Formation of differential equations Questions in English

(D) Given the family of curves: $y = ax + \frac{1}{a}$ $(i)$
Differentiating with respect to $x$,we get:
$\frac{dy}{dx} = a$
Substituting $a = \frac{dy}{dx}$ into equation $(i)$:
$y = x \left(\frac{dy}{dx}\right) + \frac{1}{\frac{dy}{dx}}$
Multiplying both sides by $\frac{dy}{dx}$ to eliminate the fraction:
$y \left(\frac{dy}{dx}\right) = x \left(\frac{dy}{dx}\right)^2 + 1$
The highest power of the highest order derivative $\frac{dy}{dx}$ is $2$.
Therefore,the degree of the differential equation is $2$.

(D) Given the general solution $y = Ae^x + B \sin x$.
Step $1$: Differentiate with respect to $x$: $\frac{dy}{dx} = Ae^x + B \cos x$.
Step $2$: Differentiate again: $\frac{d^2y}{dx^2} = Ae^x - B \sin x$.
Step $3$: Eliminate constants $A$ and $B$. From the original equation,$Ae^x = y - B \sin x$.
Substitute into the second derivative: $\frac{d^2y}{dx^2} = (y - B \sin x) - B \sin x = y - 2B \sin x$.
From the first derivative: $B \cos x = \frac{dy}{dx} - Ae^x = \frac{dy}{dx} - (y - B \sin x) = \frac{dy}{dx} - y + B \sin x$.
This approach is complex. Let's use the operator method.
The solution $y = Ae^x + B \sin x$ implies $y$ is a linear combination of $e^x$ and $\sin x$.
The characteristic roots are $1, i, -i$.
The differential operator is $(D-1)(D^2+1)y = 0$.
Expanding this: $(D^3 - D^2 + D - 1)y = 0$.
This gives $\frac{d^3y}{dx^3} - \frac{d^2y}{dx^2} + \frac{dy}{dx} - y = 0$.
However,the question specifies a second-order equation $f(x) \frac{d^2y}{dx^2} + g(x) \frac{dy}{dx} + h(x) y = 0$.
For $y = Ae^x + B \sin x$,we have $y' = Ae^x + B \cos x$ and $y'' = Ae^x - B \sin x$.
$y'' - y = -B \sin x - B \sin x = -2B \sin x$.
$y' - y = B \cos x - B \sin x$.
By eliminating $A$ and $B$,the resulting equation is $(\cos x + \sin x) y'' - (\cos x - \sin x) y' - (\cos x + \sin x) y = 0$.
Thus $f(x) = \cos x + \sin x$,$g(x) = -\cos x + \sin x$,$h(x) = -\cos x - \sin x$.
Summing these: $f(x) + g(x) + h(x) = (\cos x + \sin x) + (-\cos x + \sin x) + (-\cos x - \sin x) = -\cos x + \sin x$.

(B) The equation of a family of circles with radius $a$ and center $(h, k)$ is given by $(x - h)^2 + (y - k)^2 = a^2$.
Since there are two arbitrary constants $h$ and $k$,we differentiate twice with respect to $x$.
Differentiating with respect to $x$: $2(x - h) + 2(y - k)y_1 = 0$,which implies $(x - h) = -(y - k)y_1$.
Differentiating again: $1 + y_1^2 + (y - k)y_2 = 0$,so $(y - k) = -\frac{1 + y_1^2}{y_2}$.
Substituting $(y - k)$ back into the first derivative equation: $(x - h) = -(-\frac{1 + y_1^2}{y_2})y_1 = \frac{y_1(1 + y_1^2)}{y_2}$.
Now substitute $(x - h)$ and $(y - k)$ into the original circle equation: $(\frac{y_1(1 + y_1^2)}{y_2})^2 + (-\frac{1 + y_1^2}{y_2})^2 = a^2$.
This simplifies to $\frac{y_1^2(1 + y_1^2)^2}{y_2^2} + \frac{(1 + y_1^2)^2}{y_2^2} = a^2$.
Factoring out $(1 + y_1^2)^2$: $\frac{(1 + y_1^2)^2 (y_1^2 + 1)}{y_2^2} = a^2$.
Thus,$(1 + y_1^2)^3 = a^2 y_2^2$.

(B) Given the general solution: $y = A e^{-x} + B \cos x$ ... $(i)$
Differentiating with respect to $x$: $\frac{dy}{dx} = -A e^{-x} - B \sin x$ ... $(ii)$
Differentiating again: $\frac{d^2y}{dx^2} = A e^{-x} - B \cos x$ ... $(iii)$
From $(i)$ and $(ii)$,we eliminate $B$: $y \sin x + \frac{dy}{dx} \cos x = A e^{-x} (\sin x - \cos x)$ ... $(iv)$
From $(i)$ and $(iii)$,we eliminate $B$: $y + \frac{d^2y}{dx^2} = 2 A e^{-x}$ ... $(v)$
From $(iv)$ and $(v)$,we eliminate $A$: $y \sin x + \frac{dy}{dx} \cos x = \frac{1}{2} (y + \frac{d^2y}{dx^2}) (\sin x - \cos x)$
Multiplying by $2$: $2y \sin x + 2 \frac{dy}{dx} \cos x = (y + \frac{d^2y}{dx^2}) (\sin x - \cos x)$
Rearranging terms: $(\cos x - \sin x) \frac{d^2y}{dx^2} + 2 \cos x \frac{dy}{dx} + (\sin x + \cos x) y = 0$.

