Differential Equations

(x² $-$ y²) dx + 2xy dy = 0

${{dy} \over {dx}} = {{{y^2} - {x^2}} \over {2xy}}$

Put   $y = vx \Rightarrow {{dy} \over {dx}} = v + x{{dv} \over {dx}}$

Solving we get,

$\int {{{2v} \over {{v^2} + 1}}dv = \int { - {{dx} \over x}} } $

ln(v² + 1) = $-$ ln x + C

(y² + x²) = Cx

1 + 1 = C $ \Rightarrow $ C = 2

y² + x² = 2x

${{dy} \over {dx}} + 3{\sec ^2}x.y = {\sec ^2}x$

I.F. = ${e^{3\int {{{\sec }^2}xdx} }} = {e^{3\tan x}}$

or   $y.e{}^{3\tan x} = \int {{{\sec }^2}x.{e^{3\tan x}}} $

or   $y.{e^{3\tan x}} = {1 \over 3}{e^{3\tan x}} + C$

Given

$y\left( {{\pi \over 4}} \right) = {4 \over 3}$

$ \therefore $   ${4 \over 3}.{e^3} = {1 \over 3}{e^3} + C$

$ \therefore $   C = e³

If f(xy) = f(x) f(y) $\forall $ x, y $ \in $ R and f(0) $ \ne $ 0

put x = y = 0

$ \Rightarrow $  f(0) = [f(0)]²

$ \Rightarrow $  f(0) = 1

put y = 0 $ \Rightarrow $ f(0) = f(x) f(0)

$ \Rightarrow $ f(x) = 1

given that ${{dy} \over {dx}}$ = f(x)

$ \therefore $  ${{dy} \over {dx}}$ = 1 $ \Rightarrow $ y = x + k

given that y(0) = 1

$ \therefore $  k = 1

hence y = x + 1

y$\left( {{1 \over 4}} \right)$ + y$\left( {{3 \over 4}} \right)$ = $\left( {{1 \over 4} + 1} \right)$ + $\left( {{3 \over 4} + 1} \right)$ = 3

Given,

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

$ \Rightarrow $  ${{dy} \over {dx}} + \left( {{2 \over x}} \right)y = x$

This is a linear differential equation.

$ \therefore $  I.F $ = {e^{\int {{2 \over x}dx} }}$

$ = {e^{2\ln x}}$

$ = {x^2}$

$ \therefore $  Solution is,

$y \cdot {x^2} = \int {x \cdot {x^2}dx} $

$ \Rightarrow $  $y{x^2} = {{{x^4}} \over 4} + C$

given  $y\left( 1 \right) = 1$

$ \therefore $  $1.1 = {4 \over 4} + C$

$ \Rightarrow $  $C = {3 \over 4}$

$ \therefore $  Equation is

$y{x^2} = {{{x^4}} \over 4} + {3 \over 4}$

$ \therefore $  $y\left( {{1 \over 2}} \right)$   means   $x = {1 \over 2}$

$ \therefore $  $y \cdot {\left( {{1 \over 2}} \right)^2} = {1 \over 4} \times {\left( {{1 \over 2}} \right)^4} + {3 \over 4}$

$ \Rightarrow $  ${y \over 4} = {1 \over {64}} + {3 \over 4}$

$ \Rightarrow $   ${y \over 4} = {{1 + 48} \over {64}}$

$ \Rightarrow $  y = ${{49} \over {16}}$

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.

Equation of ellipse,

${{{x^2}} \over {{a^2}}} + {{{y^2}} \over {{b^2}}} = 1$

As ellipse passes through (0, 3)

$\therefore\,\,\,$ ${{{0^2}} \over {{a^2}}} + {{{3^2}} \over {{b^2}}} = 1$

$ \Rightarrow $    b² = 9

$\therefore\,\,\,$ Equation of ellipse becomes,

${{{x^2}} \over {{a^2}}} + {{{y^2}} \over 9} = 1$

Differentiating w.r.t    x, we get,

${{2x} \over {a{}^2}}$ + ${{2y} \over 9}$ . ${{dy} \over {dx}} = 0$

$ \Rightarrow $    ${x \over {{a^2}}}$ = $-$ ${y \over 9}.{{dy} \over {da}}$

$ \Rightarrow $    ${x \over {{a^2}}} = - {y \over 9}.y'......$ (1)

We got earlier,

${{{x^2}} \over {{a^2}}} + {{{y^2}} \over 9}$ = 1

$ \Rightarrow $   ${x \over {{a^2}}}.x + {{{y^2}} \over 9} = 1$

putting value of equation (1) here,

$ {-{y\,y'} \over 9}.x + {{{y^2}} \over 9} = 1$

$ \Rightarrow $    $-$ xyy' + y² = 9

$ \Rightarrow $   xyy' $-$ y² + 9 = 0

Given,

sin x ${{dy} \over {dx}} + y\cos y = 4x$

$ \Rightarrow \,\,\,\,{{dy} \over {dx}}\,$ + y cot x = 4x cosec x

This is a linear differential equation of form,

${{dy} \over {dx}}\,$ + py = Q

Where p = cot x and Q = 4x cosec x

So, Integrating factor (I. F)

= ${e^{\int {pdx} }}$

= ${e^{\int {\cot dx} }}$

= ${e^{\ln \left| {\sin \,x} \right|}}$

= sin x as $x \in \left( {0,\pi } \right)$

Solution of the differential equation is

y sin x = $\int {} $ 4x cosecx sinx dx + c

$ \Rightarrow \,\,\,\,$ y sinx = $\int {4x\,dx\, + c} $

$ \Rightarrow \,\,\,\,$ y sinx = 4.${{{x^2}} \over 2} + c$

$ \Rightarrow \,\,\,\,$ y sinx = 2x² + c . . . . . (1)

