Formation of differential equations Questions in English

(D) The equation of the family of parabolas with focus at the origin $(0,0)$ and the $X$-axis as the axis of symmetry is given by $y^2 = 4a(x+a)$,where $a$ is an arbitrary parameter.
Expanding this,we have $y^2 = 4ax + 4a^2$.
Differentiating with respect to $x$,we get:
$2y \frac{dy}{dx} = 4a \Rightarrow a = \frac{y}{2} \frac{dy}{dx}$.
Substituting the value of $a$ back into the original equation $y^2 = 4a(x+a)$:
$y^2 = 4 \left( \frac{y}{2} \frac{dy}{dx} \right) \left( x + \frac{y}{2} \frac{dy}{dx} \right)$
$y^2 = 2y \frac{dy}{dx} \left( x + \frac{y}{2} \frac{dy}{dx} \right)$
Dividing by $y$ (assuming $y \neq 0$):
$y = 2x \frac{dy}{dx} + y \left( \frac{dy}{dx} \right)^2$
$y \left( \frac{dy}{dx} \right)^2 + 2x \frac{dy}{dx} - y = 0$.
Since the equation involves only the first derivative $\frac{dy}{dx}$,the order of the differential equation is $1$.

(B) The equation of a circle with center $(h, k)$ and radius $r = 5$ is $(x - h)^2 + (y - k)^2 = 25$.
Since there are two arbitrary constants $h$ and $k$,we differentiate twice.
Differentiating with respect to $x$: $2(x - h) + 2(y - k)y' = 0$,which gives $(x - h) = -(y - k)y'$.
Differentiating again: $1 = -[(y')^2 + (y - k)y'']$,which gives $(y - k) = -\frac{1 + (y')^2}{y''}$.
Substituting $(y - k)$ back into the first derivative equation: $(x - h) = \left(\frac{1 + (y')^2}{y''}\right)y'$.
Substituting these into the original circle equation: $\left(\frac{1 + (y')^2}{y''}\right)^2 (y')^2 + \left(\frac{1 + (y')^2}{y''}\right)^2 = 25$.
Simplifying,we get $(1 + (y')^2)^2 \left(\frac{(y')^2 + 1}{(y'')^2}\right) = 25$,which leads to $(1 + (y')^2)^3 = 25(y'')^2$.
The order $m$ is the highest derivative present,which is $y''$,so $m = 2$.
The degree $l$ is the power of the highest derivative,which is $2$,so $l = 2$.
Thus,$2l + 3m = 2(2) + 3(2) = 4 + 6 = 10$.

(A) The equation of the family of all concentric circles centered at $(h, k)$ is given by $(x-h)^2 + (y-k)^2 = r^2$.
Here,$(h, k)$ are fixed constants (the center),and $r$ is the radius,which is the only arbitrary parameter.
Since there is only one arbitrary parameter $(r)$,the order of the differential equation is equal to the number of arbitrary parameters.
Therefore,the order of the differential equation is $1$.

(A) The equation of a parabola with axis parallel to the $y$-axis and axis of symmetry $x=1$ is given by $(x-1)^2 = 4a(y-k)$,where $a$ and $k$ are arbitrary constants.
Alternatively,we can write this as $y = A(x-1)^2 + B$,where $A$ and $B$ are arbitrary constants.
Differentiating with respect to $x$ once: $\frac{dy}{dx} = 2A(x-1)$.
Differentiating again with respect to $x$: $\frac{d^2y}{dx^2} = 2A$.
From the first derivative,$A = \frac{1}{2(x-1)} \frac{dy}{dx}$.
Substituting this into the second derivative: $\frac{d^2y}{dx^2} = 2 \left( \frac{1}{2(x-1)} \frac{dy}{dx} \right) = \frac{1}{x-1} \frac{dy}{dx}$.
Rearranging gives: $(x-1) \frac{d^2y}{dx^2} - \frac{dy}{dx} = 0$.

