Formation of differential equations Questions in English

(B) The general equation of a circle touching the $Y$-axis at the origin $(0,0)$ and having its center on the $X$-axis is $(x-a)^2 + (y-0)^2 = a^2$,where $a$ is the radius of the circle.
Expanding this,we get $x^2 - 2ax + a^2 + y^2 = a^2$,which simplifies to $x^2 + y^2 = 2ax$.
To eliminate the arbitrary constant $a$,we differentiate both sides with respect to $x$:
$\frac{d}{dx}(x^2 + y^2) = \frac{d}{dx}(2ax)$
$2x + 2y \frac{dy}{dx} = 2a$.
Substituting the value of $2a = \frac{x^2+y^2}{x}$ into the differentiated equation:
$2x + 2y \frac{dy}{dx} = \frac{x^2+y^2}{x}$.
Multiplying both sides by $x$:
$2x^2 + 2xy \frac{dy}{dx} = x^2 + y^2$.
Rearranging the terms,we get $x^2 - y^2 + 2xy \frac{dy}{dx} = 0$.

(C) Given the solution $y = e^x (A \cos x + B \sin x)$.
First,differentiate with respect to $x$ using the product rule:
$\frac{dy}{dx} = e^x (A \cos x + B \sin x) + e^x (-A \sin x + B \cos x)$
$\frac{dy}{dx} = e^x ((A + B) \cos x + (B - A) \sin x)$
Now,differentiate again with respect to $x$:
$\frac{d^2 y}{dx^2} = e^x ((A + B) \cos x + (B - A) \sin x) + e^x (-(A + B) \sin x + (B - A) \cos x)$
$\frac{d^2 y}{dx^2} = e^x ((A + B + B - A) \cos x + (B - A - A - B) \sin x)$
$\frac{d^2 y}{dx^2} = e^x (2B \cos x - 2A \sin x)$
Now,substitute $\frac{dy}{dx}$ and $y$ into the equation $\frac{d^2 y}{dx^2} - 2 \frac{dy}{dx} + 2y = 0$:
$e^x (2B \cos x - 2A \sin x) - 2e^x ((A + B) \cos x + (B - A) \sin x) + 2e^x (A \cos x + B \sin x)$
$= e^x [ (2B - 2A - 2B + 2A) \cos x + (-2A - 2B + 2A + 2B) \sin x ]$
$= e^x [ 0 \cos x + 0 \sin x ] = 0$.
Thus,the correct option is $C$.

(C) Given the family of curves $y = C_1 e^{C_2 x}$.
First,differentiate with respect to $x$:
$y^{\prime} = C_1 C_2 e^{C_2 x}$
Since $y = C_1 e^{C_2 x}$,we can substitute this into the derivative:
$y^{\prime} = y C_2$
$C_2 = \frac{y^{\prime}}{y}$
Now,differentiate $y^{\prime} = C_1 C_2 e^{C_2 x}$ again with respect to $x$:
$y^{\prime \prime} = C_1 C_2^2 e^{C_2 x}$
Substitute $y = C_1 e^{C_2 x}$ into the second derivative:
$y^{\prime \prime} = (C_1 e^{C_2 x}) C_2^2 = y C_2^2$
Substitute $C_2 = \frac{y^{\prime}}{y}$ into the equation:
$y^{\prime \prime} = y \left( \frac{y^{\prime}}{y} \right)^2$
$y^{\prime \prime} = y \frac{(y^{\prime})^2}{y^2}$
$y^{\prime \prime} = \frac{(y^{\prime})^2}{y}$
$y y^{\prime \prime} = (y^{\prime})^2$
Thus,the correct option is $C$.

(A) The equation of a parabola with vertex at the origin and axis along the positive $Y$-axis is given by $x^2 = 4ay$,where $a$ is an arbitrary constant.
To find the differential equation,we differentiate both sides with respect to $x$:
$\frac{d}{dx}(x^2) = \frac{d}{dx}(4ay)$
$2x = 4a \frac{dy}{dx}$
$x = 2a \frac{dy}{dx}$
From this,we get $2a = \frac{x}{dy/dx}$.
Substitute $2a$ back into the original equation $x^2 = 2a(2y)$:
$x^2 = \left( \frac{x}{dy/dx} \right) (2y)$
$x^2 \frac{dy}{dx} = 2xy$
$x \frac{dy}{dx} = 2y$
$x \frac{dy}{dx} - 2y = 0$.
Thus,the correct option is $A$.

(C) Given equation: $y^2 = (x + c)^3$
Differentiating both sides with respect to $x$:
$2y \frac{dy}{dx} = 3(x + c)^2$
From this,we have $(x + c)^2 = \frac{2y}{3} \frac{dy}{dx}$.
Cubing both sides:
$(x + c)^6 = \left(\frac{2y}{3} \frac{dy}{dx}\right)^3$
Since $(x + c)^3 = y^2$,we have $(x + c)^6 = (y^2)^2 = y^4$.
Substituting this into the equation:
$y^4 = \frac{8y^3}{27} \left(\frac{dy}{dx}\right)^3$
Dividing both sides by $y^3$ (assuming $y \neq 0$):
$y = \frac{8}{27} \left(\frac{dy}{dx}\right)^3$
$27y = 8 \left(\frac{dy}{dx}\right)^3$

(C) The general equation of a parabola with its axis parallel to the $Y$-axis is given by $(x-h)^2 = 4a(y-k)$,where $(h, k)$ is the vertex and $a$ is a constant.
This equation contains three arbitrary constants: $h$,$k$,and $a$.
To find the differential equation,we differentiate with respect to $x$ three times.
First differentiation: $2(x-h) = 4a \frac{dy}{dx} \implies (x-h) = 2a y_1$.
Second differentiation: $1 = 2a y_2$.
Third differentiation: $0 = 2a y_3$.
Since $2a \neq 0$,we must have $y_3 = 0$.