(A) Given $y=A e^x+B \sin 2 x$.
Differentiating with respect to $x$:
$\frac{d y}{d x}=A e^x+2 B \cos 2 x$ ...$(1)$
Differentiating again:
$\frac{d^2 y}{d x^2}=A e^x-4 B \sin 2 x$ ...$(2)$
From $(1)$,$A e^x = \frac{d y}{d x} - 2 B \cos 2 x$. Substituting this into $(2)$:
$\frac{d^2 y}{d x^2} = \frac{d y}{d x} - 2 B \cos 2 x - 4 B \sin 2 x$.
By testing the options or eliminating constants $A$ and $B$,we find that the differential equation is $(\cos 2 x-\sin 2 x) \frac{d^2 y}{d x^2}+4 \sin 2 x \frac{d y}{d x}-4(\sin 2 x+\cos 2 x) y=0$.

(B) Given the equation: $y=x[a \cos (\log x)+b \sin (\log x)]$.
First,differentiate with respect to $x$:
$\frac{d y}{d x} = [a \cos (\log x) + b \sin (\log x)] + x [-a \sin (\log x) \cdot \frac{1}{x} + b \cos (\log x) \cdot \frac{1}{x}]$
$\frac{d y}{d x} = \frac{y}{x} - a \sin (\log x) + b \cos (\log x)$.
Multiply by $x$: $x \frac{d y}{d x} = y - ax \sin (\log x) + bx \cos (\log x)$.
Differentiate again with respect to $x$:
$x \frac{d^2 y}{d x^2} + \frac{d y}{d x} = \frac{d y}{d x} - [a \sin (\log x) + ax \cos (\log x) \cdot \frac{1}{x}] + [b \cos (\log x) - bx \sin (\log x) \cdot \frac{1}{x}]$
$x \frac{d^2 y}{d x^2} = -a \sin (\log x) - a \cos (\log x) + b \cos (\log x) - b \sin (\log x)$.
Multiply by $x$: $x^2 \frac{d^2 y}{d x^2} = -ax \sin (\log x) - ax \cos (\log x) + bx \cos (\log x) - bx \sin (\log x)$.
Rearranging terms: $x^2 \frac{d^2 y}{d x^2} = -[ax \cos (\log x) + bx \sin (\log x)] - [ax \sin (\log x) - bx \cos (\log x)]$.
Using $y = ax \cos (\log x) + bx \sin (\log x)$ and $x \frac{d y}{d x} - y = -ax \sin (\log x) + bx \cos (\log x)$,we get:
$x^2 \frac{d^2 y}{d x^2} = -y - (x \frac{d y}{d x} - y) = -x \frac{d y}{d x}$.
Wait,simplifying the original expression $y = x[a \cos(\log x) + b \sin(\log x)]$:
Let $t = \log x$,then $x = e^t$. $y = e^t [a \cos t + b \sin t]$.
This is a linear differential equation with constant coefficients in $t$.
The characteristic equation is $(m-1)^2 + 1 = 0 \Rightarrow m^2 - 2m + 2 = 0$.
Converting back to $x$: $x^2 \frac{d^2 y}{d x^2} - x \frac{d y}{d x} + 2y = 0$.

(A) The equation of the family of circles with fixed radius $r$ and center $(h, 3)$ is given by:
$(x-h)^2 + (y-3)^2 = r^2$ --- $(1)$
where $h$ is an arbitrary parameter.
Differentiating both sides with respect to $x$:
$2(x-h) + 2(y-3)\frac{dy}{dx} = 0$
$x-h = -(y-3)\frac{dy}{dx}$
Substituting $(x-h)$ into equation $(1)$:
$[-(y-3)\frac{dy}{dx}]^2 + (y-3)^2 = r^2$
$(y-3)^2 \left(\frac{dy}{dx}\right)^2 + (y-3)^2 = r^2$
Dividing both sides by $(y-3)^2$:
$\left(\frac{dy}{dx}\right)^2 + 1 = \frac{r^2}{(y-3)^2}$

(B) Given the equation of the family of curves: $a x^2+b y^2=1$.
Differentiating with respect to $x$,we get: $2 a x+2 b y \frac{d y}{d x}=0$,which simplifies to $a x+b y \frac{d y}{d x}=0$.
From this,we have $a = -\frac{b y}{x} \frac{d y}{d x}$.
Substituting $a$ into the original equation: $(-\frac{b y}{x} \frac{d y}{d x}) x^2 + b y^2 = 1 \Rightarrow -b x y \frac{d y}{d x} + b y^2 = 1$.
Alternatively,differentiating $a x+b y \frac{d y}{d x}=0$ again with respect to $x$: $a + b \left( y \frac{d^2 y}{d x^2} + (\frac{d y}{d x})^2 \right) = 0$.
Since $a = -\frac{b y}{x} \frac{d y}{d x}$,we substitute this: $-\frac{b y}{x} \frac{d y}{d x} + b y \frac{d^2 y}{d x^2} + b (\frac{d y}{d x})^2 = 0$.
Dividing by $b$ (assuming $b \neq 0$): $-\frac{y}{x} \frac{d y}{d x} + y \frac{d^2 y}{d x^2} + (\frac{d y}{d x})^2 = 0$.
Multiplying by $x$: $-y \frac{d y}{d x} + x y \frac{d^2 y}{d x^2} + x (\frac{d y}{d x})^2 = 0$.
Thus,the differential equation is $x y \frac{d^2 y}{d x^2} + x (\frac{d y}{d x})^2 - y \frac{d y}{d x} = 0$.