Given that, $y\left( {{\pi \over 2}} \right) = 0$

$\therefore\,\,\,$ x = $y\left( {{\pi \over 2}} \right) = 0$ and y = 0

Put this x = ${\pi \over 2}$ and y = 0 at equation (1)

0.1 = 2. $\left( {{\pi \over 2}} \right)$² + c

$ \Rightarrow \,\,\,$ c =$ - {{{\pi ^2}} \over 2}$

So, differential equation is

y sin x = 2x² $-$ ${{{\pi ^2}} \over 2}\,\,.....(2)$

Now we have to find y $\left( {{\pi \over 6}} \right).$

So, put x = ${{\pi \over 6}}$ at equation (2)

y . sin ${{\pi \over 6}}$ = 2 ${\left( {{\pi \over 6}} \right)^2} - {{{\pi ^2}} \over 2}$

$ \Rightarrow \,\,\,\,\,\,\,y.{1 \over 2}$ = 2. ${{{\pi ^2}} \over {36}} - {{{\pi ^2}} \over 2}$

$ \Rightarrow \,\,\,\,\,{y \over 2} = {{{\pi ^2}} \over {18}} - {{{\pi ^2}} \over 2}$

$ \Rightarrow \,\,\,\,{y \over 2} = {{{\pi ^2} - 9{\pi ^2}} \over {18}}$

$ \Rightarrow \,\,\,\,\,y = - {{8{\pi ^2}} \over 9}$

(x² $-$ y²) dx + 2xydy = 0

$ \Rightarrow $ ${{dy} \over {dx}}$ = ${{{y^2} - {x^2}} \over {2xy}}$

Let y = vx

${{dy} \over {dx}}$ = v + x ${{dv} \over {dx}}$

$ \Rightarrow $  v + x${{dv} \over {dx}}$ = ${{{v^2}{x^2} - {x^2}} \over {2v{x^2}}}$ $ \Rightarrow $ v + x${{dv} \over {dx}}$ = ${{{v^2} - 1} \over {2v}}$

$ \Rightarrow $  x${{dv} \over {dx}}$ = ${{ - {v^2} - 1} \over {2v}}$

$ \Rightarrow $  ${{2vdv} \over {{v^2} + 1}}$ = $-$ ${{dx} \over x}$

After intergrating, we get

$\ln \left| {{v^2} + 1} \right|$ = $-$ ln$\left| x \right|$ + lnc

${{y{}^2} \over {{x^2}}}$ + 1 = ${c \over x}$

As curve passes through the point (1, 1), so 1 + 1 = c

$ \Rightarrow $  c = 2

x² + y² $-$ 2x = 0, which is a circle of radius one.

When x $ \in $ [0, 1], then ${{dy} \over {dx}}$ + 2y = 1 

$ \Rightarrow $ y = ${1 \over 2}$ + C₁e^$-$2x

$ \because $ y(0) = 0 $ \Rightarrow $ y(x) = ${1 \over 2}$ $-$ ${1 \over 2}$e^$-$2x

Here, y(1) = ${1 \over 2}$ $-$ ${1 \over 2}$ e^$-$2 = ${{{e^2} - 1} \over {2{e^2}}}$

When $x \notin \left[ {0,1} \right]$, then ${{dy} \over {dx}}$ + 2y = 0 $ \Rightarrow $ y = c₂ e^$-$2x

$ \because $  y(1) = ${{{e^2} - 1} \over 2}$ $ \Rightarrow $ ${{{e^2} - 1} \over 2}$ = c²e^$-$2 $ \Rightarrow $ C₂ = ${{{e^2} - 1} \over 2}$

$ \therefore $  y(x) $\left( {{{{e^2} - 1} \over 2}} \right){e^{ - 2x}} \Rightarrow y\left( {{3 \over 2}} \right)$ = ${{{e^2} - 1} \over {2{e^3}}}$

It is given that

$2x = {y^{1/5}} + {y^{ - 1/5}}$

$ \Rightarrow 2x = {y^{1/5}} + 1/{y^{1/5}}$

Therefore, $2x = a + {1 \over a} \Rightarrow {a^2} - 2ax + 1 = 0$

$a = {{2x \pm \sqrt {4{x^2} - 4} } \over 2}$

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

$ \Rightarrow a = x \pm \sqrt {{x^2} - 1} $

$ \Rightarrow {y^{1/5}} = x \pm \sqrt {{x^2} - 1} $

$ \Rightarrow y = {(x \pm \sqrt {{x^2} - 1} )^5}$

Therefore,

${{dy} \over {dx}} = 5{(x \pm \sqrt {{x^2} - 1} )^4}\left( {1 \pm {{2x} \over {2\sqrt {{x^2} - 1} }}} \right)$

$ = 5{(x + \sqrt {{x^2} - 1} )^4}\left( {{{\sqrt {{x^2} - 1} \pm x} \over {\sqrt {{x^2} - 1} }}} \right)$

$ \Rightarrow {{dy} \over {dx}} = {{ - 5y} \over {\sqrt {{x^2} - 1} }}$ ...... (1)

$ \Rightarrow {{{d^2}y} \over {d{x^2}}} = {{\left[ {\sqrt {{x^2} - 1} \left( { - 5{{dy} \over {dx}}} \right) - 5( - 5y){1 \over 2}{{2x} \over {\sqrt {{x^2} - 1} }}} \right]} \over {({x^2} - 1)}}$