(D) Given equation: $y = e^x(a \cos x + b \sin x)$
Differentiating with respect to $x$ using the product rule:
$\frac{dy}{dx} = e^x(a \cos x + b \sin x) + e^x(-a \sin x + b \cos x)$
Since $y = e^x(a \cos x + b \sin x)$,we can write:
$\frac{dy}{dx} = y + e^x(b \cos x - a \sin x)$
Differentiating again with respect to $x$:
$\frac{d^2y}{dx^2} = \frac{dy}{dx} + [e^x(b \cos x - a \sin x) + e^x(-b \sin x - a \cos x)]$
Substitute $e^x(b \cos x - a \sin x) = \frac{dy}{dx} - y$:
$\frac{d^2y}{dx^2} = \frac{dy}{dx} + (\frac{dy}{dx} - y) - e^x(a \cos x + b \sin x)$
Since $e^x(a \cos x + b \sin x) = y$,we have:
$\frac{d^2y}{dx^2} = \frac{dy}{dx} + \frac{dy}{dx} - y - y$
$\frac{d^2y}{dx^2} = 2\frac{dy}{dx} - 2y$
Rearranging the terms:
$\frac{d^2y}{dx^2} - 2\frac{dy}{dx} + 2y = 0$

(A) Given the equation $y=c^2+\frac{c}{x}$.
First,differentiate with respect to $x$:
$\frac{dy}{dx} = 0 - \frac{c}{x^2} = -\frac{c}{x^2}$.
From this,we can express the constant $c$ in terms of $x$ and $\frac{dy}{dx}$:
$c = -x^2\frac{dy}{dx}$.
Now,substitute this value of $c$ back into the original equation:
$y = (-x^2\frac{dy}{dx})^2 + \frac{-x^2\frac{dy}{dx}}{x}$.
Simplifying the expression:
$y = x^4\left(\frac{dy}{dx}\right)^2 - x\frac{dy}{dx}$.
Rearranging the terms to form the differential equation:
$x^4\left(\frac{dy}{dx}\right)^2 - x\frac{dy}{dx} - y = 0$.

(A) The equation of a line with $x$-intercept $a$ and $y$-intercept $b$ is given by $\frac{x}{a} + \frac{y}{b} = 1$.
Since there are two arbitrary constants $a$ and $b$,we differentiate the equation twice.
Differentiating with respect to $x$:
$\frac{1}{a} + \frac{1}{b} \frac{d y}{d x} = 0$
Differentiating again with respect to $x$:
$0 + \frac{1}{b} \frac{d^2 y}{d x^2} = 0$
Since $b \neq 0$,we have $\frac{d^2 y}{d x^2} = 0$.

(A) Let the centre of the circle be $(0, k)$. Since the circle passes through the origin $(0, 0)$,the radius of the circle is $r = \sqrt{(0-0)^2 + (k-0)^2} = |k|$.
The equation of the circle is $(x-0)^2 + (y-k)^2 = k^2$.
Expanding this,we get $x^2 + y^2 - 2ky + k^2 = k^2$,which simplifies to $x^2 + y^2 = 2ky$.
To eliminate the arbitrary constant $k$,we differentiate with respect to $x$:
$2x + 2y \frac{dy}{dx} = 2k \frac{dy}{dx}$.
From the circle equation,$k = \frac{x^2+y^2}{2y}$.
Substituting this value of $k$ into the differentiated equation:
$2x + 2y \frac{dy}{dx} = 2 \left( \frac{x^2+y^2}{2y} \right) \frac{dy}{dx}$.
$2x + 2y \frac{dy}{dx} = \frac{x^2+y^2}{y} \frac{dy}{dx}$.
Multiplying by $y$:
$2xy + 2y^2 \frac{dy}{dx} = (x^2+y^2) \frac{dy}{dx}$.
Rearranging the terms:
$(x^2+y^2) \frac{dy}{dx} - 2y^2 \frac{dy}{dx} = 2xy$.
$(x^2-y^2) \frac{dy}{dx} = 2xy$.
Thus,$(x^2-y^2) \frac{dy}{dx} - 2xy = 0$.

(A) Let the equation of the ellipse be $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$.
Given that the major axis is twice the minor axis,we have $2a = 2(2b)$,which implies $a = 2b$.
Substituting $a = 2b$ into the equation,we get $\frac{x^2}{(2b)^2} + \frac{y^2}{b^2} = 1$.
This simplifies to $\frac{x^2}{4b^2} + \frac{y^2}{b^2} = 1$,or $x^2 + 4y^2 = 4b^2$.
Differentiating both sides with respect to $x$,we get $\frac{d}{dx}(x^2 + 4y^2) = \frac{d}{dx}(4b^2)$.
This gives $2x + 8y \frac{dy}{dx} = 0$.
Dividing by $2$,we obtain $x + 4y \frac{dy}{dx} = 0$.

(B) Let $(h, 0)$ be the center of the circle and $r$ be the radius. The equation of the circle is $(x-h)^2 + y^2 = r^2$.
Since the radius $r$ is not fixed,this family of circles has two parameters $h$ and $r$. However,for a specific radius $r$,the family is $(x-h)^2 + y^2 = r^2$.
Differentiating with respect to $x$: $2(x-h) + 2y \frac{dy}{dx} = 0$,which gives $x-h = -y \frac{dy}{dx}$.
Substituting this into the circle equation: $(-y \frac{dy}{dx})^2 + y^2 = r^2$,so $y^2 (\frac{dy}{dx})^2 + y^2 = r^2$.
Differentiating again with respect to $x$: $2y \frac{dy}{dx} (\frac{dy}{dx})^2 + y^2 (2 \frac{dy}{dx} \frac{d^2y}{dx^2}) + 2y \frac{dy}{dx} = 0$.
Dividing by $2y \frac{dy}{dx}$ (assuming $y \neq 0$ and $\frac{dy}{dx} \neq 0$): $(\frac{dy}{dx})^2 + y \frac{d^2y}{dx^2} + 1 = 0$.