(A) The general solution $f(x, y, c_1, c_2) = 0$ contains two arbitrary constants,so the order of the corresponding differential equation is $k = 2$.
Substituting $k = 2$ into the given equation $x^k + y^k = c^2$,we get $x^2 + y^2 = c^2$.
Differentiating both sides with respect to $x$:
$\frac{d}{dx}(x^2 + y^2) = \frac{d}{dx}(c^2)$
$2x + 2y \frac{dy}{dx} = 0$
Dividing by $2$:
$x + y \frac{dy}{dx} = 0$
Rearranging the terms:
$\frac{dy}{dx} + \frac{x}{y} = 0$.

(A) Given equation: $y^2 = 4a(x + a)$ ... $(i)$
On differentiating with respect to $x$,we get:
$2y \frac{dy}{dx} = 4a(1 + 0) = 4a$
$2y y^{\prime} = 4a$
$a = \frac{1}{2} y y^{\prime}$
Now,substitute the value of $a$ in equation $(i)$:
$y^2 = 4 \left( \frac{1}{2} y y^{\prime} \right) \left( x + \frac{1}{2} y y^{\prime} \right)$
$y^2 = 2y y^{\prime} \left( \frac{2x + y y^{\prime}}{2} \right)$
$y^2 = y y^{\prime} (2x + y y^{\prime})$
Dividing both sides by $y$ (assuming $y \neq 0$):
$y = y^{\prime} (2x + y y^{\prime})$
$y = 2x y^{\prime} + y (y^{\prime})^2$
$y - y (y^{\prime})^2 = 2x y^{\prime}$

(D) Given the equation: $\sqrt{1+y^2}=C x e^{\tan ^{-1} x}$
Differentiating both sides with respect to $x$:
$\frac{1}{2\sqrt{1+y^2}} \cdot 2y \frac{dy}{dx} = C \left( e^{\tan ^{-1} x} + x e^{\tan ^{-1} x} \cdot \frac{1}{1+x^2} \right)$
$\frac{y}{\sqrt{1+y^2}} \frac{dy}{dx} = C e^{\tan ^{-1} x} \left( 1 + \frac{x}{1+x^2} \right)$
Since $C e^{\tan ^{-1} x} = \frac{\sqrt{1+y^2}}{x}$,substitute this into the equation:
$\frac{y}{\sqrt{1+y^2}} \frac{dy}{dx} = \frac{\sqrt{1+y^2}}{x} \left( \frac{1+x^2+x}{1+x^2} \right)$
$\frac{xy}{\sqrt{1+y^2}} \frac{dy}{dx} = \frac{\sqrt{1+y^2}(1+x+x^2)}{1+x^2}$
$xy(1+x^2) dy = (1+y^2)(1+x+x^2) dx$
$xy(1+x^2) dy - (1+y^2)(1+x+x^2) dx = 0$

(D) Given equation is $xy = ae^x + be^{-x} + x^2$ $(i)$
Differentiating both sides with respect to $x$,we get:
$y + xy' = ae^x - be^{-x} + 2x$ (ii)
Differentiating again with respect to $x$,we get:
$y' + y' + xy'' = ae^x + be^{-x} + 2$
$2y' + xy'' = ae^x + be^{-x} + 2$ (iii)
From $(i)$,we have $ae^x + be^{-x} = xy - x^2$.
Substituting this into (iii):
$xy'' + 2y' = (xy - x^2) + 2$
Rearranging the terms,we get:
$xy'' + 2y' - xy + x^2 - 2 = 0$

(B) Given the family of curves: $y = ae^{4x} + be^{-x}$ ... $(i)$
Differentiating with respect to $x$:
$\frac{dy}{dx} = 4ae^{4x} - be^{-x}$ ... (ii)
Differentiating again with respect to $x$:
$\frac{d^2y}{dx^2} = 16ae^{4x} + be^{-x}$ ... (iii)
To eliminate the constants $a$ and $b$,we can solve the system of linear equations $(i)$,(ii),and (iii). Adding $(i)$ and (ii):
$y + \frac{dy}{dx} = 5ae^{4x} \implies ae^{4x} = \frac{1}{5}(y + \frac{dy}{dx})$
Subtracting (ii) from $(i)$ multiplied by $4$:
$4y - \frac{dy}{dx} = 5be^{-x} \implies be^{-x} = \frac{1}{5}(4y - \frac{dy}{dx})$
Substituting these into (iii):
$\frac{d^2y}{dx^2} = 16[\frac{1}{5}(y + \frac{dy}{dx})] + \frac{1}{5}(4y - \frac{dy}{dx})$
$\frac{d^2y}{dx^2} = \frac{16y}{5} + \frac{16}{5}\frac{dy}{dx} + \frac{4y}{5} - \frac{1}{5}\frac{dy}{dx} = 4y + 3\frac{dy}{dx}$
Thus,the differential equation is $f = \frac{d^2y}{dx^2} - 3\frac{dy}{dx} - 4y = 0$.
To find $\frac{df}{dx}$,differentiate $f$ with respect to $x$:
$\frac{df}{dx} = \frac{d}{dx}(\frac{d^2y}{dx^2} - 3\frac{dy}{dx} - 4y) = \frac{d^3y}{dx^3} - 3\frac{d^2y}{dx^2} - 4\frac{dy}{dx}$.

(B) Given the equation: $y = ax^2 + bx + c$.
Here,the number of arbitrary constants is $3$ $(a, b, c)$.
To find the differential equation,we differentiate with respect to $x$ repeatedly.
First derivative: $\frac{dy}{dx} = 2ax + b$.
Second derivative: $\frac{d^2y}{dx^2} = 2a$.
Third derivative: $\frac{d^3y}{dx^3} = 0$.
Since the third derivative is zero,the differential equation is $\frac{d^3y}{dx^3} = 0$.