(A) The equation of a circle passing through the origin with its center on the $X$-axis is given by $(x-a)^2 + y^2 = a^2$,where $a$ is a parameter.
Expanding this,we get $x^2 - 2ax + a^2 + y^2 = a^2$,which simplifies to $x^2 + y^2 = 2ax$.
Differentiating both sides with respect to $x$,we get $2x + 2y \frac{dy}{dx} = 2a$.
Substituting the value of $a = x + y \frac{dy}{dx}$ into the equation $x^2 + y^2 = 2ax$,we get $x^2 + y^2 = 2x(x + y \frac{dy}{dx})$.
This simplifies to $x^2 + y^2 = 2x^2 + 2xy \frac{dy}{dx}$.
Rearranging the terms,we get $y^2 - x^2 = 2xy \frac{dy}{dx}$,which can be written as $(x^2 - y^2) dx + 2xy dy = 0$.

(A) Given the equation $y^2 = 4a(x+a)$.
Differentiating both sides with respect to $x$,we get:
$2y \frac{dy}{dx} = 4a$
$y \frac{dy}{dx} = 2a$
So,$a = \frac{y}{2} \frac{dy}{dx}$.
Substitute the value of $a$ back into the original equation:
$y^2 = 4 \left( \frac{y}{2} \frac{dy}{dx} \right) \left( x + \frac{y}{2} \frac{dy}{dx} \right)$
$y^2 = 2y \frac{dy}{dx} \left( x + \frac{y}{2} \frac{dy}{dx} \right)$
Dividing by $y$ (assuming $y \neq 0$):
$y = 2x \frac{dy}{dx} + y \left( \frac{dy}{dx} \right)^2$.

(B) Given the family of curves $y = \tan(ax + b)$.
First,differentiate with respect to $x$:
$y_1 = \frac{dy}{dx} = \sec^2(ax + b) \cdot a$
Since $\sec^2 \theta = 1 + \tan^2 \theta$,we have $y_1 = a(1 + y^2)$.
Thus,$a = \frac{y_1}{1 + y^2}$.
Now,differentiate $y_1 = a(1 + y^2)$ with respect to $x$ again:
$y_2 = a(2y y_1) \cdot y_1 + a(1 + y^2) \cdot 0$ (Wait,differentiating $y_1 = a(1 + y^2)$ gives $y_2 = a(2y y_1)$).
Substitute $a = \frac{y_1}{1 + y^2}$ into the equation $y_2 = 2ay y_1$:
$y_2 = 2 \left( \frac{y_1}{1 + y^2} \right) y y_1$
$y_2 = \frac{2y y_1^2}{1 + y^2}$
Rearranging gives: $(1 + y^2) y_2 - 2y y_1^2 = 0$.

(A) Given the equation $Ax^3+Bxy=4$.
Rearranging for $y$,we get $Bxy = 4-Ax^3$,so $y = \frac{4}{Bx} - \frac{Ax^2}{B}$.
Let $C_1 = \frac{4}{B}$ and $C_2 = -\frac{A}{B}$. Then $y = C_1 x^{-1} + C_2 x^2$.
Now,differentiate with respect to $x$: $\frac{dy}{dx} = -C_1 x^{-2} + 2C_2 x$.
Differentiate again: $\frac{d^2y}{dx^2} = 2C_1 x^{-3} + 2C_2$.
Substitute these into the differential equation $F(x) \frac{d^2y}{dx^2} + G(x) \frac{dy}{dx} - 2y = 0$:
$F(x)(2C_1 x^{-3} + 2C_2) + G(x)(-C_1 x^{-2} + 2C_2 x) - 2(C_1 x^{-1} + C_2 x^2) = 0$.
Grouping terms by $C_1$ and $C_2$:
$C_1(2F(x)x^{-3} - G(x)x^{-2} - 2x^{-1}) + C_2(2F(x) + 2xG(x) - 2x^2) = 0$.
For this to hold for arbitrary constants $C_1, C_2$,the coefficients must be zero:
$2F(x)x^{-3} - G(x)x^{-2} - 2x^{-1} = 0 \implies 2F(x) - xG(x) - 2x^2 = 0$.
$2F(x) + 2xG(x) - 2x^2 = 0 \implies F(x) + xG(x) - x^2 = 0$.
At $x=1$:
$2F(1) - G(1) - 2 = 0$ (Eq $1$)
$F(1) + G(1) - 1 = 0$ (Eq $2$)
Adding Eq $1$ and Eq $2$: $3F(1) - 3 = 0 \implies F(1) = 1$.
Substituting $F(1)=1$ into Eq $2$: $1 + G(1) - 1 = 0 \implies G(1) = 0$.
Therefore,$F(1) + G(1) = 1 + 0 = 1$.

