Mathematics M.C.Q (1 Marks)

This is one parameter family of circles, so its differential equation is of order one.

This is one parameter family of curves

So its differential equation is of order one.

$\therefore$ $m = 3$ and $n = 2$.

Here $r$ is arbitrary constant order of differential equation $= 1.$

There are three arbitrary constants, $h, k$ and $a.$

$\therefore $ order $= 3$.

==> $\frac{{{d^2}y}}{{d{x^2}}} = - 36\sin 3x = - 9 \times 4\sin 3x = - 9y$

==> $\frac{{{d^2}y}}{{d{x^2}}} + 9y = 0$.

we get $2x + 2y\frac{{dy}}{{dx}} = 0$ ==> $x+y\frac{dy}{dx}=0$

==> $y + 1 = \frac{{dy}}{{dx}}(x - 1)$ ==> $y = (x - 1)\frac{{dy}}{{dx}} - 1$.

Eliminating $a$ from $(i)$ and $(ii)$, required equation is

$y{\rm{ }}\left[ {1 - {{\left( {\frac{{dy}}{{dx}}} \right)}^2}} \right] = 2x\frac{{dy}}{{dx}}$.

==>$\frac{{{d^2}v}}{{d{r^2}}} = \frac{2}{r}\left( {\frac{A}{{{r^2}}}} \right)$

==> $\frac{{{d^2}v}}{{d{r^2}}} = \frac{2}{r}\left( { - \frac{{dv}}{{dr}}} \right)$

==>$\frac{{{d^2}v}}{{d{r^2}}} + \frac{2}{r}\frac{{dv}}{{dr}} = 0$.

==> $\frac{{{d^2}x}}{{d{t^2}}} = - {n^2}a\cos (nt + b)$ ==> $\frac{{{d^2}x}}{{d{t^2}}} = - {n^2}x$

==> $\frac{{{d^2}x}}{{d{t^2}}} + {n^2}x = 0$.

where $A, B, C$ are arbitrary constants.

Differentiating $(i)$ $ w.r.t$. $x$, we get $\frac{{dy}}{{dx}} = 2Ax + B .....(ii)$

Which on differentiating $w.r.t.$ $x$ gives$\frac{{{d^2}y}}{{d{x^2}}} = 2A .....(iii)$

Differentiating $w.r.t.$ $x$ again, we get $\frac{{{d^3}y}}{{d{x^3}}} = 0$.

==> $\frac{{dy}}{{dx}} = - y + {e^{ - x}}$ ==> $\frac{{dy}}{{dx}} + y = {e^{ - x}}$.

Hence differential equation is, $y = x\frac{{dy}}{{dx}} + \frac{{dy}}{{dx}} - {\left( {\frac{{dy}}{{dx}}} \right)^3}$.

where $m$ is arbitrary constant

Differentiate $(i)$ $w.r.t$. $x$, we get

$\frac{{dy}}{{dx}} = m$ ==> $\frac{{dy}}{{dx}} = \frac{y}{x}$, $(By (i)).$

Differentiating, we get $\frac{{dy}}{{dx}} = ma{e^{mx}} - mb{e^{ - mx}}$.

Differentiating again, we get $\frac{{{d^2}y}}{{d{x^2}}} = {m^2}a{e^{mx}} + {m^2}b{e^{ - mx}}$

$ = {m^2}(a{e^{mx}} + b{e^{ - mx}}) = {m^2}y$ or $\frac{{{d^2}y}}{{d{x^2}}} - {m^2}y = 0$.

Differentiate $(i),$ we get

${y_1} = a\,\left[ {\cos \left( {\frac{1}{x} + b} \right) - x\sin \left( {\frac{1}{x} + b} \right){\rm{ }}\left( {\frac{{ - 1}}{{{x^2}}}} \right)} \right]$

$ = a\left[ {\cos \,\left( {\frac{1}{x} + b} \right) + \frac{1}{x}\sin \left( {\frac{1}{x} + b} \right)} \right].....(ii)$

Again, differentiate $(ii)$, we get ${y_2} = \frac{{ - a}}{{{x^3}}}\cos \left( {\frac{1}{x} + b} \right)$

$ = \frac{{ - ax}}{{{x^4}}}\cos \left( {\frac{1}{x} + b} \right)$$ = \frac{{ - y}}{{{x^4}}}$

==>$x^4y_2+y = 0$.