(B) Since the circles touch the $y$-axis at the origin,their centers must lie on the $x$-axis. Let the center be $(h, 0)$ and the radius be $h$.
The equation of the circle is $(x-h)^2 + y^2 = h^2$,which simplifies to $x^2 - 2hx + h^2 + y^2 = h^2$,or $x^2 + y^2 - 2hx = 0$ $(1)$.
Differentiating both sides with respect to $x$,we get $2x + 2y\frac{dy}{dx} - 2h = 0$,which simplifies to $h = x + y\frac{dy}{dx}$.
Substituting this value of $h$ into equation $(1)$,we get $x^2 + y^2 - 2(x + y\frac{dy}{dx})x = 0$.
Expanding this,we get $x^2 + y^2 - 2x^2 - 2xy\frac{dy}{dx} = 0$.
Rearranging the terms,we obtain $y^2 - x^2 - 2xy\frac{dy}{dx} = 0$,which is equivalent to $x^2 - y^2 + 2xy\frac{dy}{dx} = 0$.

(D) Given the family of lines: $y = mx + \frac{4}{m}$ $(1)$
Differentiating with respect to $x$: $\frac{dy}{dx} = m$
Substituting the value of $m$ into equation $(1)$:
$y = \left(\frac{dy}{dx}\right)x + \frac{4}{\left(\frac{dy}{dx}\right)}$
Multiplying both sides by $\frac{dy}{dx}$:
$y\left(\frac{dy}{dx}\right) = x\left(\frac{dy}{dx}\right)^2 + 4$
Rearranging the terms:
$x\left(\frac{dy}{dx}\right)^2 - y\left(\frac{dy}{dx}\right) + 4 = 0$

(A) The equation of a parabola with vertex at the origin and axis along the positive $Y$-axis is given by $x^2 = 4ay$,where $a > 0$ is an arbitrary constant.
To find the differential equation,we differentiate with respect to $x$:
$2x = 4a \frac{dy}{dx}$
$\Rightarrow a = \frac{2x}{4(dy/dx)} = \frac{x}{2(dy/dx)}$.
Substituting the value of $a$ back into the original equation:
$x^2 = 4 \left( \frac{x}{2(dy/dx)} \right) y$
$x^2 = \frac{2xy}{dy/dx}$
$x^2 \frac{dy}{dx} = 2xy$
Dividing by $x$ (since $x \neq 0$):
$x \frac{dy}{dx} = 2y$
$x \frac{dy}{dx} - 2y = 0$.

(A) The equation of a parabola with focus at the origin $(0,0)$ and the $X$-axis as its axis is given by $(y-0)^2 = 4a(x+a)$,which simplifies to $y^2 = 4a(x+a)$.
Differentiating both sides with respect to $x$,we get $2y\frac{dy}{dx} = 4a$,which implies $a = \frac{y}{2}\frac{dy}{dx}$.
Substituting the value of $a$ into the original equation $y^2 = 4a(x+a)$,we get:
$y^2 = 4\left(\frac{y}{2}\frac{dy}{dx}\right)\left(x + \frac{y}{2}\frac{dy}{dx}\right)$
$y^2 = 2y\frac{dy}{dx}\left(x + \frac{y}{2}\frac{dy}{dx}\right)$
Dividing by $y$ (assuming $y \neq 0$):
$y = 2x\frac{dy}{dx} + y\left(\frac{dy}{dx}\right)^2$
Rearranging the terms,we get $y\left(\frac{dy}{dx}\right)^2 + 2x\frac{dy}{dx} - y = 0$,which is equivalent to $y\left(\frac{dy}{dx}\right)^2 = y - 2x\frac{dy}{dx}$ or $-y\left(\frac{dy}{dx}\right)^2 = 2x\frac{dy}{dx} - y$.

(A) Given the equation: $y = A \cos \omega t + B \sin \omega t$
Differentiating with respect to $t$:
$\frac{dy}{dt} = -A \omega \sin \omega t + B \omega \cos \omega t$
Differentiating again with respect to $t$:
$\frac{d^2y}{dt^2} = -A \omega^2 \cos \omega t - B \omega^2 \sin \omega t$
Factoring out $-\omega^2$:
$\frac{d^2y}{dt^2} = -\omega^2 (A \cos \omega t + B \sin \omega t)$
Since $y = A \cos \omega t + B \sin \omega t$,we substitute $y$ into the equation:
$\frac{d^2y}{dt^2} = -\omega^2 y$
Rearranging the terms gives:
$\frac{d^2y}{dt^2} + \omega^2 y = 0$

(C) The equation of a parabola with its axis parallel to the $y$-axis is given by $(x-h)^2 = 4a(y-k)$. Since the axis is the $y$-axis,the vertex lies on the $y$-axis,so $h=0$. Thus,the equation is $x^2 = 4a(y-k)$.
Differentiating with respect to $x$:
$2x = 4a \frac{dy}{dx} \implies 4a = \frac{2x}{dy/dx}$.
Differentiating again with respect to $x$:
$2 = 4a \frac{d^2y}{dx^2}$.
Substituting the value of $4a$ from the first derivative into the second derivative equation:
$2 = \left( \frac{2x}{dy/dx} \right) \frac{d^2y}{dx^2}$.
Simplifying:
$1 = \frac{x}{dy/dx} \cdot \frac{d^2y}{dx^2}$
$\implies \frac{dy}{dx} = x \frac{d^2y}{dx^2}$
$\implies x \frac{d^2y}{dx^2} - \frac{dy}{dx} = 0$.

(D) The equation of a line with $x$-intercept $a$ and $y$-intercept $b$ is given by $\frac{x}{a} + \frac{y}{b} = 1$.
Multiplying by $ab$,we get $bx + ay = ab$.
Differentiating both sides with respect to $x$,we get $b + a \frac{dy}{dx} = 0$,which implies $\frac{dy}{dx} = -\frac{b}{a}$.
Differentiating again with respect to $x$,since $-\frac{b}{a}$ is a constant,we get $\frac{d^{2} y}{d x^{2}} = 0$.