(C) The equation of a family of circles with center $(0, b)$ on the $Y$-axis and radius $a$ is given by $x^2 + (y - b)^2 = a^2$.
Differentiating with respect to $x$:
$2x + 2(y - b)y_1 = 0$
$x + (y - b)y_1 = 0$ ... $(i)$
Differentiating again with respect to $x$:
$1 + (y - b)y_2 + y_1^2 = 0$
$(y - b)y_2 = -(1 + y_1^2)$
$y - b = -\frac{1 + y_1^2}{y_2}$ ... $(ii)$
Substituting $(ii)$ into $(i)$:
$x + \left(-\frac{1 + y_1^2}{y_2}\right)y_1 = 0$
$x - \frac{y_1(1 + y_1^2)}{y_2} = 0$
$xy_2 = y_1(1 + y_1^2)$

(D) Given equation: $y = a e^{2x} + b x e^{2x} = e^{2x}(a + bx)$.
First derivative with respect to $x$: $\frac{dy}{dx} = 2e^{2x}(a + bx) + b e^{2x} = 2y + b e^{2x}$.
Rearranging gives $b e^{2x} = \frac{dy}{dx} - 2y$ ... $(i)$.
Second derivative with respect to $x$: $\frac{d^2y}{dx^2} = 2\frac{dy}{dx} + 2b e^{2x}$ ... $(ii)$.
Substitute $(i)$ into $(ii)$:
$\frac{d^2y}{dx^2} = 2\frac{dy}{dx} + 2(\frac{dy}{dx} - 2y)$.
$\frac{d^2y}{dx^2} = 2\frac{dy}{dx} + 2\frac{dy}{dx} - 4y$.
$\frac{d^2y}{dx^2} - 4\frac{dy}{dx} + 4y = 0$.
Thus,the differential equation is $y^{\prime \prime} - 4y^{\prime} + 4y = 0$.

(B) Given the equation: $y = A \cos 3x + B \sin 3x$
Differentiating with respect to $x$:
$\frac{dy}{dx} = -3A \sin 3x + 3B \cos 3x$
Differentiating again with respect to $x$:
$\frac{d^2y}{dx^2} = -9A \cos 3x - 9B \sin 3x$
Factor out $-9$:
$\frac{d^2y}{dx^2} = -9(A \cos 3x + B \sin 3x)$
Since $y = A \cos 3x + B \sin 3x$,we substitute $y$ into the equation:
$\frac{d^2y}{dx^2} = -9y$
Rearranging gives:
$\frac{d^2y}{dx^2} + 9y = 0$

(A) The equation of the family of hyperbolas with centres at the origin and axes along the coordinate axes is given by $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ (or $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ for the general conic form). Differentiating with respect to $x$: $\frac{2x}{a^2} - \frac{2y}{b^2} y_1 = 0 \Rightarrow \frac{x}{a^2} = \frac{y y_1}{b^2} \Rightarrow \frac{y y_1}{x} = \frac{b^2}{a^2} = k$ (constant).
Differentiating again with respect to $x$: $\frac{d}{dx} \left( \frac{y y_1}{x} \right) = 0$.
Using the quotient rule: $\frac{x(y y_2 + y_1^2) - y y_1}{x^2} = 0$.
Since $x \neq 0$,we have $x y y_2 + x y_1^2 - y y_1 = 0$.

(D) Given the equation: $ax + by = 1$.
Differentiating with respect to $x$:
$a + b \frac{dy}{dx} = 0$
$\frac{dy}{dx} = -\frac{a}{b}$
Differentiating again with respect to $x$:
$\frac{d^2y}{dx^2} = 0$.
Thus,the required differential equation is $\frac{d^2y}{dx^2} = 0$.

(A) Given the equation: $y = a e^x + b e^{-x} + c \cos x + d \sin x$
Taking the first derivative: $y' = a e^x - b e^{-x} - c \sin x + d \cos x$
Taking the second derivative: $y'' = a e^x + b e^{-x} - c \cos x - d \sin x$
Taking the third derivative: $y''' = a e^x - b e^{-x} + c \sin x - d \cos x$
Taking the fourth derivative: $y^{(4)} = a e^x + b e^{-x} + c \cos x + d \sin x$
Comparing the fourth derivative with the original equation,we get $y^{(4)} = y$,which can be written as $y^{(4)} - y = 0$.

