dy} \over {dx}} = \pm {{\sqrt {1 - {x^2}} } \over x}$

[on replacing h by x]

$ \Rightarrow dy = \pm {{\sqrt {1 - {x^2}} } \over x}dx$

On puting x = sin$\theta $, dx = cos$\theta $d$\theta $, we get

$dy = \pm {{\sqrt {1 - {{\sin }^2}\theta } } \over {\sin \theta }}\cos \theta d\theta $

$ = \pm {{{{\cos }^2}\theta } \over {\sin \theta }}d\theta $

$ = \pm (\cos ec\theta - \sin \theta )d\theta $

$ \Rightarrow y = \pm [1n(\cos ec\theta - \cot \theta ) + \cos \theta ] + C$

$ \Rightarrow y = \pm \left[ {1n\left( {{{1 - \cos \theta } \over {\sin \theta }}} \right) + \cos \theta } \right] + C$

$ \Rightarrow y = \pm \left[ {1n\left( {{{1 - \sqrt {1 - {{\sin }^2}\theta } } \over {\sin \theta }}} \right) + \sqrt {1 - {{\sin }^2}\theta } } \right] + C$

$ \Rightarrow y = \pm \left[ {1n\left( {{{1 - \sqrt {1 - {x^2}} } \over x}} \right) + \sqrt {1 - {x^2}} } \right] + C$

[$ \because $ x = sin$\theta $]

$ = \pm \left[ { - 1n{{1 + \sqrt {1 - {x^2}} } \over x} + \sqrt {1 - {x^2}} } \right] + C$

[on rationalization]

$ \because $ The curve is in the first quadrant so y must be positive, so

$y = 1n\left( {{{1 + \sqrt {1 - {x^2}} } \over x}} \right) - \sqrt {1 - {x^2}} + C$

As curve passes through (1, 0), so

$0 = 0 - 0 + c \Rightarrow c = 0$, so required curve is

$y = 1n\left( {{{1 + \sqrt {1 - {x^2}} } \over x}} \right) - \sqrt {1 - {x^2}} $

and required differential eaquation is

${{dy} \over {dx}} = - {{\sqrt {1 - {x^2}} } \over x}$

$ \Rightarrow xy' + \sqrt {1 - {x^2}} = 0$

Hence, options (a) and (c) are correct.

$g(x) = \int_{\sin x}^{\sin (2x)} {{{\sin }^{ - 1}}} (t)\,dt$

g'(x) = 2cos 2x sin^$-$1(sin2 x) $-$ cos x sin^$-$1(sin x)

$g'\left( {{\pi \over 2}} \right) = - 2{\sin ^{ - 1}}(0) = 0$

$g'\left( { - {\pi \over 2}} \right) = - 2{\sin ^{ - 1}}(0) = 0$

No option is matching.

Options (A) and (B): The given differential equation is

$ \left[x^2+4 x+4+y(x+2)\right] \frac{d y}{d x}-y^2=0 \quad(x>0) $

which is further simplified as follows:

$ \left[(x+2)^2+y(x+2)\right] \frac{d y}{d x}-y^2=0 $

Substituting $x+2=t$, we get

$ \frac{d x}{d y}=\frac{d t}{d y} $

Now,

$ (x+2)^2+y(x+2)-y^2 \frac{d x}{d y}=0 $

That is,

$ \begin{aligned} &t^2+y t-y^2 \frac{d t}{d y} =0 \\\\ &y^2 \frac{d t}{d y}-y t-t^2 =0 \\\\ &\frac{1}{t^2} \frac{d t}{d y}-\frac{1}{y t} =\frac{1}{y^2} \end{aligned} $

Let $\frac{1}{t}=z$; therefore,

$ \frac{d t}{d y}\left(-\frac{1}{t^2}\right)=\frac{d z}{d y} $

Now,

$\begin{aligned} \frac{-d z}{d y}-\frac{z}{y} & =\frac{1}{y^2} \Rightarrow \frac{d z}{d y}+\frac{z}{y}=\frac{-1}{y^2} \\\\ \Rightarrow d(z y) & =\int-\frac{1}{y} d y \\\\ \Rightarrow z y & =-\ln |y|+\ln c \\\\ \Rightarrow \frac{y}{t} & =-\ln |y|+\ln c\end{aligned}$

$ \Rightarrow \frac{y}{(x+2)}=-\ln |y|+\ln c ~~~~~...(1) $

which passes through the point $(1,3)$. Therefore, from Eq. (1), we get

$ \begin{aligned} &\frac{z}{3} =-\ln 3+c \Rightarrow c=\ln 3 e \\\\ &\frac{y}{x+2} =-\ln |y|+\ln 3 e=\ln \left(\frac{3 e}{|y|}\right) \\\\ &\frac{3 e}{|y|} =e^{y /(x+2)} \\\\ &3 e =|y| e^{y /(x+2)} \end{aligned} $

Substituting $y=(x+2)$, we get $3 e=|x+2| e^1$

$ |x+2|=3 \Rightarrow x+2=-3,3 \Rightarrow x=-5,1 $

Therefore, $x=1$ (since $x \neq-5)$.