(B) Given equation is $y^{2}=(2 x+c)^{5}$ ...$(i)$
Differentiating both sides with respect to $x$:
$2 y \frac{dy}{dx} = 5(2 x+c)^{4} \times 2$
$y \frac{dy}{dx} = 5(2 x+c)^{4}$
From this,we find $(2 x+c)$:
$(2 x+c)^{4} = \frac{y}{5} \frac{dy}{dx}$
$(2 x+c) = \left(\frac{y}{5} \frac{dy}{dx}\right)^{1/4}$
Substituting this into equation $(i)$:
$y^{2} = \left[\left(\frac{y}{5} \frac{dy}{dx}\right)^{1/4}\right]^{5}$
$y^{2} = \left(\frac{y}{5} \frac{dy}{dx}\right)^{5/4}$
Raising both sides to the power of $4$:
$(y^{2})^{4} = \left(\frac{y}{5} \frac{dy}{dx}\right)^{5}$
$y^{8} = \frac{y^{5}}{5^{5}} \left(\frac{dy}{dx}\right)^{5}$
$y^{8} = \frac{y^{5}}{3125} \left(\frac{dy}{dx}\right)^{5}$
Dividing by $y^{5}$ (assuming $y \neq 0$):
$y^{3} = \frac{1}{3125} \left(\frac{dy}{dx}\right)^{5}$
$3125 y^{3} = \left(\frac{dy}{dx}\right)^{5}$
$\left(\frac{dy}{dx}\right)^{5} - 3125 y^{3} = 0$

(D) Given $y = e^{ax}$.
Taking the natural logarithm on both sides,we get $\log y = \log(e^{ax})$.
Since $\log(e^{ax}) = ax$,we have $\log y = ax$ ...$(1)$.
Differentiating both sides with respect to $x$,we get $\frac{1}{y} \frac{dy}{dx} = a$.
Now,substitute the value of $a$ from the derivative into equation $(1)$:
$\log y = \left( \frac{1}{y} \frac{dy}{dx} \right) x$.
Multiplying both sides by $y$,we obtain $y \log y = x \frac{dy}{dx}$,which is $x \frac{dy}{dx} = y \log y$.

(A) Given the equation: $\tan^{-1} x + \tan^{-1} y = c$
Differentiating both sides with respect to $x$:
$\frac{d}{dx}(\tan^{-1} x) + \frac{d}{dx}(\tan^{-1} y) = \frac{d}{dx}(c)$
$\frac{1}{1+x^2} + \frac{1}{1+y^2} \cdot \frac{dy}{dx} = 0$
Rearranging the terms to solve for $\frac{dy}{dx}$:
$\frac{1}{1+y^2} \cdot \frac{dy}{dx} = -\frac{1}{1+x^2}$
$\frac{dy}{dx} = -\frac{1+y^2}{1+x^2}$
Thus,the correct option is $A$.

(C) Given the equation $y = mx + \frac{2}{m} \dots (1)$
Differentiating with respect to $x$,we get $\frac{dy}{dx} = m$.
Substituting $m = \frac{dy}{dx}$ into equation $(1)$:
$y = x \left(\frac{dy}{dx}\right) + \frac{2}{\frac{dy}{dx}}$
Multiplying both sides by $\frac{dy}{dx}$:
$y \left(\frac{dy}{dx}\right) = x \left(\frac{dy}{dx}\right)^{2} + 2$.

(A) Given $y=e^{x}(A \cos x+B \sin x)$ ... $(1)$
Differentiating with respect to $x$:
$\frac{dy}{dx} = e^{x}(A \cos x + B \sin x) + e^{x}(-A \sin x + B \cos x)$
$\frac{dy}{dx} = y + e^{x}(-A \sin x + B \cos x)$ ... $(2)$
Differentiating again with respect to $x$:
$\frac{d^{2}y}{dx^{2}} = \frac{dy}{dx} + e^{x}(-A \sin x + B \cos x) + e^{x}(-A \cos x - B \sin x)$
Substitute $e^{x}(-A \sin x + B \cos x) = \frac{dy}{dx} - y$ from $(2)$:
$\frac{d^{2}y}{dx^{2}} = \frac{dy}{dx} + (\frac{dy}{dx} - y) - e^{x}(A \cos x + B \sin x)$
Since $e^{x}(A \cos x + B \sin x) = y$:
$\frac{d^{2}y}{dx^{2}} = 2\frac{dy}{dx} - y - y$
$\frac{d^{2}y}{dx^{2}} - 2\frac{dy}{dx} + 2y = 0$

(B) Given the function $y = a(x-a)^{2} \quad ...(1)$
Differentiating with respect to $x$,we get $\frac{dy}{dx} = 2a(x-a) \quad ...(2)$
From $(2)$,$a = \frac{1}{2} \frac{dy/dx}{x-a}$. Substituting $a$ into $(1)$:
$y = \frac{1}{2} \frac{dy/dx}{x-a} (x-a)^2 = \frac{1}{2} (x-a) \frac{dy}{dx}$
Thus,$(x-a) = \frac{2y}{dy/dx}$.
Substituting $(x-a)$ back into $(2)$:
$\frac{dy}{dx} = 2a \left( \frac{2y}{dy/dx} \right) \implies a = \frac{(dy/dx)^2}{4y}$.
Now,substitute $a$ and $(x-a)$ into $y = a(x-a)^2$:
$y = \left( \frac{(dy/dx)^2}{4y} \right) \left( \frac{2y}{dy/dx} \right)^2$
$y = \frac{(dy/dx)^2}{4y} \cdot \frac{4y^2}{(dy/dx)^2} = y$.
To find the differential equation,we eliminate $a$ from $y = a(x-a)^2$ and $y' = 2a(x-a)$.
From $y' = 2a(x-a)$,we have $a = x - \frac{y'}{2a}$ is not correct.
Correct approach: $x-a = \frac{y'}{2a}$. Substitute into $y = a(x-a)^2$:
$y = a \left( \frac{y'}{2a} \right)^2 = \frac{(y')^2}{4a}$.
Since $a = x - (x-a) = x - \frac{y'}{2a}$,we have $2a^2 - 2ax + y' = 0$.
Solving for $a$ and substituting leads to the relation $8y^3 = (y')^2 (2x - y')$. Option $B$ is the correct form.