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

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

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

==> $\sqrt {1 - {x^2}} \,\,dy + \sqrt {1 - {y^2}} \,\,dx = 0$.

$\frac{{dx}}{{dt}} = \cos t$; $\frac{{dy}}{{dt}} = - p\sin pt$; $\frac{{dy}}{{dx}} = \frac{{ - p\sin pt}}{{\cos t}}$

$\frac{{{d^2}y}}{{d{x^2}}} = \frac{{ - \cos t\,{p^2}\cos pt(dt/dx) - p\sin pt\sin t(dt/dx)}}{{{{\cos }^2}t}}$

==> $(1 - {x^2})\frac{{{d^2}y}}{{d{x^2}}} - x\frac{{dy}}{{dx}} + {p^2}y = 0$

or $(1 - {x^2}){y_2} - x{y_1} + {p^2}y = 0$.

==> $\frac{{dy}}{{dx}} = 3A{e^{3x}} + 5B{e^{5x}}$==>$\frac{{{d^2}y}}{{d{x^2}}} = 9A{e^{3x}} + 25B{e^{5x}}$

==> $\frac{{{d^2}y}}{{d{x^2}}} - 8\frac{{dy}}{{dx}} + 15y = 0$ (By inspection)

$y=c_{1} e^{x}+c_{2} e^{x}$        .........$(1)$

Differentiating with respect to $\mathrm{x}$, we get:

$\frac{d y}{d x}=c_{1} e^{x}-c_{2} e^{-x}$

Again, differentiating with respect to $\mathrm{x}$, we get:

$\frac{d^{2} y}{d x^{2}}=c_{1} e^{x}+c_{2} e^{-x}$

$\Rightarrow \frac{d^{2} y}{d x^{2}}=y$

$\Rightarrow \frac{d^{2} y}{d x^{2}}-y=0$

This is the required differential equation of the given equation of curve.

Hence, the correct answer is $\mathrm{A}$.

Differentiating with respect to $\mathrm{x}$, we get:

$\frac{d y}{d x}=1$          .............$(1)$

Again, differentiating with respect to $\mathrm{x}$, we get:

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

Now, on substituting the values of $\mathrm{y}, \frac{d^{2} y}{d x^{2}},$ and $\frac{d y}{d x}$ from equation $(1)$ and $(2)$ in each of the given alternatives, we find that only the differential equation given in alternative $C$ is correct.

$\frac{d^{2} y}{d x^{2}}-x \frac{d y}{d x}+x y=0-x^{2} \cdot 1+x x$

$=-x^{2}+x^{2}$

$=0$

Hence, the correct answer is $\mathrm{B}.$

$\frac{{{{\sec }^2}y}}{{\tan y}}dy = - 3\frac{{{e^x}}}{{1 - {e^x}}}dx$

$\int {\frac{{{{\sec }^2}y}}{{\tan y}}} dy = - 3\int {\frac{{{e^x}}}{{1 - {e^x}}}dx} $

==> $\log (\tan y) = 3\log (1 - {e^x}) + \log c$ ==> $\tan y = c{(1 - {e^x})^3}$.

Now on integrating both sides, we get

${\tan ^{ - 1}}y = {\tan ^{ - 1}}x + {\tan ^{ - 1}}c$==> ${\tan ^{ - 1}}y = {\tan ^{ - 1}}\left( {\frac{{x + c}}{{1 - cx}}} \right)$
 

==> $y = \frac{{x + c}}{{1 - cx}}$ ==> $y - x = c(1 + xy)$.

==> $\cos ydy = \left( {{e^x}\log x + \frac{{{e^x}}}{x}} \right){\rm{ }}dx$

On integrating, $\sin y = {e^x}\log x + c$.

On integrating both sides, we get

==> $y = - \log (\sin x + \cos x) + \log c$

==> $y = \log \left( {\frac{c}{{\sin x + \cos x}}} \right)$ ==> ${e^y}(\sin x + \cos x) = c$.

$({x^2} - y{x^2})\frac{{dy}}{{dx}} + {y^2} + x{y^2} = 0$==>$\frac{{1 - y}}{{{y^2}}}dy + \frac{{1 + x}}{{{x^2}}}dx = 0$

==> $\left( {\frac{1}{{{y^2}}} - \frac{1}{y}} \right)dy + \left( {\frac{1}{{{x^2}}} + \frac{1}{x}} \right)dx = 0$

On integrating, we get the required solution

$\log \left( {\frac{x}{y}} \right) = \frac{1}{x} + \frac{1}{y} + c$.