Therefore, $({x^2} - 1){{{d^2}y} \over {d{x^2}}} = - 5\sqrt {{x^2} - 1} {{dy} \over {dx}} + 5y{x \over {\sqrt {{x^2} - 1} }}$

$ \Rightarrow ({x^2} - 1){{{d^2}y} \over {d{x^2}}} = 25y - x{{dy} \over {dx}}$

$ \Rightarrow ({x^2} - 1){{{d^2}y} \over {d{x^2}}} + 1x{{dy} \over {dx}} - 25y = 0$

Therefore, $\lambda$ = 1, k = $-$25; hence,

$\lambda + k = - 24$

Given,

y dx = $\left( {x + 3{y^2}} \right)dy$

$ \Rightarrow $$\,\,\,$ y ${{dx} \over {dy}}$ = x + 3y²

$ \Rightarrow $$\,\,\,$ ${{dx} \over {dy}}$ $-$ ${x \over y} = 3y$

If   =    ${e^{ - \int {{1 \over y}dy} }}$ = ${e^{ - \ln y}}$ = ${1 \over y}$

$\therefore\,\,\,$ Soluation is ,

x . ${1 \over y}$ = $\int {3y.{1 \over y}dy} $

$ \Rightarrow $$\,\,\,$ ${x \over y}$ = 3y + c

This curve passing through (1, 1)

$\therefore\,\,\,$ 1 = 3 + c

$ \Rightarrow $$\,\,\,$ c = $-$ 2

$\therefore\,\,\,$ Curve is, x = 3y² $-$ 2y

Now put every point in this equation, and see which point satisfy this equation.

Following this method you can see ($-$ ${1 \over 3}$, ${1 \over 3}$) point satisfy this equation.

$\left( {2 + \sin x} \right){{dy} \over {dx}} + \left( {y + 1} \right)\cos x = 0$

$ \Rightarrow $ ${d \over {dx}}\left( {2 + \sin x} \right)\left( {y + 1} \right) = 0$

On integrating, we get

(2 + sin x) (y + 1) = C

At x = 0, y = 1 we have

(2 + sin 0) (1 + 1) = C

$ \Rightarrow $ C = 4

$ \Rightarrow $ $\left( {y + 1} \right) = {4 \over {2 + \sin x}}$

$ \Rightarrow $ y = ${4 \over {2 + \sin x}} - 1$

Now $y\left( {{\pi \over 2}} \right) = {4 \over {2 + \sin {\pi \over 2}}} - 1$

= ${4 \over 3} - 1$ = ${1 \over 3}$

${{dy} \over {dx}} = {1 \over {8\sqrt x + \sqrt {9 + \sqrt x } \sqrt {4 + \sqrt {9 + \sqrt x } } }}$

$ \Rightarrow y = \sqrt {4 + \sqrt { + \sqrt x } } + c$

Now, $y(0) = \sqrt 7 + c$

$ \Rightarrow c = 0$

$y(256) = \sqrt {4 + \sqrt {9 + \sqrt {16} } } = \sqrt {4 + 5} = 3$

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

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.

Given,

${{dy} \over {dx}} + {y \over 2}\sec x = {{\tan x} \over {2y}}$

$ \Rightarrow $   $2y{{dy} \over {dx}} + {y^2}\sec x = \tan x$

Now, let

y² $=$ t

$ \Rightarrow $   2y${{dy} \over {dx}} = {{dt} \over {dx}}$

$ \therefore $    New equation,

${{dt} \over {dx}} + t\sec x = \tan x$

$ \therefore $   I.F $=$ ${e^{\int {\sec xdx} }}$

$=$   ${e^{\ln \left( {\sec x + \tan x} \right)}}$

$=$   sec x + tan x

$ \therefore $   Solution is,

t(sec x + tan x) $=$ $\int {\tan x} $ (sec x + tan x) dx

$ \Rightarrow $   t(sec x + tan x) $=$ sec x + tan x $-$ x + c

$ \Rightarrow $    t $=$ 1 $-$ ${x \over {\sec x + \tan x}} + c$

$ \Rightarrow $   y² $=$ 1 $-$ ${x \over {\sec x + \tan x}} + c$

Given,

y(0) $=$ 1

$ \therefore $   1 $=$ 1 $-$ 0 + c

$ \Rightarrow $   c $=$ 0

$ \therefore $   y² $=$ 1 $-$ ${x \over {\sec x + \tan x}}$

$\mathop {\lim }\limits_{t \to x} {{{t^2}f\left( x \right) - {x^2}f\left( t \right)} \over {t - x}} = 1$

It is in ${0 \over 0}$ form

So, applying L' Hospital rule,

$\mathop {\lim }\limits_{t \to x} {{2tf\left( x \right) - {x^2}f'\left( x \right)} \over 1} = 1$

$ \Rightarrow $   2xf(x) $-$ x²f '(x) = 1

$ \Rightarrow $   f '(x) $-$ ${2 \over x}$f(x) $=$ ${1 \over {{x^2}}}$

$ \therefore $   I.F = ${e^{\int {{{ - 2} \over x}dx} }} = {e^{ - 2\log x}} = {1 \over {{x^2}}}$

$ \therefore $   Solution of equation,

f(x)${1 \over {{x^2}}}$ = $\int {{1 \over x}\left( { - {1 \over {{x^2}}}} \right)} \,dx$