(B) Let $(h, 8)$ be the centre of the circle.
Since the circle touches the $X$-axis,its radius is $r = 8$.
The equation of the circle is $(x-h)^{2} + (y-8)^{2} = 8^{2} = 64$ ... $(1)$
Differentiating equation $(1)$ with respect to $x$,we get:
$2(x-h) + 2(y-8) \frac{d y}{d x} = 0$
$(x-h) = -(y-8) \frac{d y}{d x}$
Substituting the value of $(x-h)$ into equation $(1)$:
$[-(y-8) \frac{d y}{d x}]^{2} + (y-8)^{2} = 64$
$(y-8)^{2} (\frac{d y}{d x})^{2} + (y-8)^{2} = 64$
$(y-8)^{2} [1 + (\frac{d y}{d x})^{2}] = 64$
Thus,the correct option is $B$.

(B) The equation of a circle with center $(h, k)$ and radius $r$ is given by $(x - h)^2 + (y - k)^2 = r^2$.
Given the center is $A(-1, 2)$,the equation becomes $(x - (-1))^2 + (y - 2)^2 = r^2$.
Expanding this,we get $(x + 1)^2 + (y - 2)^2 = r^2$.
$x^2 + 2x + 1 + y^2 - 4y + 4 = r^2$.
$x^2 + y^2 + 2x - 4y + 5 - r^2 = 0$.
Let $c = 5 - r^2$,where $c$ is an arbitrary constant.
Thus,the general solution is $x^2 + y^2 + 2x - 4y + c = 0$.

(A) The general equation of a parabola with its axis along the $y$-axis is given by $(x - 0)^2 = 4a(y - k)$,where $a$ and $k$ are arbitrary constants.
This simplifies to $x^2 = 4ay - 4ak$.
Differentiating both sides with respect to $x$,we get:
$2x = 4a \frac{dy}{dx}$
$\Rightarrow x = 2a \frac{dy}{dx}$
$\Rightarrow \frac{1}{2a} = \frac{1}{x} \frac{dy}{dx}$
Differentiating again with respect to $x$ to eliminate the arbitrary constant $a$:
$\frac{d}{dx} \left( \frac{1}{x} \cdot \frac{dy}{dx} \right) = \frac{d}{dx} \left( \frac{1}{2a} \right)$
Using the product rule on the left side:
$\frac{1}{x} \cdot \frac{d^2y}{dx^2} + \frac{dy}{dx} \left( - \frac{1}{x^2} \right) = 0$
Multiplying the entire equation by $x^2$:
$x \frac{d^2y}{dx^2} - \frac{dy}{dx} = 0$

(B) Since the circle touches the $y$-axis at the origin,its center must lie on the $x$-axis. Let the center be $(a, 0)$ and the radius be $a$.
The equation of the circle is $(x-a)^2 + (y-0)^2 = a^2$.
Expanding this,we get $x^2 - 2ax + a^2 + y^2 = a^2$,which simplifies to $x^2 + y^2 - 2ax = 0$ ... $(i)$.
Differentiating equation $(i)$ with respect to $x$,we get $2x + 2y \frac{dy}{dx} - 2a = 0$,which implies $2a = 2x + 2y \frac{dy}{dx}$ ... $(ii)$.
Substituting the value of $2a$ from $(ii)$ into $(i)$,we get $x^2 + y^2 - x(2x + 2y \frac{dy}{dx}) = 0$.
This simplifies to $x^2 + y^2 - 2x^2 - 2xy \frac{dy}{dx} = 0$.
Thus,the differential equation is $y^2 - x^2 - 2xy \frac{dy}{dx} = 0$,or $x^2 - y^2 + 2xy \frac{dy}{dx} = 0$.

(B) Given equation is $(x-h)^{2}+(y-k)^{2}=a^{2}$ ... $(i)$
Differentiating with respect to $x$:
$2(x-h) + 2(y-k) \frac{dy}{dx} = 0$
$(x-h) + (y-k) \frac{dy}{dx} = 0$ ... (ii)
Differentiating again with respect to $x$:
$1 + \left(\frac{dy}{dx}\right)^{2} + (y-k) \frac{d^{2}y}{dx^{2}} = 0$
$(y-k) = -\frac{1 + (dy/dx)^{2}}{d^{2}y/dx^{2}}$ ... (iii)
From (ii),$(x-h) = -(y-k) \frac{dy}{dx}$. Substituting (iii) into this:
$(x-h) = \frac{[1 + (dy/dx)^{2}]}{d^{2}y/dx^{2}} \cdot \frac{dy}{dx}$ ... (iv)
Substituting (iii) and (iv) into $(i)$:
$\left[ \frac{[1 + (dy/dx)^{2}] \cdot (dy/dx)}{d^{2}y/dx^{2}} \right]^{2} + \left[ -\frac{1 + (dy/dx)^{2}}{d^{2}y/dx^{2}} \right]^{2} = a^{2}$
$\frac{[1 + (dy/dx)^{2}]^{2}}{ (d^{2}y/dx^{2})^{2} } \left[ (dy/dx)^{2} + 1 \right] = a^{2}$
$\left[1 + \left(\frac{dy}{dx}\right)^{2}\right]^{3} = a^{2} \left(\frac{d^{2}y}{dx^{2}}\right)^{2}$