(D) Given the general solution $y=(a+b) e^{cx+d}$.
Let $A = (a+b)e^d$. Then the equation simplifies to $y = A e^{cx}$.
Here,$A$ and $c$ are the only independent arbitrary constants.
Differentiating with respect to $x$:
$y^{(1)} = A c e^{cx} = c y$.
Differentiating again with respect to $x$:
$y^{(2)} = c y^{(1)}$.
From the first derivative,we have $c = \frac{y^{(1)}}{y}$.
Substituting this into the second derivative equation:
$y^{(2)} = \left(\frac{y^{(1)}}{y}\right) y^{(1)}$.
$y y^{(2)} = (y^{(1)})^2$.
$y y^{(2)} - (y^{(1)})^2 = 0$.

(B) The given solution is $y=e^{2 x}(c \cosh \sqrt{2} x+d \sinh \sqrt{2} x)$.
This is of the form $y=e^{\alpha x}(c \cosh \beta x+d \sinh \beta x)$,which corresponds to the auxiliary equation roots $m = \alpha \pm \beta$.
Here,$\alpha = 2$ and $\beta = \sqrt{2}$.
The roots are $m = 2 \pm \sqrt{2}$.
The characteristic equation is $(m - (2 + \sqrt{2}))(m - (2 - \sqrt{2})) = 0$.
$(m - 2 - \sqrt{2})(m - 2 + \sqrt{2}) = 0$.
$(m - 2)^2 - (\sqrt{2})^2 = 0$.
$m^2 - 4m + 4 - 2 = 0$.
$m^2 - 4m + 2 = 0$.
Replacing $m^2$ with $y^{\prime \prime}$ and $m$ with $y^{\prime}$,we get the differential equation $y^{\prime \prime} - 4y^{\prime} + 2y = 0$.

(D) Given the equation $y=A e^x+B e^{-2 x}$.
First,differentiate with respect to $x$:
$\frac{d y}{d x} = A e^x - 2B e^{-2x}$.
Second,differentiate again with respect to $x$:
$\frac{d^2 y}{d x^2} = A e^x + 4B e^{-2x}$.
Now,we eliminate the constants $A$ and $B$. From the first derivative,$A e^x = \frac{d y}{d x} + 2B e^{-2x}$.
Substitute this into the second derivative equation:
$\frac{d^2 y}{d x^2} = (\frac{d y}{d x} + 2B e^{-2x}) + 4B e^{-2x} = \frac{d y}{d x} + 6B e^{-2x}$.
Alternatively,consider the characteristic equation for the roots $m_1 = 1$ and $m_2 = -2$. The differential equation is $(D-1)(D+2)y = 0$,where $D = \frac{d}{dx}$.
Expanding this: $(D^2 + 2D - D - 2)y = 0$,which simplifies to $\frac{d^2 y}{d x^2} + \frac{d y}{d x} - 2y = 0$.

(C) The given equation is $y=(a+b) \sin (x+c)-d e^{x+e+f}$.
Let $A = (a+b)$ and $B = d e^{e+f}$. Then the equation becomes $y = A \sin(x+c) - B e^x$.
There are $3$ independent arbitrary constants: $A$,$c$,and $B$.
The order of the differential equation obtained by eliminating $n$ independent arbitrary constants is $n$.
Since there are $3$ independent arbitrary constants,the order of the differential equation is $3$.

(A) The equation of the family of circles passing through $(0,0)$ and having its center on the $X$-axis is given by $(x-r)^2 + y^2 = r^2$,where $r$ is the radius of the circle and $(r, 0)$ is the center.
Expanding the equation: $x^2 - 2xr + r^2 + y^2 = r^2$,which simplifies to $x^2 + y^2 - 2xr = 0$ or $r = \frac{x^2 + y^2}{2x}$.
Differentiating both sides with respect to $x$:
$2x + 2y \frac{dy}{dx} - 2r = 0$
$x + y \frac{dy}{dx} = r$
Substituting the value of $r$ from the original equation:
$x + y \frac{dy}{dx} = \frac{x^2 + y^2}{2x}$
$2x^2 + 2xy \frac{dy}{dx} = x^2 + y^2$
$2xy \frac{dy}{dx} + x^2 - y^2 = 0$