(A) Given equation is $l x^2 + m y^2 - (x + y) = 0 \ldots (i)$
To eliminate the two arbitrary constants $l$ and $m$,we differentiate the equation twice with respect to $x$.
Differentiating $(i)$ with respect to $x$:
$2lx + 2myy^{\prime} - (1 + y^{\prime}) = 0 \ldots (ii)$
Differentiating $(ii)$ with respect to $x$:
$2l + 2m(y y^{\prime \prime} + (y^{\prime})^2) - y^{\prime \prime} = 0 \ldots (iii)$
We have a system of three linear equations in $l, m, -1$:
$l(x^2) + m(y^2) + (-1)(x+y) = 0$
$l(2x) + m(2yy^{\prime}) + (-1)(1+y^{\prime}) = 0$
$l(2) + m(2(y^{\prime 2} + yy^{\prime \prime})) + (-1)(y^{\prime \prime}) = 0$
For a non-trivial solution for $l, m, -1$,the determinant of the coefficients must be zero:
$\left|\begin{array}{ccc}x^2 & y^2 & -(x+y) \\ 2x & 2yy^{\prime} & -(1+y^{\prime}) \\ 2 & 2(y^{\prime 2} + yy^{\prime \prime}) & -y^{\prime \prime}\end{array}\right| = 0$
Multiplying the third column by $-1$,we get:
$\left|\begin{array}{ccc}x^2 & y^2 & x+y \\ 2x & 2yy^{\prime} & 1+y^{\prime} \\ 2 & 2(y^{\prime 2} + yy^{\prime \prime}) & y^{\prime \prime}\end{array}\right| = 0$

(C) The equation of a family of circles with constant radius $r$ and center $(a, b)$ is $(x-a)^2 + (y-b)^2 = r^2$.
Differentiating with respect to $x$: $2(x-a) + 2(y-b)y' = 0 \Rightarrow x-a = -(y-b)y'$.
Substituting this into the circle equation: $(y-b)^2 (y')^2 + (y-b)^2 = r^2 \Rightarrow (y-b)^2 [1 + (y')^2] = r^2 \Rightarrow y-b = \frac{\pm r}{\sqrt{1+(y')^2}}$.
Differentiating $x-a = -(y-b)y'$ again: $1 = -[y'' (y-b) + (y')^2]$.
Substituting $y-b$: $1 + (y')^2 = -y'' \left( \frac{\pm r}{\sqrt{1+(y')^2}} \right)$.
Rearranging: $1 + (y')^2 = \mp \frac{r y''}{\sqrt{1+(y')^2}} \Rightarrow (1+(y')^2)^{3/2} = \mp r y''$.
Squaring both sides: $(1+(y')^2)^3 = r^2 (y'')^2$.

(C) The general equation of a family of circles with a constant radius $r$ and center $(a, b)$ is given by $(x-a)^2 + (y-b)^2 = r^2$.
Since there are two arbitrary constants $a$ and $b$,the order of the differential equation is $k = 2$.
Differentiating with respect to $x$: $2(x-a) + 2(y-b)y' = 0 \Rightarrow (x-a) + (y-b)y' = 0$.
Differentiating again: $1 + (y')^2 + (y-b)y'' = 0 \Rightarrow (y-b) = -\frac{1+(y')^2}{y''}$.
Substituting this back into the first derivative equation: $(x-a) = -y' \left(-\frac{1+(y')^2}{y''}\right) = \frac{y'(1+(y')^2)}{y''}$.
Substituting $(x-a)$ and $(y-b)$ into the original equation: $\left(\frac{y'(1+(y')^2)}{y''}\right)^2 + \left(-\frac{1+(y')^2}{y''}\right)^2 = r^2$.
Simplifying: $\frac{(1+(y')^2)^2}{(y'')^2} ( (y')^2 + 1 ) = r^2 \Rightarrow (1+(y')^2)^3 = r^2(y'')^2$.
The highest order derivative is $y''$,so the order $k = 2$.
The power of the highest order derivative is $2$,so the degree $l = 2$.
Thus,$k^2 + l^2 = 2^2 + 2^2 = 4 + 4 = 8$.

(D) The general equation of a circle with center $(a, 0)$ on the $X$-axis and passing through the origin $(0, 0)$ is given by $(x-a)^2 + y^2 = a^2$.
Expanding this,we get $x^2 - 2ax + a^2 + y^2 = a^2$,which simplifies to $x^2 + y^2 - 2ax = 0$.
To eliminate the arbitrary constant $a$,we differentiate with respect to $x$: $2x + 2y\frac{dy}{dx} - 2a = 0$.
This gives $a = x + y\frac{dy}{dx}$.
Substituting this value of $a$ back into the equation $x^2 + y^2 = 2ax$,we get $x^2 + y^2 = 2x(x + y\frac{dy}{dx})$.
Expanding the right side,$x^2 + y^2 = 2x^2 + 2xy\frac{dy}{dx}$.
Rearranging the terms,we get $y^2 - x^2 - 2xy\frac{dy}{dx} = 0$.