$ \Rightarrow $   ${{f(x)} \over {{x^2}}} = {1 \over {3{x^3}}} + C$

Given that,

f(1) = 1

$ \therefore $   ${1 \over 1}$ = ${1 \over 3}$ + C

$ \Rightarrow $   C $=$ ${2 \over 3}$

$ \therefore $   f(x) $=$ ${2 \over 3}$ x² + ${1 \over {3x}}$

$ \therefore $   f$\left( {{3 \over 2}} \right) = {2 \over 3} \times {\left( {{3 \over 2}} \right)^2} + {1 \over 3} \times {2 \over 3} = {{31} \over {18}}$

$y\left( {1 + xy} \right)dx = xdy$

${{xdy - ydx} \over {{y^2}}} = xdx \Rightarrow \int { - d\left( {{x \over y}} \right) = \int {xdx} } $

$\,\,\,\,\,\,\,\,\,\,\,\,\,\,$ $ - {x \over y} = {{{x^2}} \over 2} + C\,\,$

as $\,\,\,y\left( 1 \right) = - 1 \Rightarrow C = {1 \over 2}$

Hence, $y = {{ - 2x} \over {{x^2} + 1}} \Rightarrow f\left( {{{ - 1} \over 2}} \right) = {4 \over 5}$

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.

$f'(x) = 2 - {{f(x)} \over x}$

$ \Rightarrow f'(x) = {1 \over x}f(x) = 2$ is linear differential equation.

Hence, $I.F. = {e^{\int {{1 \over x}dx} }} = {e^{\ln x}} = x$

Thus the solution is given by

$x\,.\,f(x) = \int {2x\,dx + \lambda } $

i.e., $xf(x) = {x^2} + \lambda $

As f(1) $\ne$ 1, we have $\lambda$ $\ne$ 0

$\therefore$ $f(x) = x + {\lambda \over x}$, $\lambda$ $\ne$ 0

Thus, $f'(x) = 1 - {\lambda \over {{x^2}}}$, $\lambda$ $\ne$ 0

Now, $\mathop {\lim }\limits_{x \to {0^ + }} f'\left( {{1 \over x}} \right) = \mathop {\lim }\limits_{x \to {0^ + }} (1 - \lambda {x^2}) = 1$

$\mathop {\lim }\limits_{x \to {0^ + }} xf\left( {{1 \over x}} \right) = \mathop {\lim }\limits_{x \to {0^ + }} x\left( {{1 \over x} + \lambda x} \right) = \mathop {\lim }\limits_{x \to {0^ + }} (1 + \lambda {x^2}) = 1$

$\mathop {\lim }\limits_{x \to {0^ + }} {x^2}f'(x) = \mathop {\lim }\limits_{x \to {0^ + }} {x^2}\left( {1 - {\lambda \over {{x^2}}}} \right) = \mathop {\lim }\limits_{x \to {0^ + }} ({x^2} - \lambda ) = - \lambda $

Again, $\mathop {\lim }\limits_{x \to {0^ + }} f(x) \to \infty $

Hence the function is not bounded.

Note that $\lambda$ can be $-$ve or +ve.

Given, ${{dy} \over {dx}} + \left( {{1 \over {x\,\log \,x}}} \right)y = 2$

$I.F. = {e^{\int {{1 \over {x\log x}}dx} }} = {e^{\log \left( {\log x} \right)}} = \log x$

$y.\log x = \int {2\,\log xdx + c} $

$y\log x = 2\left[ {x\log x - x} \right] + c$

Put $x=1,y.0=-2+c$ $ \Rightarrow c = 2$

Put $x=e$

$y\log e = 2e\left( {\log e - 1} \right) + c \Rightarrow y\left( e \right) = c = 2$

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 differential equation is

${{dp\left( t \right)} \over {dt}} = {1 \over 2}p\left( t \right) - 200$

By separating the variable, we get

$dp\left( t \right) = \left[ {{1 \over 2}p\left( t \right) - 200} \right]dt$

$ \Rightarrow {{dp\left( t \right)} \over {{1 \over 2}p\left( t \right) - 200}} = dt$

$\,\,\,\,\,\,\,\,\,\,\,\,\,\,$ Integrating on both the sides,

$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\int {{{d\left( {p\left( t \right)} \right)} \over {{1 \over 2}p\left( t \right) - 200}}} = \int {dt} $

Let ${1 \over 2}p\left( t \right) - 200 = s \Rightarrow {{dp\left( t \right)} \over 2} = ds$

So, $\int {{{d\,p\left( t \right)} \over {\left( {{1 \over 2}p\left( t \right) - 200} \right)}}} = \int {dt} $

$ \Rightarrow \int {{{2ds} \over s} = \int {dt} } \Rightarrow 2\log s = t + c$

$ \Rightarrow 2\log \left( {{{p\left( t \right)} \over 2} - 200} \right) = t + c$

$ \Rightarrow {{p\left( t \right)} \over 2} - 200 = {e^{{1 \over 2}}}k$

Using given condition $p\left( t \right) = 400 - 300{e^{t/2}}$

Given, $\frac{d y}{d x}+\left(\frac{x}{x^2-1}\right) y=\frac{x^4+2 x}{\sqrt{1-x^2}}$

The above differential equation is a linear differential equation.