(C) Given $y = (\sin^{-1} x)^2 + A \cos^{-1} x + B$ ... $(i)$
Differentiating with respect to $x$:
$y' = 2(\sin^{-1} x) \cdot \frac{1}{\sqrt{1 - x^2}} - \frac{A}{\sqrt{1 - x^2}}$
$y' \sqrt{1 - x^2} = 2 \sin^{-1} x - A$ ... $(ii)$
Differentiating again with respect to $x$:
$y'' \sqrt{1 - x^2} + y' \cdot \frac{-2x}{2\sqrt{1 - x^2}} = \frac{2}{\sqrt{1 - x^2}}$
Multiply throughout by $\sqrt{1 - x^2}$:
$y'' (1 - x^2) - x y' = 2$
Comparing this with the given equation $(a - x^2) y'' - x y' = b$,we get $a = 1$ and $b = 2$.
Therefore,$\frac{b + a}{b - a} = \frac{2 + 1}{2 - 1} = \frac{3}{1} = 3$.

(D) Given equation is $y = ax + b$.
Differentiating with respect to $x$,we get $\frac{dy}{dx} = a$.
Differentiating again with respect to $x$,we get $\frac{d^2 y}{dx^2} = 0$.
Since the equation $y = ax + b$ contains two arbitrary constants $a$ and $b$,it represents the general solution of the second-order differential equation $\frac{d^2 y}{dx^2} = 0$.

(B) Given the equation $x^2+y^2=1$.
On differentiating both sides with respect to $x$,we get:
$2x + 2y y^{\prime} = 0$
Dividing by $2$,we have $x + y y^{\prime} = 0$.
Now,differentiating again with respect to $x$ using the product rule on $y y^{\prime}$:
$\frac{d}{dx}(x) + \frac{d}{dx}(y y^{\prime}) = 0$
$1 + (y y^{\prime \prime} + (y^{\prime}) \cdot y^{\prime}) = 0$
$1 + y y^{\prime \prime} + (y^{\prime})^2 = 0$
Thus,the correct differential equation is $y y^{\prime \prime} + (y^{\prime})^2 + 1 = 0$.

(C) The equation of the family of lines passing through the origin is given by $y = mx$,where $m$ is an arbitrary constant.
To find the differential equation,differentiate both sides with respect to $x$:
$\frac{dy}{dx} = m$
Substitute the value of $m = \frac{y}{x}$ into the differentiated equation:
$\frac{dy}{dx} = \frac{y}{x}$
Multiplying both sides by $x$,we get:
$y = x \frac{dy}{dx}$

(D) Given the family of curves: $x^2 = c(y+c)^2$.
Taking the square root on both sides,we get $x = \sqrt{c}(y+c)$ (assuming $x > 0$ for simplicity,as the parameter $c$ absorbs the sign).
Differentiating with respect to $x$: $1 = \sqrt{c} \frac{dy}{dx}$,which implies $\sqrt{c} = \frac{dx}{dy}$.
Substituting $\sqrt{c} = \frac{dx}{dy}$ into the equation $x = \sqrt{c}(y+c)$:
$x = \frac{dx}{dy} \left( y + \left( \frac{dx}{dy} \right)^2 \right)$.
Multiplying by $\left( \frac{dy}{dx} \right)^3$:
$x \left( \frac{dy}{dx} \right)^3 = \left( \frac{dx}{dy} \cdot \frac{dy}{dx} \right) \left( y \left( \frac{dy}{dx} \right)^2 + 1 \right)$.
Since $\frac{dx}{dy} \cdot \frac{dy}{dx} = 1$,we have:
$x \left( \frac{dy}{dx} \right)^3 = y \left( \frac{dy}{dx} \right)^2 + 1$.
Rearranging the terms,we get $x \left( \frac{dy}{dx} \right)^3 - y \left( \frac{dy}{dx} \right)^2 - 1 = 0$.
Thus,option $D$ is correct.

(A) The general equation of a parabola with its axis 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 to eliminate the three constants.
First derivative: $\frac{dy}{dx} = 2Ax + B$.
Second derivative: $\frac{d^2y}{dx^2} = 2A$.
Third derivative: $\frac{d^3y}{dx^3} = 0$.
Thus,the required differential equation is $\frac{d^3y}{dx^3} = 0$.