On integrating both sides, we get

$\log y + y = \log x + x + \log A$

==> $\log \left( {\frac{y}{{Ax}}} \right) = x - y$ ==>$y = Ax{e^{x - y}}$.

==> $\frac{{{e^y}dy}}{{{e^y} + 1}} + \frac{{\cos x}}{{\sin x}}dx = 0$

On integrating both the functions, we get

$\log ({e^y} + 1) + \log (\sin x) = \log c$ ==>$({e^y} + 1)\sin x = c$.

Now put $x + y = v$ and $\frac{{dy}}{{dx}} = \frac{{dv}}{{dx}} - 1$

Therefore $\frac{{dy}}{{dx}} = \sin (x + y)$ reduces to $\frac{{dv}}{{1 + \sin v}} = dx$

Now on integrating both the sides, we get

$\tan v - \sec v = x + c$ or $\tan (x + y) - \sec (x + y) = x + c$.

==> $\left( {\frac{y}{{y + 1}}} \right)dy = \left( {\frac{{{e^x}}}{{{e^x} + 1}}} \right)\,dx$ ==> $\left[ {1 - \frac{1}{{y + 1}}} \right]dy = \left( {\frac{{{e^x}}}{{{e^x} + 1}}} \right)\,dx$

==> $\int_{}^{} {\left\{ {1 - \frac{1}{{y + 1}}} \right\}} dy = \int_{}^{} {\frac{{{e^x}}}{{{e^x} + 1}}} dx$

==> $y = \log (y + 1) + \log ({e^x} + 1) + \log c$ or ${e^y} = c(y + 1)({e^x} + 1)$.

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

Now on integrating both sides, we get

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

or $2\log (y - 1) + \log (1 - {x^2}) = \log {y^2}{c^2}$

Hence the solution is ${(y - 1)^2}(1 - {x^2}) = {c^2}{y^2}$.

==> $y = - \int_{}^{} {\sqrt {a + x} } dx + \int_{}^{} {\frac{a}{{\sqrt {a + x} }}} dx$
$\left\{ \because \int_{{}}^{{}}{\frac{x}{\sqrt{a+x}}}dx=\int_{{}}^{{}}{\frac{x+a-a}{\sqrt{a+x}}}dx \right\}$

==> $y = - \frac{2}{3}{(a + x)^{3/2}} + 2a\sqrt {a + x} + c$

==> $3y = - \sqrt {a + x} (2(a + x) - 6a) + 3c$

==> $3y = - 2\sqrt {a + x} (x - 2a) + 3c$

==> $3y + 2\sqrt {a + x} (x - 2a) = 3c$.

==> $\frac{{\cos y}}{{\sin y}}dy = - \frac{{\sin x}}{{\cos x}}dx$ ==> $\cot ydy = - \tan xdx$

On integrating, we get

$\log \sin y = \log \cos x + \log c$ ==>$\sin y = c\cos x$.

==> $\int_{}^{} {\frac{{{e^{2y}} - 1}}{{{e^y}}}} dy = \int_{}^{} {\frac{{1 - {x^2}}}{x}dx} $ ==> ${e^y} + {e^{ - y}} = \log x - \frac{{{x^2}}}{2} + c$.

Therefore, the differential equation reduces to

$\frac{{dv}}{{dx}} = (1 + \cos v) + \sin v$

$ = 2{\cos ^2}\frac{v}{2} + 2\sin \frac{v}{2}\cos \frac{v}{2} = 2{\cos ^2}\frac{v}{2}\left( {1 + \tan \frac{v}{2}} \right)$

==> $\int_{}^{} {\frac{{{{\sec }^2}(v/2)dv}}{{2[1 + \tan (v/2)]}}} = \int_{}^{} {dx} $

==> $\log \left[ {1 + \tan \left( {\frac{{x + y}}{2}} \right)} \right] = x + c$.

to $\frac{{dv}}{{dx}} = \frac{{v + 2}}{{2v + 5}}$ ==>$\int_{}^{} {\frac{{2v + 5}}{{v + 2}}} dv = \int_{}^{} {dx} $

==> $\int_{}^{} {\left[ {2 + \frac{1}{{(v + 2)}}} \right]} dv = \int_{}^{} {dx} $ ==> $2v + \log (v + 2) = x + c$

or $2(x - y) + \log (x - y + 2) = x + c$.