Integrating factor I.F. $=e^{\int \frac{x}{x^2-1} d x}$

$\begin{aligned} & \Rightarrow \text { I.F. }=e^{\frac{1}{2} \int \frac{-2 x d x}{1-x^2}} \\ & \Rightarrow \text { I.F. }=e^{\frac{1}{2} \ln \left(1-x^2\right)} \quad\left(\int \frac{f^{\prime}(x) d x}{f(x)}=\ln f(x)+c\right) \\ & \Rightarrow \text { I.F. }=e^{\ln \sqrt{1-x^2}} \\ & \Rightarrow \text { I.F. }=\sqrt{1-x^2} \end{aligned}$

Now, the solution of the given differential equation is

$\begin{aligned} & \Rightarrow y \cdot \text { (I.F. })=\int\left(\frac{x^4+2 x}{\sqrt{1-x^2}}\right)(\text { L.F. }) d x+\text { C } \\ & \Rightarrow y \sqrt{1-x^2}=\int\left(\frac{x^4+2 x}{\sqrt{1-x^2}}\right) \sqrt{1-x^2} d x+\text { C } \\ & \Rightarrow y \sqrt{1-x^2}=\frac{x^5}{5}+2 \times \frac{x^2}{2}+\text { C } \\ & \Rightarrow y=f(x)=\frac{\frac{x^5}{5}+x^2+\mathrm{C}}{\sqrt{1-x^2}} \end{aligned}$

Given, $f(0)=0$

$\begin{array}{ll} \Rightarrow & 0=\frac{0+0+C}{\sqrt{1-0}} \\ \Rightarrow & C=0 \\ \therefore & f(x)=\frac{\frac{x^5}{5}+x^2}{\sqrt{1-x^2}}=\frac{x^5}{5 \sqrt{1-x^2}}+\frac{x^2}{\sqrt{1-x^2}} \end{array}$

Let $\mathrm{I}=\int_{\frac{-\sqrt{3}}{2}}^{\frac{\sqrt{3}}{2}} f(x) d x$

$\Rightarrow \mathrm{I}=\int_{\frac{-\sqrt{3}}{2}}^{\frac{\sqrt{3}}{2}}\left(\frac{x^5}{5 \sqrt{1-x^2}}+\frac{x^2}{\sqrt{1-x^2}}\right) d x$

Here, $\frac{x^5}{5 \sqrt{1-x^2}}$ is an odd function and $\frac{x^2}{\sqrt{1-x^2}}$ is an even function.

$\begin{aligned} & \therefore \mathrm{I}=0+2 \int_0^{\frac{\sqrt{3}}{2}} \frac{x^2}{\sqrt{1-x^2}} d x \\ & \Rightarrow \mathrm{I}=2 \int_0^{\frac{\sqrt{3}}{2}}\left(\frac{x^2-1+1}{\sqrt{1-x^2}}\right) d x \\ & \Rightarrow \mathrm{I}=2 \int_0^{\frac{\sqrt{3}}{2}}\left(-\sqrt{1-x^2}+\frac{1}{\sqrt{1-x^2}}\right) d x \\ & \Rightarrow \mathrm{I}=2\left[-\left(\frac{x}{2} \sqrt{1-x^2}+\frac{1}{2} \sin ^{-1} x\right)+\sin ^{-1} x\right]_0^{\frac{\sqrt{3}}{2}} \\ & \Rightarrow \mathrm{I}=2\left[-\frac{x}{2} \sqrt{1-x^2}+\frac{1}{2} \sin ^{-1} x\right]_0^{\frac{\sqrt{3}}{2}} \\ & \Rightarrow \mathrm{I}=2\left[-\frac{\sqrt{3}}{4} \times \frac{1}{2}+\frac{1}{2} \times \frac{\pi}{3}\right] \\ & \Rightarrow \mathrm{I}=\frac{\pi}{3}-\frac{\sqrt{3}}{4} \end{aligned}$

Hint:

(i) The solution of linear differential equation $\frac{d y}{d x}+\mathrm{P}(x) \cdot y=\mathrm{Q}(x)$ is

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

Where, $\quad \text { L.F. } =e^{\int p(x) d x}$

(ii) Recall the property

$\int_{-a}^a f(x) d x=\left\{\begin{array}{l} 2 \int_0^a f(x) d x, \text { When } f(x) \text { is an even } \\ \text { function. } \\ 0, \text { When } f(x) \text { is an odd function. } \end{array}\right.$

Given, Rate of change is ${{dp} \over {dx}} = 100 - 12\sqrt x $

$ \Rightarrow dP = \left( {100 - 12\sqrt x } \right)dx$

By intergrating $\int {dP = \int {\left( {100 - 12\sqrt x } \right)} } dx$

$\int {dP} = \int {\left( {100 - 12\sqrt x } \right)} dx$

$P = 100x - 8{x^{3/2}} + C$

Given, when $x=0$ then $P=2000$

$ \Rightarrow C = 2000$

Now when $x$$=25$

then $P = 100 \times 25 - 8 \times {\left( {25} \right)^{3/2}} + 2000$

$\,\,\,\,\,\,\,\,\,\,$ $=4500-1000$

$ \Rightarrow P = 3500$

Given, the slope of curve at $(x,y)$ is

$\begin{aligned} & \frac{y}{x}+\sec \left(\frac{y}{x}\right), x>0 . \\ \therefore \quad & \frac{d y}{d x}=\frac{y}{x}+\sec \left(\frac{y}{x}\right) \quad \text{... (i)} \end{aligned}$

Put $y=v x \Rightarrow \frac{d y}{d x}=v+x \frac{d v}{d x}$ in the equation (i)