(D) Given equation is $y=e^x(a \cos x+b \sin x)$.
Differentiating with respect to $x$ using the product rule:
$\frac{d y}{d x} = e^x(a \cos x + b \sin x) + e^x(-a \sin x + b \cos x)$
$\frac{d y}{d x} = y + e^x(-a \sin x + b \cos x) \quad \dots(I)$
Differentiating again with respect to $x$:
$\frac{d^2 y}{d x^2} = \frac{d y}{d x} + \frac{d}{d x}[e^x(-a \sin x + b \cos x)]$
$\frac{d^2 y}{d x^2} = \frac{d y}{d x} + e^x(-a \sin x + b \cos x) + e^x(-a \cos x - b \sin x)$
From equation $(I)$,$e^x(-a \sin x + b \cos x) = \frac{d y}{d x} - y$.
Substituting this into the second derivative:
$\frac{d^2 y}{d x^2} = \frac{d y}{d x} + (\frac{d y}{d x} - y) - e^x(a \cos x + b \sin x)$
Since $e^x(a \cos x + b \sin x) = y$,we have:
$\frac{d^2 y}{d x^2} = 2 \frac{d y}{d x} - y - y$
$\frac{d^2 y}{d x^2} - 2 \frac{d y}{d x} + 2 y = 0$.

(D) Statement $(I)$:
Given $y=(\alpha+\beta+\gamma) x$. Let $k = \alpha+\beta+\gamma$,where $k$ is a single arbitrary constant.
Then $y = kx$.
Differentiating with respect to $x$,we get $\frac{dy}{dx} = k$.
Since there is only one independent arbitrary constant,the order of the differential equation is $1$.
Thus,Statement $(I)$ is false.
Statement $(II)$:
Given $y = \alpha x + \beta \sin x + \gamma e^x$.
This equation contains $3$ independent arbitrary constants $(\alpha, \beta, \gamma)$.
Differentiating with respect to $x$ three times:
$(1) \frac{dy}{dx} = \alpha + \beta \cos x + \gamma e^x$
$(2) \frac{d^2y}{dx^2} = -\beta \sin x + \gamma e^x$
$(3) \frac{d^3y}{dx^3} = -\beta \cos x + \gamma e^x$
Since we have $3$ independent arbitrary constants,we can eliminate them to form a differential equation of order $3$.
Thus,Statement $(II)$ is true.

(C) Given,$y = e^x(A \cos x + B \sin x)$ ...$(i)$
Differentiating with respect to $x$ using the product rule:
$y^{\prime} = e^x(A \cos x + B \sin x) + e^x(-A \sin x + B \cos x)$
$y^{\prime} = y + e^x(-A \sin x + B \cos x)$
$y^{\prime} - y = e^x(-A \sin x + B \cos x)$ ...(ii)
Differentiating again with respect to $x$:
$y^{\prime \prime} - y^{\prime} = e^x(-A \sin x + B \cos x) + e^x(-A \cos x - B \sin x)$
Substitute $e^x(-A \sin x + B \cos x) = y^{\prime} - y$ from (ii) and $e^x(-A \cos x - B \sin x) = -y$ from $(i)$:
$y^{\prime \prime} - y^{\prime} = (y^{\prime} - y) - y$
$y^{\prime \prime} - y^{\prime} = y^{\prime} - 2y$
$y^{\prime \prime} - 2y^{\prime} + 2y = 0$
Thus,the correct option is $C$.

(A) Given the family of parabolas is $y^2 = 4a(x+a) \quad ...(i)$
Differentiating with respect to $x$,we get:
$2y \frac{dy}{dx} = 4a$
$\Rightarrow a = \frac{1}{2} y \frac{dy}{dx} \quad ...(ii)$
Substituting the value of $a$ from equation $(ii)$ into equation $(i)$:
$y^2 = 4 \left( \frac{1}{2} y \frac{dy}{dx} \right) \left( x + \frac{1}{2} y \frac{dy}{dx} \right)$
$y^2 = 2y \frac{dy}{dx} \left( x + \frac{1}{2} y \frac{dy}{dx} \right)$
Dividing by $y$ (assuming $y \neq 0$):
$y = 2x \frac{dy}{dx} + y \left( \frac{dy}{dx} \right)^2$
Rearranging the terms,we get:
$y \left( \frac{dy}{dx} \right)^2 + 2x \frac{dy}{dx} - y = 0$