(A) The general equation of a circle passing through the origin with its center on the $y$-axis is given by $x^{2} + (y-a)^{2} = a^{2}$,where $a$ is an arbitrary constant.
Expanding this,we get $x^{2} + y^{2} - 2ay + a^{2} = a^{2}$,which simplifies to $x^{2} + y^{2} - 2ay = 0$.
To eliminate the arbitrary constant $a$,we differentiate with respect to $x$:
$2x + 2y \frac{dy}{dx} - 2a \frac{dy}{dx} = 0$.
Dividing by $2$,we get $x + y \frac{dy}{dx} - a \frac{dy}{dx} = 0$,which implies $a = \frac{x + y \frac{dy}{dx}}{\frac{dy}{dx}} = x \frac{dx}{dy} + y$.
Substituting $a$ back into the original equation $x^{2} + y^{2} = 2ay$:
$x^{2} + y^{2} = 2(x \frac{dx}{dy} + y)y = 2xy \frac{dx}{dy} + 2y^{2}$.
Rearranging,$x^{2} - y^{2} = 2xy \frac{dx}{dy}$.
Since $\frac{dx}{dy} = \frac{1}{\frac{dy}{dx}}$,we have $x^{2} - y^{2} = \frac{2xy}{\frac{dy}{dx}}$.
Thus,$(x^{2} - y^{2}) \frac{dy}{dx} = 2xy$,or $(x^{2} - y^{2}) \frac{dy}{dx} - 2xy = 0$.

(D) The general equation of a family of circles with center on the $x$-axis and passing through the origin is given by $(x-a)^2 + y^2 = a^2$,which simplifies to $x^2 - 2ax + a^2 + y^2 = a^2$,or $x^2 + y^2 - 2ax = 0$.
Differentiating with respect to $x$,we get $2x + 2y \frac{dy}{dx} - 2a = 0$,which implies $a = x + y \frac{dy}{dx}$.
Substituting the value of $a$ back into the original equation: $x^2 + y^2 - 2x(x + y \frac{dy}{dx}) = 0$.
$x^2 + y^2 - 2x^2 - 2xy \frac{dy}{dx} = 0 \Rightarrow y^2 - x^2 - 2xy \frac{dy}{dx} = 0$.
However,for a general family of circles with center $(a, 0)$ and radius $r$,the equation is $(x-a)^2 + y^2 = r^2$. Differentiating twice eliminates two constants $a$ and $r$. The standard form for circles with center on the $x$-axis is $y \frac{d^2y}{dx^2} + (\frac{dy}{dx})^2 + 1 = 0$.

(B) Given the family of lines: $y = mx + \frac{4}{m}$ $(i)$
Differentiating both sides with respect to $x$:
$\frac{dy}{dx} = m$
Substitute $m = \frac{dy}{dx}$ into equation $(i)$:
$y = x\left(\frac{dy}{dx}\right) + \frac{4}{\frac{dy}{dx}}$
Multiply the entire equation by $\frac{dy}{dx}$ to clear the denominator:
$y\left(\frac{dy}{dx}\right) = x\left(\frac{dy}{dx}\right)^{2} + 4$
Rearranging the terms to form the differential equation:
$x\left(\frac{dy}{dx}\right)^{2} - y\frac{dy}{dx} + 4 = 0$
Thus,the required differential equation is $x\left(\frac{dy}{dx}\right)^{2} - y\frac{dy}{dx} + 4 = 0$.

(D) The general equation of a parabola with its axis parallel to the $y$-axis is given by $(x-h)^{2} = 4a(y-k)$,where $h, k,$ and $a$ are arbitrary constants.
Since there are $3$ arbitrary constants,we need to differentiate the equation $3$ times.
Let $(x-h)^{2} = A(y-k)$,where $A = 4a$.
Differentiating with respect to $x$: $2(x-h) = A y_{1}$.
Differentiating again with respect to $x$: $2 = A y_{2}$.
Differentiating a third time with respect to $x$: $0 = A y_{3}$.
Since $A = 4a \neq 0$,we must have $y_{3} = 0$.
None of the given options match $y_{3} = 0$.

(B) Given the equation $y=c^{2}+\frac{c}{x}$.
Differentiating both sides with respect to $x$,we get:
$\frac{d y}{d x} = -\frac{c}{x^{2}}$
From this,we can express $c$ in terms of $x$ and $\frac{d y}{d x}$:
$c = -x^{2} \frac{d y}{d x}$
Now,substitute this value of $c$ back into the original equation:
$y = (-x^{2} \frac{d y}{d x})^{2} + \frac{-x^{2} \frac{d y}{d x}}{x}$
$y = x^{4} (\frac{d y}{d x})^{2} - x \frac{d y}{d x}$
Rearranging the terms to form the differential equation:
$x^{4} (\frac{d y}{d x})^{2} - x \frac{d y}{d x} - y = 0$
Thus,the correct option is $B$.

(C) Given $y=e^x(a+bx+x^2)$.
First,differentiate with respect to $x$:
$\frac{dy}{dx} = e^x(a+bx+x^2) + e^x(b+2x)$
$\frac{dy}{dx} = y + e^x(b+2x) \quad ...(i)$
Now,differentiate again with respect to $x$:
$\frac{d^2 y}{dx^2} = \frac{dy}{dx} + e^x(b+2x) + e^x(2)$
Substitute $e^x(b+2x) = \frac{dy}{dx} - y$ from equation $(i)$:
$\frac{d^2 y}{dx^2} = \frac{dy}{dx} + (\frac{dy}{dx} - y) + 2e^x$
$\frac{d^2 y}{dx^2} = 2\frac{dy}{dx} - y + 2e^x$
Rearranging the terms,we get:
$\frac{d^2 y}{dx^2} - 2\frac{dy}{dx} - 2e^x + y = 0$.

(C) The general equation of a circle passing through the origin with its center on the $X$-axis is given by $(x-a)^2 + (y-0)^2 = a^2$,where $(a, 0)$ is the center and $a$ is the radius.
Expanding this,we get $x^2 - 2ax + a^2 + y^2 = a^2$,which simplifies to $x^2 + y^2 - 2ax = 0 ... (i)$.
Differentiating equation $(i)$ with respect to $x$,we get $2x + 2y \frac{dy}{dx} - 2a = 0$,which implies $2a = 2x + 2y \frac{dy}{dx} ... (ii)$.
Substituting the value of $2a$ from equation $(ii)$ into equation $(i)$,we get $x^2 + y^2 - x(2x + 2y \frac{dy}{dx}) = 0$.
Simplifying this,$x^2 + y^2 - 2x^2 - 2xy \frac{dy}{dx} = 0$.
Thus,$y^2 - x^2 - 2xy \frac{dy}{dx} = 0$,which gives $y^2 = x^2 + 2xy \frac{dy}{dx}$.