$\begin{aligned} & \Rightarrow \quad v+x \frac{d v}{d x}=\frac{v x}{x}+\sec \left(\frac{v x}{x}\right) \\ & \Rightarrow \quad v+x \frac{d v}{d x}=v+\sec (v) \\ & \Rightarrow \quad x \frac{d v}{d x}=\sec v \\ & \Rightarrow \quad \cos v d v=\frac{d x}{x} \\ & \Rightarrow \quad \int \cos v d v=\int \frac{d x}{x} \\ & \Rightarrow \quad \sin v=\ln x+c \\ & \Rightarrow \quad \sin \left(\frac{y}{x}\right)=\ln x+c \quad \text{... (i)} \end{aligned}$

Put $x=1$ and $y=\frac{\pi}{6}$ in the equation (ii)

$\begin{aligned} \Rightarrow & & \sin \frac{\pi}{6} & =\log 1+c \\ \Rightarrow & & \frac{1}{2} & =0+c \\ \Rightarrow & & c & =\frac{1}{2} \end{aligned}$

Put $c=\frac{1}{2}$ in the equation (ii)

$\Rightarrow \quad \sin \frac{y}{x}=\log x+\frac{1}{2}$

Hints:

(i) the slope of a curve at $(x, y)$ is equal to $\frac{d y}{d x}$

(ii) For the solution of homogeneous differential equation $\frac{d y}{d x}=f\left(\frac{y}{x}\right)$ put $y=v x$ i.e. $\frac{d y}{d x}=v+x \frac{d v}{d x}$.

Given differential equation is

${{dp\left( t \right)} \over {dt}} = 0.5p\left( t \right) - 450$

$ \Rightarrow {{dp\left( t \right)} \over {dt}} = {1 \over 2}p\left( t \right) - 450$

$ \Rightarrow {{dp\left( t \right)} \over {dt}} = {{p\left( t \right) - 900} \over 2}$

$ \Rightarrow 2{{dp\left( t \right)} \over {dt}} = \left[ {900 - p\left( t \right)} \right]$

$ \Rightarrow 2{{dp\left( t \right)} \over {900 - p\left( t \right)}} = - dt$

Integrate both sides, we get

$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, - 2\int {{{dp\left( t \right)} \over {900 - p\left( t \right)}}} = \int {dt} $

$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,Let\,\,900 - p\left( t \right) = u$

$ \Rightarrow - dp\left( t \right) = du$

$\therefore$ $\,\,\,\,\,\,$ We have,

$2\int {{{du} \over u}} = \int {dt \Rightarrow 2\,\ln \,u = t + c} $

$ \Rightarrow 2\ln \left[ {900 - p\left( t \right)} \right] = t + c$

$\,\,\,\,\,\,\,\,\,\,\,\,\,\,$ when $t = 0,p\left( 0 \right) = 850$

$\,\,\,\,\,\,\,\,\,\,\,\,\,\,$ $2\ln \left( {50} \right) = c$

$ \Rightarrow 2\left[ {\ln \left( {{{900 - p\left( t \right)} \over {50}}} \right)} \right] = t$

$ \Rightarrow 900 - p\left( t \right) = 50{e^{{t \over 2}}}$

$ \Rightarrow p\left( t \right) = 900 - 50{e^{{t \over 2}}}$

Let $p\left( {{t_1}} \right) = 0$

$0 = 900 - 50{e^{{{{t_1}} \over 2}}}$

$\therefore$ $\,\,\,\,\,\,{t_1} = 2\ln 18$

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,

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

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

${{dV\left( t \right)} \over {dt}} = - k\left( {T - t} \right)$

$ \Rightarrow \int {dVt} = - k\int {\left( {T - t} \right)} dt$

$V\left( t \right) = {{k{{\left( {T - t} \right)}^2}} \over 2} + c$

$V\left( 0 \right) = I \Rightarrow I = {{K{T^2}} \over 2} + C$

$ \Rightarrow C = I - {{K{T^2}} \over 2}$

$\therefore$ $\,\,\,\,\,V\left( T \right) = 0 + C = I - {{K{T^2}} \over 2}$

${{dy} \over {dx}} = y + 3 \Rightarrow \int {{{dy} \over {y + 3}}} = \int {dx} $

$ \Rightarrow \ell n\left| {y + 3} \right| = x + c$

Since $y\left( 0 \right) = 2,\,\,\,$ $\,\,\,\,\,\,\,$ $\therefore$ $\,\,\,\,\,\,\,$ $\ell n5 = c$

$ \Rightarrow \ell n\left| {y + 3 = x + \ell n5} \right|$

When $x = \ell n2,$ then

$\ell n\left| {y + 3} \right| = \ell n2 + \ell n5$

$ \Rightarrow \ln \left| {y + 3} \right| = \ell n\,10$

$\therefore$ $\,\,\,\,\,$ $y + 3 = \pm 10 \Rightarrow y = 7, - 13$

$\cos xdy = y\left( {\sin x - y} \right)dx$

${{dy} \over {dx}} = y\tan x - {y^2}\,\,\sec \,x$

${1 \over {{y^2}}}{{dy} \over {dx}} - {1 \over y}\,\tan \,x = - \sec \,x\,....\left( i \right)$

Let $\,\,\,\,{1 \over y} = t \Rightarrow - {1 \over {{y^2}}}{{dy} \over {dx}} = {{dt} \over {dx}}$

From equation $(i)$

$ - {{dt} \over {dx}} - \tan x = - \sec x$

$ \Rightarrow {{dt} \over {dx}} + \left( {\tan x} \right)t = \sec x$

${\rm I}.F. = {e^{\int {\tan xdx} }} = {\left( e \right)^{\log \left| {\sec x} \right|}}\sec x$

Solution $:$ $\,\,t\left( {{\rm I}.F} \right) = \int {\left( {{\rm I}.F} \right)\sec xdx} $

$ \Rightarrow {1 \over y}\sec x = \tan x + c$

We have $y = {c_1}{e^{{c_2}x}}$

$ \Rightarrow y' = {c_1}{c_2}{e^{{c_2}x}} = {c_2}y$

$ \Rightarrow {{y'} \over y} = {c_2}$

$ \Rightarrow {{y''y\left( {y'} \right){}^2} \over {{y^2}}} = 0$

$ \Rightarrow y''y = {\left( {y'} \right)^2}$

(A) We have $f'(x) > 0,\forall x \in (0,\pi /2)$. Therefore,

$f(0) < 0$ and $f(\pi /2) > 0$

Hence, there is no one solution.