(C) Given the equation: $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$.
Differentiating with respect to $x$:
$\frac{2x}{a^2} + \frac{2y}{b^2} \frac{dy}{dx} = 0 \implies \frac{y}{b^2} \frac{dy}{dx} = -\frac{x}{a^2} \implies \frac{y}{x} \frac{dy}{dx} = -\frac{b^2}{a^2}$.
Differentiating again with respect to $x$:
$\frac{d}{dx} \left( \frac{y}{x} \frac{dy}{dx} \right) = \frac{d}{dx} \left( -\frac{b^2}{a^2} \right) = 0$.
Using the product rule:
$\frac{y}{x} \frac{d^2y}{dx^2} + \frac{dy}{dx} \left( \frac{x \frac{dy}{dx} - y}{x^2} \right) = 0$.
Multiplying by $x^2$:
$xy \frac{d^2y}{dx^2} + x \left( \frac{dy}{dx} \right)^2 - y \frac{dy}{dx} = 0$.

(B) Given the equation of the family of curves: $r^2 = a^2 \cos 2\theta$.
Differentiating both sides with respect to $\theta$:
$\frac{d}{d\theta}(r^2) = \frac{d}{d\theta}(a^2 \cos 2\theta)$.
Using the chain rule: $2r \frac{dr}{d\theta} = a^2 (-\sin 2\theta) \cdot 2$.
Simplifying: $r \frac{dr}{d\theta} = -a^2 \sin 2\theta$.
From the original equation,$a^2 = \frac{r^2}{\cos 2\theta}$.
Substituting $a^2$ into the differentiated equation:
$r \frac{dr}{d\theta} = -\left(\frac{r^2}{\cos 2\theta}\right) \sin 2\theta$.
$r \frac{dr}{d\theta} = -r^2 \tan 2\theta$.
Dividing by $r$ (assuming $r \neq 0$):
$\frac{dr}{d\theta} = -r \tan 2\theta$.

(B) Given the general solution $y = c(x - c)^2$.
Step $1$: Differentiate with respect to $x$:
$y' = 2c(x - c)$.
Step $2$: From the original equation,$c = \frac{y}{(x - c)^2}$.
Alternatively,from $y' = 2c(x - c)$,we have $c = \frac{y'}{2(x - c)}$.
Equating the two expressions for $c$:
$\frac{y}{(x - c)^2} = \frac{y'}{2(x - c)} \implies 2y = y'(x - c) \implies x - c = \frac{2y}{y'}$.
Step $3$: Substitute $x - c$ back into the expression for $y'$:
$y' = 2c \left(\frac{2y}{y'}\right) \implies c = \frac{(y')^2}{4y}$.
Step $4$: Substitute $c$ and $x - c$ into the original equation $y = c(x - c)^2$:
$y = \left(\frac{(y')^2}{4y}\right) \left(\frac{2y}{y'}\right)^2$.
$y = \frac{(y')^2}{4y} \cdot \frac{4y^2}{(y')^2} = y$.
This confirms the relation. To find the differential equation,we use $c = \frac{y'}{2(x - c)}$ and $x - c = \frac{2y}{y'}$.
Thus $c = x - \frac{2y}{y'}$.
Substitute $c$ into $y' = 2c(x - c)$:
$y' = 2(x - \frac{2y}{y'})(\frac{2y}{y'}) = 2(\frac{xy' - 2y}{y'})(\frac{2y}{y'}) = \frac{4y(xy' - 2y)}{(y')^2}$.
$(y')^3 = 4y(xy' - 2y)$.
Thus,option $B$ is correct.