(D) The given line is $5x + 2y + 7 = 0$.
Its slope is $m_1 = -\frac{5}{2}$.
Any line perpendicular to this line will have a slope $m_2 = -\frac{1}{m_1} = \frac{2}{5}$.
The equation of a family of lines with slope $\frac{2}{5}$ is $y = \frac{2}{5}x + c$,which can be written as $2x - 5y + 5c = 0$.
Let $5c = k$,so the equation is $2x - 5y + k = 0$.
Differentiating both sides with respect to $x$,we get $2 - 5\frac{dy}{dx} = 0$.
Multiplying by $dx$,we get $2dx - 5dy = 0$,or $5dy - 2dx = 0$.

(D) Given the general solution: $A x^2+B y^2=1$
Differentiating with respect to $x$:
$2 A x + 2 B y \frac{d y}{d x} = 0 \implies A x + B y \frac{d y}{d x} = 0 \dots (i)$
Differentiating again with respect to $x$:
$A + B \left[ \left( \frac{d y}{d x} \right)^2 + y \frac{d^2 y}{d x^2} \right] = 0 \dots (ii)$
From $(i)$,$A = -\frac{B y}{x} \frac{d y}{d x}$. Substituting this into $(ii)$:
$-\frac{B y}{x} \frac{d y}{d x} + B \left( \frac{d y}{d x} \right)^2 + B y \frac{d^2 y}{d x^2} = 0$
Dividing by $B$ (assuming $B \neq 0$):
$-\frac{y}{x} \frac{d y}{d x} + \left( \frac{d y}{d x} \right)^2 + y \frac{d^2 y}{d x^2} = 0$
Multiplying by $x$:
$-y \frac{d y}{d x} + x \left( \frac{d y}{d x} \right)^2 + x y \frac{d^2 y}{d x^2} = 0$
Thus,$x y \frac{d^2 y}{d x^2} + x \left( \frac{d y}{d x} \right)^2 - y \frac{d y}{d x} = 0$.

(B) Given the general solution: $y = c_{1} \cos(ax) + c_{2} \sin(ax)$.
First,differentiate with respect to $x$: $\frac{dy}{dx} = -a c_{1} \sin(ax) + a c_{2} \cos(ax)$.
Next,differentiate again with respect to $x$: $\frac{d^{2}y}{dx^{2}} = -a^{2} c_{1} \cos(ax) - a^{2} c_{2} \sin(ax)$.
Factor out $-a^{2}$: $\frac{d^{2}y}{dx^{2}} = -a^{2} (c_{1} \cos(ax) + c_{2} \sin(ax))$.
Since $y = c_{1} \cos(ax) + c_{2} \sin(ax)$,we substitute $y$ into the equation: $\frac{d^{2}y}{dx^{2}} = -a^{2} y$.
Rearranging gives: $\frac{d^{2}y}{dx^{2}} + a^{2} y = 0$.

(B) Given the function $y = e^{4x} + 2e^{-x}$.
First,find the first derivative: $\frac{dy}{dx} = 4e^{4x} - 2e^{-x}$.
Next,find the second derivative: $\frac{d^2y}{dx^2} = 16e^{4x} + 2e^{-x}$.
Substitute these into the given equation $\frac{d^2y}{dx^2} + A\frac{dy}{dx} + By = 0$:
$(16e^{4x} + 2e^{-x}) + A(4e^{4x} - 2e^{-x}) + B(e^{4x} + 2e^{-x}) = 0$.
Group the terms by $e^{4x}$ and $e^{-x}$:
$(16 + 4A + B)e^{4x} + (2 - 2A + 2B)e^{-x} = 0$.
For this to hold for all $x$,the coefficients must be zero:
$16 + 4A + B = 0$ (Equation $1$)
$2 - 2A + 2B = 0 \Rightarrow 1 - A + B = 0 \Rightarrow B = A - 1$ (Equation $2$)
Substitute $B = A - 1$ into Equation $1$:
$16 + 4A + (A - 1) = 0 \Rightarrow 5A + 15 = 0 \Rightarrow A = -3$.
Then $B = -3 - 1 = -4$.
Thus,$A = -3$ and $B = -4$.

(A) The general equation of a circle passing through the origin with its center on the $x$-axis is given by $x^{2} + y^{2} - 2hx = 0$,where $h$ is a parameter.
From this equation,we have $2h = \frac{x^{2} + y^{2}}{x}$.
Differentiating the equation $x^{2} + y^{2} - 2hx = 0$ with respect to $x$,we get:
$2x + 2y \frac{dy}{dx} - 2h = 0$.
Substituting the value of $2h$ from the first equation:
$2x + 2y \frac{dy}{dx} - \left( \frac{x^{2} + y^{2}}{x} \right) = 0$.
Multiplying the entire equation by $x$:
$2x^{2} + 2xy \frac{dy}{dx} - x^{2} - y^{2} = 0$.
Simplifying the expression:
$x^{2} - y^{2} + 2xy \frac{dy}{dx} = 0$,which can be rewritten as $y^{2} = x^{2} + 2xy \frac{dy}{dx}$.

(A) The equation of a straight line with slope $m$ and $y$-intercept $c$ is given by $y = mx + c$.
Given that the slope is equal to the $y$-intercept,we have $m = c$.
Substituting this into the line equation,we get $y = cx + c = c(x+1)$.
Differentiating both sides with respect to $x$,we get $\frac{dy}{dx} = c$.
Substituting the value of $c$ from the first equation into the derivative,we have $\frac{dy}{dx} = \frac{y}{x+1}$.
Rearranging the terms,we get $(x+1) \frac{dy}{dx} - y = 0$.