(B) Let us consider that $(a,b,c)$ is direction ratio of the intersected line. Therefore,

$ak + 4b + c = 0$

$4a + kb + 2c = 0$

${a \over {8 - k}} = {b \over {4 - 2k}} = {c \over {{k^2} - 16}}$

We need to have

$2(8 - k) + 2(4 - 2k) + ({k^2} - 16) = 0$

$ \Rightarrow k = 2,4$.

(C) Let us consider

$f(x) = |x + 2| + |x + 1| + |x - 1| + |x - 2|$

Therefore, $k$ can take values: 2, 3, 4, 5.

(D) $\int {{{dy} \over {y + 1}} = \int {dx} } $

$ \Rightarrow f(x) = 2{e^x} - 1$

$ \Rightarrow f(\ln 2) = 3$

(A) We have ${(x - 3)^2}{{dy} \over {dx}} + y = 0$

$\int {{{dx} \over {{{(x - 3)}^2}}} = - \int {{{dy} \over y}} } $

$ \Rightarrow {1 \over {x - 3}} = \ln |y| + c$

So the domain is $R \to \{ 3\} $.

(B) On substituting $x = t + 3$, we get

$\int\limits_{ - 2}^2 {(t + 2)(t + 1)t(t - 1)(t - 2)dt = \int\limits_{ - 2}^2 {t({t^2} - 1)({t^2} - 4)dt = 0} } $ (being odd function)

(C) $f(x) = {5 \over 4} - {\left( {\sin x - {1 \over 2}} \right)^2}$

The maximum value occurs when $\sin x = 1/2$.

(D) $f'(x) > 0$ if $\cos x > \sin x$.

${{dy} \over {dx}} = {{x + y} \over x} = 1 + {y \over x}$

Putting $y=$ $vx$

and ${{dy} \over {dx}} = v + x{{dv} \over {dx}}$

we get

$v + x{{dv} \over {dx}} = 1 + v$

$ \Rightarrow \int {{{dx} \over x}} = \int {dv} $

$ \Rightarrow v = \ln {\mkern 1mu} x + c$

$ \Rightarrow y = x\ln x + cx$

As $\,\,\,\,y\left( 1 \right) = 1$

$\therefore$ $c=1$

So solution is $y=xlnx+x$

The given differential equation is

$x\sqrt {{x^2} - 1} dy - y\sqrt {{y^2} - 1} dx = 0$

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

$ \Rightarrow {\sec ^{ - 1}}y = {\sec ^{ - 1}}x + c$

$y = \sec [{\sec ^{ - 1}}x + C]$ [$\because$ $y(2) = {2 \over {\sqrt 3 }}$]

${2 \over {\sqrt 3 }} = \sec ({\sec ^{ - 1}}2 + C)$

$ = {\sec ^{ - 1}}{2 \over {\sqrt 3 }} - {\sec ^{ - 1}}2 = C$

$C = {\pi \over 6} - {\pi \over 3} = {{ - \pi } \over 6}$ [$\therefore$ $y = \sec \left[ {{{\sec }^{ - 1}}x - {\pi \over 6}} \right]$]

Statement 1 is true

Also, ${1 \over y} = \cos \left[ {\cos - {\pi \over x} - {\pi \over 6}} \right]$

$\cos \left( {{{\cos }^{ - 1}}{1 \over x}} \right)\cos {\pi \over 6} + \sin ({\cos ^{ - 1}}{1 \over x})\sin {\pi \over 6}$

$ = {1 \over y} = {{\sqrt 3 } \over {2x}} + {1 \over 2}\sqrt {1 - {1 \over {{x^2}}}} $

$\therefore$ Statement 2 is false.

General equation of circles passing through origin

and having their centres on the $x$-axis is

${x^2} + {y^2} + 2gx = 0\,\,\,\,....\left( i \right)$

On differentiating $w.r.t.x,$ we get

$2x + 2y.{{dy} \over {dx}} + 2g = 0$

$ \Rightarrow g = - \left( {x + y{{dy} \over {dx}}} \right)$

$\therefore$ equation $(i)$ be

${x^2} + {y^2} + 2\left\{ { - \left( {x + y{{dy} \over {dx}}} \right)} \right\}x = 0$

$ \Rightarrow {x^2} + {y^2} - 2{x^2} - 2x{{dy} \over {dx}}.y = 0$

$ \Rightarrow {y^2} = {x^2} + 2xy{{dy} \over {dx}}$

Given, $ \frac{d y}{d x}=\frac{\sqrt{1-y^{2}}}{y}$

$\Rightarrow \int \frac{y}{\sqrt{1-y^{2}}} d y=\int d x$

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

Here, centre $(-c, 0)$ and radius $=1$

$A{x^2} + B{y^2} = 1\,\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( 1 \right)$