That is, the solution curve intersects $y=(x+2)$ exactly at one point and not at two points. Therefore, option (A) is correct and option (B) is incorrect.

Option (C): We have

$ \frac{3 e}{\left|(x+2)^2\right|}=e^{(x+2)} $

which meets at two points for $x<0$ and for $x>0$, there is no intersection point.

Hence, option $(\mathrm{C})$ is incorrect.

Option (D): We have

$ \frac{3 e}{(x+3)^2}=e^{\frac{(x+3)^2}{(x+2)}}=e^{\frac{(x+2)^2+1+2(x+2)}{(x+2)}}=e^{2+\frac{1}{(x+2)}+(x+2)} $

Therefore, there is no intersection point for $x>0$. Hence option (D) is correct.

Given: $\left(1+e^x\right) y^{\prime}+y e^x=1$

$\left(1+e^x\right) \frac{d y}{d x}+y e^x=1$

$\frac{d y}{d x}+\frac{e^x}{1+e^x} y=\frac{1}{1+e^x}$, which is liner differential equation in $y$.

Comparing above equation with $\frac{d y}{d x}+\mathrm{P} y=Q$, we get $Q$, we get

$\begin{aligned} \quad \mathrm{P} & =\frac{e^x}{1+e^x} \text { and } \mathrm{Q}=\frac{1}{1+e^x} \\ \text { So, I.F. } & =e^{\int \mathrm{P} \cdot d x} \\ \text { I.F. } & =e^{\int \frac{e^x}{1+e^x} d x} \\ \text { I.F. } & =e^{\ln \left(1+e^x\right)} \end{aligned}$

$\left\{\because \int \frac{f^{\prime}(x)}{f(x)} d x=\ln f(x)\right\}$

$\text { I.F }=1+e^x$

So, solution of given differential equation is given by

$y \text { (I.F.) }=\int \text { Q.(I.F.) } d x$

$\begin{aligned} y \cdot\left(1+e^x\right) & =\int \frac{1}{1+e^x} \cdot\left(1+e^x\right) d x \\ y\left(1+e^x\right) & =\int 1 d x \\ y\left(1+e^x\right) & =x+\mathrm{C} \\ \because \quad y(0) & =2 \\ \Rightarrow \quad 2\left(1+e^0\right) & =0+\mathrm{C} \\ c & =4 \end{aligned}$

So, $y(x)=\frac{x+4}{1+e^x}\quad \text{... (i)}$

Put $x=-4$ in the equation, we get

$y(-4)=0$

Put $x=-2$ in the above equation (i), we get

$y(-2)=\frac{2}{1+e^{-2}} \neq 0$

For critical points, $y^{\prime}=0$

From e.q, (i), $y\left(1+e^x\right)=x+4$

Differentiating the above equation w.r.t. $x$, we get

$\begin{aligned} y^{\prime}\left(1+e^x\right)+y\left(e^x\right) & =1 \\ 0\left(1+e^x\right)+y e^x & =1 \quad \left\{\because y^{\prime}=0\right\} \\ y e^x-1 & =0 \end{aligned}$

$\quad \begin{aligned} \text { Now, } \text { let } g(x) & =y e^x-1 \\ g(x) & =\frac{(x+4) e^x}{1+e^x}-1 \\ g(x) & =\frac{(x+3) e^x-1}{1+e^x} \end{aligned}$

$\text { Now, } \quad g(-1)=\frac{2 e^{-1}-1}{1+e^{-1}}=\frac{2-e}{1+e}<0 \quad \{\because e=2 \cdot 7 \cdot 8\}$

$\text { And, } \quad g(0)=\frac{3 e^0-1}{1+e^0}=1>0$

So, there exists one value of $x$ in $(-1,0)$ for which

$\begin{aligned} g(x) & =0 \\ \Rightarrow \quad y^{\prime} & =0 \end{aligned}$

There exist a critical point of $y(x)$ in $(-1,0)$

Hint :

(i) Use solution of liner differential equation

$\begin{aligned} \frac{d y}{d x}+\mathrm{P} y & =\mathrm{Q} \text { is given by } \\ y \text { (I.F.) } & =\int \mathrm{Q} \cdot \text { (I.F) } d x, \text { where I.F }=e^{\int p \cdot d x} \end{aligned}$

(ii) For critical points, $y^{\prime}=0$

Let equation of circle whose centre lie on straight line $y=x$ be

$(x-k)^2+(y-k)^2=r^2~~~~...(i)$

Differentiating the above equation w.r.t. $x$, we get

$ \begin{aligned} & 2(x-k)+2(y-k) y^{\prime}=0 \\\\ \Rightarrow & x+y y^{\prime}=k\left(1+y^{\prime}\right)~~~~~...(ii) \\\\ \Rightarrow & k=\frac{x+y y^{\prime}}{1+y^{\prime}} \end{aligned} $