(A) Let the radius of the circle be $a$. Then the center of the circle is $(a, a)$. Hence,the equation of the circle is:
$(x-a)^2 + (y-a)^2 = a^2$
$x^2 - 2ax + a^2 + y^2 - 2ay + a^2 = a^2$
$x^2 + y^2 - 2ax - 2ay + a^2 = 0$ --- $(i)$
Differentiating equation $(i)$ with respect to $x$,we get:
$2x + 2y \frac{dy}{dx} - 2a - 2a \frac{dy}{dx} = 0$
$x + y \frac{dy}{dx} - a(1 + \frac{dy}{dx}) = 0$
$a = \frac{x + y \frac{dy}{dx}}{1 + \frac{dy}{dx}}$
Substituting the value of $a$ into equation $(i)$:
$(x-a)^2 + (y-a)^2 = a^2$
$(x-a)^2 + (y-a)^2 = (\frac{x + y \frac{dy}{dx}}{1 + \frac{dy}{dx}})^2$
Note that $(x-a) = x - \frac{x + y \frac{dy}{dx}}{1 + \frac{dy}{dx}} = \frac{x + x \frac{dy}{dx} - x - y \frac{dy}{dx}}{1 + \frac{dy}{dx}} = \frac{(x-y) \frac{dy}{dx}}{1 + \frac{dy}{dx}}$
And $(y-a) = y - \frac{x + y \frac{dy}{dx}}{1 + \frac{dy}{dx}} = \frac{y + y \frac{dy}{dx} - x - y \frac{dy}{dx}}{1 + \frac{dy}{dx}} = \frac{y-x}{1 + \frac{dy}{dx}}$
Substituting these into $(x-a)^2 + (y-a)^2 = a^2$:
$(\frac{(x-y) \frac{dy}{dx}}{1 + \frac{dy}{dx}})^2 + (\frac{y-x}{1 + \frac{dy}{dx}})^2 = (\frac{x + y \frac{dy}{dx}}{1 + \frac{dy}{dx}})^2$
$(x-y)^2 (\frac{dy}{dx})^2 + (x-y)^2 = (x + y \frac{dy}{dx})^2$
$(x-y)^2 [1 + (\frac{dy}{dx})^2] = (x + y \frac{dy}{dx})^2$

(A) Given the family of curves: $y = a + b e^{2x} + c e^{-3x}$ $(i)$
Differentiating with respect to $x$:
$y_1 = 2b e^{2x} - 3c e^{-3x}$ (ii)
Differentiating again:
$y_2 = 4b e^{2x} + 9c e^{-3x}$ (iii)
Differentiating a third time:
$y_3 = 8b e^{2x} - 27c e^{-3x}$ (iv)
To eliminate the constants $a, b, c$,we consider the linear combination $y_3 + k_1 y_2 + k_2 y_1 = 0$.
The characteristic equation for the terms $e^{2x}$ and $e^{-3x}$ is $(m-2)(m+3)m = 0$,which is $m(m^2 + m - 6) = 0$,or $m^3 + m^2 - 6m = 0$.
This corresponds to the differential equation $y_3 + y_2 - 6y_1 = 0$.

(B) Given the equation of simple harmonic motion: $x = A \cos (nt + \alpha)$ ... $(i)$
Differentiating with respect to $t$:
$\frac{dx}{dt} = -A n \sin (nt + \alpha)$ ... (ii)
Differentiating again with respect to $t$:
$\frac{d^2x}{dt^2} = -An \frac{d}{dt} \sin (nt + \alpha)$
$\frac{d^2x}{dt^2} = -An^2 \cos (nt + \alpha)$
Substituting $x = A \cos (nt + \alpha)$ into the equation:
$\frac{d^2x}{dt^2} = -n^2 x$
Therefore,the differential equation is:
$\frac{d^2x}{dt^2} + n^2 x = 0$

(B) Given the family of curves: $y=(\alpha+\beta x) e^{p x}$ $(i)$
Differentiating with respect to $x$:
$y^{\prime} = \beta e^{p x} + p(\alpha+\beta x) e^{p x}$
$y^{\prime} = \beta e^{p x} + p y$ (ii)
From (ii),we have $\beta e^{p x} = y^{\prime} - p y$.
Differentiating (ii) again with respect to $x$:
$y^{\prime \prime} = \beta p e^{p x} + p y^{\prime}$
Substitute $\beta e^{p x} = y^{\prime} - p y$ into the equation:
$y^{\prime \prime} = p(y^{\prime} - p y) + p y^{\prime}$
$y^{\prime \prime} = p y^{\prime} - p^2 y + p y^{\prime}$
$y^{\prime \prime} = 2 p y^{\prime} - p^2 y$
Rearranging the terms,we get:
$y^{\prime \prime} - 2 p y^{\prime} + p^2 y = 0$

(C) The equation of the family of parabolas with vertex at $(0,-1)$ and axis along the $Y$-axis is given by $x^2 = 4a(y+1)$ $(i)$.
On differentiating both sides with respect to $x$,we get $2x = 4a y^{\prime}$,which implies $a = \frac{x}{2y^{\prime}}$.
Substituting the value of $a$ into equation $(i)$,we get $x^2 = 4 \left( \frac{x}{2y^{\prime}} \right) (y+1)$.
Simplifying this,we get $x = \frac{2(y+1)}{y^{\prime}}$.
Thus,$x y^{\prime} = 2y + 2$,or $x y^{\prime} - 2y - 2 = 0$.

(A) Given equation is $xy = ae^x + be^{-x}$.
First,differentiate with respect to $x$ using the product rule on the left side:
$x \frac{dy}{dx} + y = ae^x - be^{-x}$.
Now,differentiate again with respect to $x$:
$x \frac{d^2 y}{dx^2} + \frac{dy}{dx} + \frac{dy}{dx} = ae^x + be^{-x}$.
Simplify the expression:
$x \frac{d^2 y}{dx^2} + 2 \frac{dy}{dx} = ae^x + be^{-x}$.
Since $ae^x + be^{-x} = xy$,substitute this back into the equation:
$x \frac{d^2 y}{dx^2} + 2 \frac{dy}{dx} = xy$.
Rearranging gives the final differential equation:
$x \frac{d^2 y}{dx^2} + 2 \frac{dy}{dx} - xy = 0$.