==> $\int_{}^{} {\frac{{y(1 + y)}}{{(1 - y)}}dy} = \int_{}^{} {\frac{{(1 - {x^2})}}{x}dx} $; Now integrate it.

On integrating we get $ - \frac{1}{2}\log (1 - {x^2}) = - \frac{1}{2}\log (1 - {y^2}) + \log c$

==> $\log (1 - {x^2}) - \log (1 - {y^2}) = - 2\log c$ ==> $\frac{{1 - {x^2}}}{{1 - {y^2}}} = {c^{ - 2}}$

Hence $(1 - {y^2}) = {c^2}(1 - {x^2})$.

On integrating both sides, we get

$\frac{{{{(\log y)}^2}}}{2} + [{x^2}( - \cos x) + \int_{}^{} {2x\cos xdx} ] = c$

==> $\frac{{{{(\log y)}^2}}}{2} - {x^2}\cos x + 2(x\sin x + \cos x) = c$

==> $\frac{{{{(\log y)}^2}}}{2} + (2 - {x^2})\cos x + 2x\sin x = c$.

==>$\int_{}^{} {(2y\log y + y)dy = \int_{}^{} {{e^x}({{\sin }^2}x + \sin 2x} )dx} $

On integrating by parts, we get ${y^2}(\log y) = {e^x}{\sin ^2}x + c$.

==> $\int_{}^{} {\frac{{ydy}}{{1 + {y^2}}} = \int_{}^{} {\frac{{(1 + x + {x^2})}}{{x(1 + {x^2})}}} } dx = \int_{}^{} {\frac{1}{x}dx} + \int_{}^{} {\frac{{dx}}{{1 + {x^2}}}} $

==> $\frac{1}{2}\log (1 + {y^2}) = \log x + {\tan ^{ - 1}}x + c$.

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

==> $\int_{}^{} {\frac{x}{{\sqrt {1 + {x^2}} }}dx + } \int_{}^{} {\frac{y}{{\sqrt {1 + {y^2}} }}dy = c} $

==> $\sqrt {1 + {x^2}} + \sqrt {1 + {y^2}} = c$.

Multiply the equation by ${e^{3x + 3y}}$ ==> ${e^{5x}}dx + {e^{5y}}dy = 0$

On integrating, we get ${e^{5x}} + {e^{5y}} = 5c' = c$.

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

==> $\int_{}^{} {\left[ {\frac{1}{{1 + {y^2}}} + \frac{y}{{1 + {y^2}}}} \right]} dy + \int_{}^{} {\left[ {\frac{1}{{1 + {x^2}}} + \frac{x}{{1 + {x^2}}}} \right]dx + c} = 0$

==> ${\tan ^{ - 1}}y + \frac{1}{2}\log (1 + {y^2}) + {\tan ^{ - 1}}x + \frac{1}{2}\log (1 + {x^2}) = c$.

==> $\frac{{dv}}{{dx}} = {v^2} + 1$ ==>$\frac{{dv}}{{{v^2} + 1}} = dx$

On integrating, we get

${\tan ^{ - 1}}v = x + c$ or $v = \tan (x + c)$ ==>$x + y = \tan (x + c)$.

==> $\int_{}^{} {\sec y\log (\sec y + \tan y)dy} $

$ = \int_{}^{} {\sec x\log (\sec x + \tan x)dx} $

Put $\log (\sec x + \tan x) = t$ and $\log (\sec y + \tan y) = z$

$\frac{{{{[\log (\sec x + \tan x)]}^2}}}{2} = \frac{{{{[\log (\sec y + \tan y)]}^2}}}{2} + c$.

==> $\int_{}^{} {\frac{{dy}}{{(y - 2)(y + 1)}}} = \int_{}^{} {\frac{{dx}}{{(x + 3)(x - 1)}}} $

==> $\frac{1}{3}\int_{}^{} {\left( {\frac{1}{{y - 2}} - \frac{1}{{y + 1}}} \right)} dy = \frac{1}{4}\int_{}^{} {\left( {\frac{1}{{x - 1}} - \frac{1}{{x + 3}}} \right)\,} dx$

==> $\frac{1}{3}\log \left| {\frac{{y - 2}}{{y + 1}}} \right| = \frac{1}{4}\left| {\frac{{x - 1}}{{x + 3}}} \right| + c$.