Differentiating the eq. (ii) w.r.t. $x$, we get

$ \begin{aligned} & \Rightarrow 1+y y^{\prime \prime}+\left(y^{\prime}\right)^2=k y^{\prime \prime} \quad\left\{\because(u v)^{\prime}=u v^{\prime}+v u^{\prime}\right\} \\\\ & \Rightarrow 1+y y^{\prime \prime}+\left(y^{\prime}\right)^2=\left(\frac{x+y y^{\prime}}{1+y^{\prime}}\right) y^{\prime \prime} \\\\ & \Rightarrow 1+y y^{\prime \prime}+\left(y^{\prime}\right)^2+y^{\prime}+y y^{\prime} y^{\prime \prime}+\left(y^{\prime}\right)^3 \\\\ & =x y^{\prime \prime}+y y^{\prime} y^{\prime \prime} \\\\ & \Rightarrow y^{\prime \prime}(y-x)+\left(y^{\prime}\right)^2\left(1+y^{\prime}\right)+1+y^{\prime}=0 \\\\ & \Rightarrow y^{\prime \prime}(y-x)+y^{\prime}\left(y^{\prime}+\left(y^{\prime}\right)^2+1\right)+1=0 \end{aligned} $

Comparing the above equation with $p y^{\prime \prime}+\mathrm{Q} y^{\prime}$ $+1=0$, we get

$ \begin{aligned} & P=y-x \text { and } \mathrm{Q}=y^{\prime}+\left(y^{\prime}\right)^2+1 \\\\ \therefore & P+Q=1-x+y+y^{\prime}+\left(y^{\prime}\right)^2 \end{aligned} $

Given, $\frac{d y}{d x}-y \tan x=2 x \sec x$

Comparing with the liner differential form

$\begin{gathered} \frac{d y}{d x}+\mathrm{P} y=\mathrm{Q} \\ \Rightarrow \mathrm{P}=-\tan x \text { and } \mathrm{Q}=2 x \sec x \end{gathered}$

Now, Integrating factor (I.F) $=e^{\int P d x}$

$\begin{aligned} & =e^{\int-\tan x d x} \\ & =e^{-\log \sec x} \\ & =\cos x \end{aligned}$

$\begin{aligned} & \text { The solution will be } y \text { (I.F) }=\int Q \text { (I.F) } d x \\ & \Rightarrow \quad y \cos x=\int 2 x \sec x \cdot \cos x d x \\ & \Rightarrow \quad y \cos x=2 \frac{x^2}{2}+c \\ & \Rightarrow \quad y=x^2 \sec x+c \sec x \\ \end{aligned}$

Now, $y(0)=0$

$\begin{aligned} & \Rightarrow \quad 0=0^2 \sec 0+\mathrm{c} \mathrm{sec} 0 \\ & \Rightarrow \quad c=0 \\ & \therefore \quad y=x^2 \sec x \\ & y^{\prime}=x^2 \sec x \cdot \tan x+2 x \sec x \quad \text{[using Product Rule in Differentiation]}\\ \end{aligned}$

Now,

$\begin{aligned} y\left(\frac{\pi}{4}\right) & =\left(\frac{\pi}{4}\right)^2 \sec \frac{\pi}{4}=\frac{\pi^2}{16} \sqrt{2}=\frac{\pi^2}{8 \sqrt{2}} \\ y^{\prime}\left(\frac{\pi}{4}\right) & =\left(\frac{\pi}{4}\right)^2 \sec \frac{\pi}{4} \tan \frac{\pi}{4}+2 \cdot \frac{\pi}{4} \sec \frac{\pi}{4} \\ & =\frac{\pi^2}{16} \cdot \sqrt{2} \cdot 1+\frac{\pi}{2} \cdot \sqrt{2} \\ & =\frac{\pi^2}{8 \sqrt{2}}+\frac{\pi}{\sqrt{2}} \end{aligned}$

$\begin{aligned} y\left(\frac{\pi}{3}\right) & =\left(\frac{\pi}{3}\right)^2 \sec \frac{\pi}{3} \\ & =\frac{\pi^2}{9} \cdot 2=\frac{2 \pi^2}{9} \\ y^{\prime}\left(\frac{\pi}{3}\right) & =\left(\frac{\pi}{3}\right)^2 \sec \frac{\pi}{3} \cdot \tan \frac{\pi}{3}+2 \cdot \frac{\pi}{3} \cdot \sec \frac{\pi}{3} \\ & =\frac{\pi^2}{9} \cdot 2 \cdot \sqrt{3}+\frac{2 \pi}{3} \cdot 2 \\ & =\frac{2 \pi^2}{3 \sqrt{3}}+\frac{4 \pi}{3} \end{aligned}$