(B) The standard equation of a parabola with focus at $(0,0)$ and axis along the $X$-axis is given by $(x+a)^2 + y^2 = (x+2a)^2$,which simplifies to $y^2 = 4a(x+a)$. However,for a parabola with focus at the origin and axis as the $X$-axis,the equation is $y^2 = 4a(x+a)$.
Let the parabola be $y^2 = 2a(x + a/2)$,or more simply $y^2 = 2ax + a^2$ where $a$ is the parameter.
Differentiating with respect to $x$:
$2y \frac{dy}{dx} = 2a \Rightarrow a = y \frac{dy}{dx}$.
Substituting $a$ into the original equation $y^2 = 2ax + a^2$:
$y^2 = 2x(y \frac{dy}{dx}) + (y \frac{dy}{dx})^2$.
Dividing by $y$ (assuming $y \neq 0$):
$y = 2x \frac{dy}{dx} + y \left(\frac{dy}{dx}\right)^2$.
Rearranging the terms:
$y \left(\frac{dy}{dx}\right)^2 = y - 2x \frac{dy}{dx}$,which is equivalent to $-y \left(\frac{dy}{dx}\right)^2 = 2x \frac{dy}{dx} - y$.

(A) The general equation of an ellipse with axes along the coordinate axes is $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$.
Differentiating with respect to $x$:
$\frac{2x}{a^2} + \frac{2y}{b^2} \cdot y^{\prime} = 0$
$\frac{y}{b^2} \cdot y^{\prime} = -\frac{x}{a^2}$
$\frac{y y^{\prime}}{x} = -\frac{b^2}{a^2}$
Differentiating again with respect to $x$:
$\frac{d}{dx} \left( \frac{y y^{\prime}}{x} \right) = \frac{d}{dx} \left( -\frac{b^2}{a^2} \right) = 0$
Using the quotient rule:
$\frac{x(y y^{\prime \prime} + (y^{\prime})^2) - y y^{\prime}(1)}{x^2} = 0$
$x y y^{\prime \prime} + x(y^{\prime})^2 - y y^{\prime} = 0$.
Thus,option $A$ is correct.

(D) Given family of curves is $y^2 - 5ax - 5a^{3/2} = 0$ ... $(i)$
Differentiating with respect to $x$:
$2yy' - 5a = 0 \Rightarrow a = \frac{2}{5}yy'$
Substituting the value of $a$ in equation $(i)$:
$y^2 - 5\left(\frac{2}{5}yy'\right)x - 5\left(\frac{2}{5}yy'\right)^{3/2} = 0$
$y^2 - 2yy'x = 5\left(\frac{2}{5}yy'\right)^{3/2}$
Squaring both sides:
$(y^2 - 2yy'x)^2 = 25 \cdot \left(\frac{2}{5}yy'\right)^3$
$(y^2 - 2yy'x)^2 = 25 \cdot \frac{8}{125} (yy')^3$
$(y^2 - 2yy'x)^2 = \frac{8}{5} (yy')^3$
The highest order derivative is $y'$,so the order $m = 1$.
The power of the highest order derivative is $3$,so the degree $n = 3$.
Therefore,$m - n = 1 - 3 = -2$.

(C) Statement $I$: The equation of a circle with center $(0, \alpha)$ and radius $k$ is $x^2 + (y - \alpha)^2 = k^2$ ... $(i)$.
Differentiating with respect to $x$: $2x + 2(y - \alpha)\frac{dy}{dx} = 0 \Rightarrow y - \alpha = -x\frac{dx}{dy}$.
Thus,$\alpha = y + x\frac{dx}{dy}$.
Substituting $\alpha$ into $(i)$: $x^2 + (-x\frac{dx}{dy})^2 = k^2 \Rightarrow x^2 + x^2(\frac{dx}{dy})^2 = k^2$.
$x^2(1 + (\frac{dx}{dy})^2) = k^2 \Rightarrow x^2(1 + \frac{1}{(dy/dx)^2}) = k^2 \Rightarrow x^2(\frac{(dy/dx)^2 + 1}{(dy/dx)^2}) = k^2$.
$x^2(dy/dx)^2 + x^2 = k^2(dy/dx)^2 \Rightarrow (x^2 - k^2)(dy/dx)^2 + x^2 = 0$. Hence,Statement $I$ is true.
Statement $II$: The circle passes through the origin and its center is on the $X$-axis,so center is $(\alpha, 0)$ and radius is $|\alpha|$.
The equation is $(x - \alpha)^2 + y^2 = \alpha^2 \Rightarrow x^2 - 2x\alpha + \alpha^2 + y^2 = \alpha^2 \Rightarrow x^2 + y^2 = 2x\alpha \Rightarrow \alpha = \frac{x^2 + y^2}{2x}$.
Differentiating $x^2 + y^2 = 2x\alpha$ with respect to $x$: $2x + 2y\frac{dy}{dx} = 2\alpha$.
Substitute $\alpha$: $x + y\frac{dy}{dx} = \frac{x^2 + y^2}{2x} \Rightarrow 2x^2 + 2xy\frac{dy}{dx} = x^2 + y^2 \Rightarrow x^2 - y^2 + 2xy\frac{dy}{dx} = 0$. Hence,Statement $II$ is true.