(A) Given the equation: $c_{1} y = (c_{2} + c_{3}) e^{x + c_{4}}$
We can simplify this by grouping the constants: $y = \left( \frac{c_{2} + c_{3}}{c_{1}} \right) e^{c_{4}} \cdot e^{x}$
Let $C = \left( \frac{c_{2} + c_{3}}{c_{1}} \right) e^{c_{4}}$,where $C$ is a single arbitrary constant.
Thus,the equation becomes $y = C e^{x}$.
Since there is only one independent arbitrary constant $C$,the order of the differential equation obtained by eliminating it is $1$.
Differentiating $y = C e^{x}$ with respect to $x$,we get $\frac{dy}{dx} = C e^{x}$.
Substituting $y = C e^{x}$ into the derivative,we get $\frac{dy}{dx} = y$.
This is a first-order differential equation,so the order is $1$.

(B) Given the family of curves $y=ax+\frac{1}{a}$.
Differentiating with respect to $x$,we get $\frac{dy}{dx}=a$.
Substituting $a=\frac{dy}{dx}$ into the original equation,we get $y=x\left(\frac{dy}{dx}\right)+\frac{1}{\frac{dy}{dx}}$.
Multiplying by $\frac{dy}{dx}$,we get $y\left(\frac{dy}{dx}\right)=x\left(\frac{dy}{dx}\right)^2+1$,which is $x\left(\frac{dy}{dx}\right)^2-y\left(\frac{dy}{dx}\right)+1=0$.
The order $m$ of this differential equation is $1$ and the degree $r$ is $2$.
Thus,$r+m = 2+1 = 3$.
The given differential equation is $\frac{dy}{dx}=\frac{y}{2x}$.
Separating variables,we get $\frac{dy}{y}=\frac{dx}{2x}$.
Integrating both sides,$\ln|y|=\frac{1}{2}\ln|x|+C$,which implies $y^2=kx$.
Using the condition $y(1)=\sqrt{r+m}=\sqrt{3}$,we get $(\sqrt{3})^2=k(1)$,so $k=3$.
Therefore,the solution is $y^2=3x$.

(A) The order of a differential equation is equal to the number of arbitrary constants present in the general equation of the family of curves.
For option $A$,the equation of a circle passing through the origin is $x^2 + y^2 + 2gx + 2fy = 0$.
Here,$g$ and $f$ are two independent arbitrary constants.
Since there are $2$ arbitrary constants,the differential equation formed will be of order $2$.
For option $B$,the equation is $y^2 = 4a(x-h)$,which involves two constants,but the condition of passing through the origin and focus on the $x$-axis restricts it.
For option $C$,the equation is $y = mx$,which has only $1$ arbitrary constant.
For option $D$,the equation $x^2 - y^2 = k^2$ has only $1$ arbitrary constant $k$.
Therefore,the correct option is $A$.

(C) Given the equation: $y=c_1 e^x+c_2 e^{\log _{e} x}+c_3 \sin ^2 x-c_4\left(\cos ^2 x-1\right)$
Using the property $e^{\log _{e} x} = x$ and $\cos^2 x - 1 = -\sin^2 x$,we simplify:
$y = c_1 e^x + c_2 x + c_3 \sin^2 x - c_4(-\sin^2 x)$
$y = c_1 e^x + c_2 x + (c_3 + c_4) \sin^2 x$
Let $C = c_3 + c_4$. Then the equation becomes:
$y = c_1 e^x + c_2 x + C \sin^2 x$
This equation contains $3$ independent arbitrary constants $(c_1, c_2, C)$.
The order of a differential equation is equal to the number of independent arbitrary constants present in its general solution.
Since there are $3$ independent constants,the order of the differential equation is $3$.

(B) The given solution is $y = a e^{2x} + b x e^{2x} + C$ ...$(i)$
Here,$a$ and $b$ are arbitrary constants,while $C$ is a fixed constant.
The order of a differential equation is equal to the number of arbitrary constants present in its general solution.
Since there are two arbitrary constants ($a$ and $b$),the order of the differential equation is $2$.
To verify,we differentiate $y$ with respect to $x$:
$y_1 = 2a e^{2x} + b(e^{2x} + 2x e^{2x}) = (2a + b)e^{2x} + 2bx e^{2x}$ ...(ii)
$y_2 = 2(2a + b)e^{2x} + 2b(e^{2x} + 2x e^{2x}) = (4a + 4b)e^{2x} + 4bx e^{2x}$ ...(iii)
By eliminating $a$ and $b$ from these equations,we obtain a second-order differential equation.

(B) The equation of a straight line at a constant distance $P$ from the origin is $x \cos \alpha + y \sin \alpha = P$.
Differentiating with respect to $x$,we get $\cos \alpha + \frac{dy}{dx} \sin \alpha = 0$,which implies $\frac{dy}{dx} = -\cot \alpha$.
From this,we have $\cos \alpha = -\frac{dy/dx}{\sqrt{1+(dy/dx)^2}}$ and $\sin \alpha = \frac{1}{\sqrt{1+(dy/dx)^2}}$.
Substituting these into the original equation: $x \left( -\frac{dy/dx}{\sqrt{1+(dy/dx)^2}} \right) + y \left( \frac{1}{\sqrt{1+(dy/dx)^2}} \right) = P$.
This simplifies to $y - x \frac{dy}{dx} = P \sqrt{1 + (\frac{dy}{dx})^2}$.
Squaring both sides,we get $(y - x \frac{dy}{dx})^2 = P^2 (1 + (\frac{dy}{dx})^2)$.
The highest derivative present is $\frac{dy}{dx}$,so the order $l = 1$.
The highest power of the derivative $\frac{dy}{dx}$ after clearing radicals is $2$,so the degree $m = 2$.
Thus,$l m^2 + l^2 m = (1)(2^2) + (1^2)(2) = 4 + 2 = 6$.