$Ax + by{{dy} \over {dx}} = 0\,\,\,\,\,\,\,\,\,\,\,\,...\left( 2 \right)$

$A + By{{{d^2}y} \over {d{x^2}}} + B{\left( {{{dy} \over {dx}}} \right)^2} = 0\,\,\,\,\,\,\,\,\,\,\,\,...\left( 3 \right)$

From $(2)$ and $(3)$

$x\left\{ { - By{{{d^2}y} \over {d{x^2}}} - B{{\left( {{{dy} \over {dx}}} \right)}^2}} \right\} + By{{dy} \over {dx}} = 0$

Dividing both sides by $-B,$ we get

$ \Rightarrow xy{{{d^2}y} \over {d{x^2}}} + x{\left( {{{dy} \over {dx}}} \right)^2} - y{{dy} \over {dx}} = 0$

Which is $DE$ of order $2$ and degree $1.$

${y^2} = 2c\left( {x + \sqrt c } \right)\,\,\,\,\,\,\,\,\,\,...\left( i \right)$

$2yy' = 2c.1\,\,\,or\,\,\,yy' = c\,\,\,...\left( {ii} \right)$

$ \Rightarrow {y^2} = 2yy'\left( {x + \sqrt {yy'} } \right)$

$\left[ \, \right.$ On putting value of $c$ from $(ii)$ in $(i)$ $\left. \, \right]$

On simplifying, we get

${\left( {y - 2xy'} \right)^2} = 4yy{'^3}\,\,\,\,\,\,\,\,\,...\left( {iii} \right)$

Hence equation $(iii)$ is of order $1$ and degree $3.$

${{xdy} \over {dx}} = y\left( {\log y - \log x + 1} \right)$

${{dy} \over {dx}} = {y \over x}\left( {\log \left( {{y \over x}} \right) + 1} \right)$

Put $\,\,\,\,y = vx$

${{dy} \over {dx}} = v + {{xdv} \over {dx}}$

$\,\,\,\,\,\,\,\,\,\,\, \Rightarrow v + {{xdv} \over {dx}} = v\left( {\log v + 1} \right)$

${{xdv} \over {dx}} = v\,\log \,v$

$\,\,\,\,\,\,\,\,\,\,\, \Rightarrow {{dv} \over {v\,\log \,v}} = {{dx} \over x}$

Put $\,\,\,\,\log \,v = z$

$ \Rightarrow {1 \over v}dv = dz$

$ \Rightarrow {{dz} \over x} = {{dx} \over x}$

$ \Rightarrow \ln \,z = \ln x + \ln \,c$

$x = cx\,\,\,\,$ or $\,\,\,\,\,\log v = cx\,\,\,$

or $\,\,\,\,$ $\log \left( {{y \over x}} \right) = cx.$

To find $y''(0)$, we need to differentiate the given equation twice with respect to $x$ and then substitute $x=0$.

Differentiating the equation $x\cos \,y + y\,cos\,x = \pi $ with respect to $x$, we get:

$ \cos\,y - x\sin\,y \cdot y' + y'\cos\,x - y\sin\,x = 0$

Now, let's differentiate this equation again with respect to $x$:

$- \sin\,y \cdot y' - \sin\,y \cdot y' - x\cos\,y \cdot (y')^2 - x\sin\,y \cdot y'' + y''\cos\,x - y'\sin\,x - y'\sin\,x - y\cos\,x = 0$

Simplifying the equation, we get:

$-2\sin\,y \cdot y' - x\cos\,y \cdot (y')^2 - x\sin\,y \cdot y'' + y''\cos\,x - 2y'\sin\,x - y\cos\,x = 0$

Now, let's substitute $x = 0$ in the original equation and the first derivative equation. This will help us find the values of $y(0)$ and $y'(0)$ respectively:

From the original equation, $x\cos \,y + y\,cos\,x = \pi$, substituting $x = 0$, we get:

$y(0) \cdot 1 = \pi$

Therefore, $y(0) = \pi$.

Now, substituting $x = 0$ in the first derivative equation, we get:

$ \cos\,y(0) + y'(0) \cdot 1 = 0$

Substituting $y(0) = \pi$, we get:

$ \cos \pi + y'(0) = 0$

Therefore, $y'(0) = 1$.

Finally, let's substitute $x = 0$, $y(0) = \pi$, and $y'(0) = 1$ in the second derivative equation:

$-2\sin\pi \cdot 1 - 0 \cdot \cos \pi \cdot (1)^2 - 0 \cdot \sin \pi \cdot y''(0) + y''(0)\cos 0 - 2 \cdot 1 \cdot \sin 0 - \pi \cdot \cos 0 = 0$

Simplifying the equation, we get:

$y''(0) - \pi = 0$

Therefore, $y''(0) = \pi$.

${x^2} + {y^2} - 2ay = 0\,\,\,\,\,\,\,\,\,\,\,...\left( 1 \right)$

Differentiate,

$2x + 2y{{dy} \over {dx}} - 2a{{dy} \over {dx}} = 0$

$\,\,\,\,\,\,\,\,\,\,\,\,\,\, \Rightarrow a = {{x + yy'} \over {y'}}$

Put in $(1),$ ${x^2} + {y^2} - 2\left( {{{x + yy'} \over {y'}}} \right) = y = 0$

$ \Rightarrow \left( {{x^2} + {y^2}} \right)y' - 2xy - 2{y^2}y' = 0$

$ \Rightarrow \left( {{x^2} - {y^2}} \right)y' = 2xy$