(A) Given the general solution: $a x^2+b y^2=1$ $\ldots$ $(i)$
Differentiating with respect to $x$: $2 a x+2 b y \frac{d y}{d x}=0 \Rightarrow a x+b y y'=0$.
Differentiating again with respect to $x$: $a+b(y')^2+b y y''=0$.
From the first derivative,$a = -b y y' / x$. Substituting this into the second derivative equation:
$-b y y' / x + b(y')^2 + b y y'' = 0$.
Dividing by $b$ (assuming $b \neq 0$): $-y y' / x + (y')^2 + y y'' = 0$.
Multiplying by $x$: $x y y'' + x (y')^2 - y y' = 0$.
The order of this differential equation is $\alpha = 2$ and the degree is $\beta = 1$.
Substituting these into the ellipse equation $\alpha x^2+\beta y^2=1$,we get $2 x^2+y^2=1$.
Rewriting in standard form: $\frac{x^2}{1/2} + \frac{y^2}{1} = 1$.
Here,$a^2 = 1/2$ and $b^2 = 1$. Since $b^2 > a^2$,the eccentricity $e$ is given by $e = \sqrt{1 - \frac{a^2}{b^2}} = \sqrt{1 - \frac{1/2}{1}} = \sqrt{1/2} = \frac{1}{\sqrt{2}}$.

(A) The general equation of a parabola whose axis is parallel to the $y$-axis is given by $y = Ax^2 + Bx + C$,where $A, B, C$ are arbitrary constants.
To find the differential equation,we differentiate with respect to $x$ three times.
First derivative: $\frac{dy}{dx} = 2Ax + B$.
Second derivative: $\frac{d^2y}{dx^2} = 2A$.
Third derivative: $\frac{d^3y}{dx^3} = 0$.
Since there are $3$ arbitrary constants,the order of the differential equation is $3$. Thus,the equation is $\frac{d^3y}{dx^3} = 0$.

(B) The equation of a hyperbola with axes parallel to the coordinate axes and center $(h, k)$ is given by $\frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1$ or $\frac{(y-k)^2}{a^2} - \frac{(x-h)^2}{b^2} = 1$.
Given the center lies on $y=2x$,we have $k=2h$.
For a hyperbola,$e^2 = 1 + \frac{b^2}{a^2} = 3$,so $b^2 = 2a^2$.
The equation becomes $(x-h)^2 - \frac{1}{2}(y-2h)^2 = \pm a^2$.
Differentiating with respect to $x$: $2(x-h) - (y-2h)y_1 = 0$,which gives $x-h = \frac{1}{2}(y-2h)y_1$.
Substituting $h = x - \frac{1}{2}(y-2h)y_1$ into the equation and simplifying leads to the differential equation $(y-2x)y_2 + y_1^2 + 2y_1 = y_1^3 + 2$.

(D) Given the family of ellipses: $\frac{x^2}{a^2} + \frac{y^2}{4} = 1$.
To find the differential equation,we differentiate both sides with respect to $x$:
$\frac{d}{dx} \left( \frac{x^2}{a^2} + \frac{y^2}{4} \right) = \frac{d}{dx} (1)$
$\frac{2x}{a^2} + \frac{2y}{4} \frac{dy}{dx} = 0$
$\frac{2x}{a^2} + \frac{y}{2} \frac{dy}{dx} = 0$
From the original equation,we have $\frac{x^2}{a^2} = 1 - \frac{y^2}{4} = \frac{4 - y^2}{4}$,so $\frac{1}{a^2} = \frac{4 - y^2}{4x^2}$.
Substitute $\frac{1}{a^2}$ into the differentiated equation:
$2x \left( \frac{4 - y^2}{4x^2} \right) + \frac{y}{2} \frac{dy}{dx} = 0$
$\frac{4 - y^2}{2x} + \frac{y}{2} \frac{dy}{dx} = 0$
Multiply by $2x$:
$(4 - y^2) + xy \frac{dy}{dx} = 0$
$xy \frac{dy}{dx} = y^2 - 4$.

(A) The general equation of a parabola whose axis of symmetry is along the $X$-axis is given by $y^2 = 4a(x - h)$,which can be rewritten as $y^2 = Ax + B$,where $A$ and $B$ are arbitrary constants.
Since there are $2$ arbitrary constants,we need to differentiate the equation twice to eliminate them.
First derivative: $2y \frac{dy}{dx} = A$.
Second derivative: $2 \left( \left( \frac{dy}{dx} \right)^2 + y \frac{d^2y}{dx^2} \right) = 0$.
Since the highest order derivative present in the differential equation is the second derivative,the order of the differential equation is $2$.

(A) Given the equation of the family of curves: $y = e^{a \sin x}$.
Taking the natural logarithm on both sides,we get: $\log y = a \sin x$.
Differentiating both sides with respect to $x$: $\frac{1}{y} \frac{dy}{dx} = a \cos x$.
From the equation $\log y = a \sin x$,we can write $a = \frac{\log y}{\sin x}$.
Substituting the value of '$a$' into the differentiated equation:
$\frac{1}{y} \frac{dy}{dx} = \left( \frac{\log y}{\sin x} \right) \cos x$.
$\frac{1}{y} \frac{dy}{dx} = \log y \cot x$.
Rearranging the terms to match the options:
$\frac{dy}{dx} = y \log y \cot x$,which can be rewritten as $\frac{dy}{dx} \tan x = y \log y$ or $y \log y = \tan x \frac{dy}{dx}$.

(B) The equation of an ellipse centered at the origin with axes as the coordinate axes is $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$.
Differentiating with respect to $x$,we get $\frac{2x}{a^{2}}+\frac{2y y^{\prime}}{b^{2}}=0$,which simplifies to $\frac{x}{a^{2}}+\frac{y y^{\prime}}{b^{2}}=0$.
This gives $\frac{b^{2}}{a^{2}} = -\frac{y y^{\prime}}{x}$.
Differentiating again with respect to $x$,we get $\frac{1}{a^{2}} + \frac{1}{b^{2}}(y y^{\prime \prime} + (y^{\prime})^{2}) = 0$.
Substituting $\frac{1}{a^{2}} = -\frac{y y^{\prime}}{b^{2}x}$,we have $-\frac{y y^{\prime}}{b^{2}x} + \frac{1}{b^{2}}(y y^{\prime \prime} + (y^{\prime})^{2}) = 0$.
Multiplying by $b^{2}x$,we get $-y y^{\prime} + x(y y^{\prime \prime} + (y^{\prime})^{2}) = 0$.
Thus,$x y y^{\prime \prime} + x(y^{\prime})^{2} - y y^{\prime} = 0$.

(B) Given equation: $y = e^{x}(A \cos x + B \sin x)$
Differentiating with respect to $x$ using the product rule:
$y' = e^{x}(A \cos x + B \sin x) + e^{x}(-A \sin x + B \cos x)$
$y' = y + e^{x}(-A \sin x + B \cos x)$
Differentiating again with respect to $x$:
$y'' = y' + [e^{x}(-A \sin x + B \cos x) + e^{x}(-A \cos x - B \sin x)]$
From the first derivative,we know $e^{x}(-A \sin x + B \cos x) = y' - y$.
Substituting this into the second derivative:
$y'' = y' + (y' - y) - e^{x}(A \cos x + B \sin x)$
Since $y = e^{x}(A \cos x + B \sin x)$,we have:
$y'' = 2y' - y - y$
$y'' - 2y' + 2y = 0$
Thus,the differential equation is $\frac{d^{2} y}{d x^{2}} - 2 \frac{d y}{d x} + 2 y = 0$.

(A) Given,$y = e^{-x} \cos 2x$.
Taking the first derivative with respect to $x$ using the product rule:
$\frac{dy}{dx} = e^{-x}(-2 \sin 2x) + \cos 2x(-e^{-x}) = -2e^{-x} \sin 2x - y$.
Rearranging gives: $\frac{dy}{dx} + y = -2e^{-x} \sin 2x$.
Taking the derivative again with respect to $x$:
$\frac{d^2y}{dx^2} + \frac{dy}{dx} = -2[e^{-x}(2 \cos 2x) + \sin 2x(-e^{-x})] = -4(e^{-x} \cos 2x) + 2(e^{-x} \sin 2x)$.
Substituting $y = e^{-x} \cos 2x$ and $-2e^{-x} \sin 2x = \frac{dy}{dx} + y$:
$\frac{d^2y}{dx^2} + \frac{dy}{dx} = -4y - (\frac{dy}{dx} + y)$.
$\frac{d^2y}{dx^2} + \frac{dy}{dx} = -4y - \frac{dy}{dx} - y$.
$\frac{d^2y}{dx^2} + 2\frac{dy}{dx} + 5y = 0$.

(A) Given curve is $y=(\cos x+y)^{1/2}$.
Squaring both sides,we get $y^2 = \cos x + y$.
Differentiating both sides with respect to $x$:
$2y \frac{dy}{dx} = -\sin x + \frac{dy}{dx}$.
Rearranging the terms:
$(2y - 1) \frac{dy}{dx} = -\sin x$.
Differentiating again with respect to $x$ using the product rule:
$(2y - 1) \frac{d^2y}{dx^2} + \frac{dy}{dx} \cdot \frac{d}{dx}(2y - 1) = -\cos x$.
$(2y - 1) \frac{d^2y}{dx^2} + \frac{dy}{dx} \cdot (2 \frac{dy}{dx}) = -\cos x$.
$(2y - 1) \frac{d^2y}{dx^2} + 2 \left(\frac{dy}{dx}\right)^2 + \cos x = 0$.

(B) Given equation: $y = ae^{bx} \dots (i)$
Differentiating with respect to $x$:
$y_1 = \frac{dy}{dx} = abe^{bx}$
Since $y = ae^{bx}$,we can write:
$y_1 = by \dots (ii)$
Differentiating $y_1 = by$ with respect to $x$:
$y_2 = \frac{d^2y}{dx^2} = by_1 \dots (iii)$
From equation $(ii)$,we have $b = \frac{y_1}{y}$.
Substituting $b$ into equation $(iii)$:
$y_2 = \left(\frac{y_1}{y}\right)y_1$
$y_2 = \frac{y_1^2}{y}$
Therefore,$yy_2 = y_1^2$.

(D) Given,$\sqrt{y}=\cos ^{-1} x \Rightarrow y=(\cos ^{-1} x)^{2}$.
On differentiating both sides with respect to $x$,we get:
$\frac{dy}{dx} = 2(\cos ^{-1} x) \times \left(\frac{-1}{\sqrt{1-x^{2}}}\right)$.
Multiplying both sides by $\sqrt{1-x^{2}}$,we get:
$\sqrt{1-x^{2}} \frac{dy}{dx} = -2 \cos ^{-1} x$.
Again,differentiating both sides with respect to $x$ using the product rule:
$\sqrt{1-x^{2}} \frac{d^{2}y}{dx^{2}} + \frac{dy}{dx} \times \left(\frac{-2x}{2\sqrt{1-x^{2}}}\right) = -2 \times \left(\frac{-1}{\sqrt{1-x^{2}}}\right)$.
$\sqrt{1-x^{2}} \frac{d^{2}y}{dx^{2}} - \frac{x}{\sqrt{1-x^{2}}} \frac{dy}{dx} = \frac{2}{\sqrt{1-x^{2}}}$.
Multiplying the entire equation by $\sqrt{1-x^{2}}$,we get:
$(1-x^{2}) \frac{d^{2}y}{dx^{2}} - x \frac{dy}{dx} = 2$.
Comparing this with the given equation $(1-x^{2}) \frac{d^{2}y}{dx^{2}} - x \frac{dy}{dx} = c$,we find $c = 2$.