(B) Given the equation $y = a e^{2x} + b e^{5x}$.
First,differentiate with respect to $x$:
$\frac{dy}{dx} = 2a e^{2x} + 5b e^{5x}$ (Equation $1$)
Next,differentiate again with respect to $x$:
$\frac{d^2y}{dx^2} = 4a e^{2x} + 25b e^{5x}$ (Equation $2$)
We want to eliminate $a$ and $b$. From Equation $1$,multiply by $5$:
$5 \frac{dy}{dx} = 10a e^{2x} + 25b e^{5x}$ (Equation $3$)
Subtract Equation $2$ from Equation $3$:
$5 \frac{dy}{dx} - \frac{d^2y}{dx^2} = 6a e^{2x} \implies a e^{2x} = \frac{5 \frac{dy}{dx} - \frac{d^2y}{dx^2}}{6}$.
Alternatively,consider the characteristic equation approach. The roots are $m_1 = 2$ and $m_2 = 5$.
The characteristic equation is $(m - 2)(m - 5) = 0$.
$m^2 - 7m + 10 = 0$.
Replacing $m^k$ with $\frac{d^k y}{dx^k}$,we get $\frac{d^2 y}{dx^2} - 7 \frac{dy}{dx} + 10 y = 0$.

(A) The equation of a family of circles touching the $Y$-axis at the origin $(0,0)$ is given by $(x-a)^2 + y^2 = a^2$,where $a$ is the radius of the circle and $(a,0)$ is the center.
Expanding this,we get $x^2 - 2ax + a^2 + y^2 = a^2$,which simplifies to $x^2 + y^2 = 2ax$ $(i)$.
Differentiating both sides with respect to $x$,we get $2x + 2y \frac{dy}{dx} = 2a$.
Thus,$a = x + y \frac{dy}{dx}$ $(ii)$.
Substituting the value of $a$ from equation $(ii)$ into equation $(i)$:
$x^2 + y^2 = 2x(x + y \frac{dy}{dx})$
$x^2 + y^2 = 2x^2 + 2xy \frac{dy}{dx}$
$2xy \frac{dy}{dx} = y^2 - x^2$
$\frac{dy}{dx} = \frac{y^2 - x^2}{2xy}$.

(A) Given equation: $A x^2 + B y^2 = 1$ $(1)$
Differentiating with respect to $x$: $2 A x + 2 B y \frac{d y}{d x} = 0 \implies A x + B y y' = 0$ $(2)$
Differentiating again with respect to $x$: $A + B (y')^2 + B y y'' = 0$ $(3)$
From $(2)$,$A = -B \frac{y y'}{x}$. Substituting into $(3)$:
$-B \frac{y y'}{x} + B (y')^2 + B y y'' = 0$
Dividing by $B$ (assuming $B \neq 0$):
$-\frac{y y'}{x} + (y')^2 + y y'' = 0$
Multiplying by $x$: $-y y' + x (y')^2 + x y y'' = 0$
Rearranging: $x y y'' + x (y')^2 = y y'$
Thus,the differential equation is $x y \frac{d^2 y}{d x^2} + x \left(\frac{d y}{d x}\right)^2 = y \frac{d y}{d x}$.

(D) Given the family of curves: $y = (a + bx + cx^2) e^x$
This can be written as $y e^{-x} = a + bx + cx^2$.
Differentiating with respect to $x$ three times:
First derivative: $y' e^{-x} - y e^{-x} = b + 2cx \implies (y' - y) e^{-x} = b + 2cx$
Second derivative: $(y'' - y') e^{-x} - (y' - y) e^{-x} = 2c \implies (y'' - 2y' + y) e^{-x} = 2c$
Third derivative: $(y''' - 2y'' + y') e^{-x} - (y'' - 2y' + y) e^{-x} = 0$
Since $e^{-x} \neq 0$,we have $y''' - 3y'' + 3y' - y = 0$.

(A) Given,$y = A e^x + B e^{2x} + C e^{3x} \quad \dots(i)$
Differentiating with respect to $x$:
$y' = A e^x + 2B e^{2x} + 3C e^{3x} \quad \dots(ii)$
Differentiating again with respect to $x$:
$y'' = A e^x + 4B e^{2x} + 9C e^{3x} \quad \dots(iii)$
Differentiating again with respect to $x$:
$y''' = A e^x + 8B e^{2x} + 27C e^{3x} \quad \dots(iv)$
Alternatively,since the roots of the auxiliary equation are $m = 1, 2, 3$,the characteristic equation is $(m-1)(m-2)(m-3) = 0$.
$(m^2 - 3m + 2)(m-3) = 0$
$m^3 - 3m^2 - 3m^2 + 9m + 2m - 6 = 0$
$m^3 - 6m^2 + 11m - 6 = 0$
Replacing $m^k$ with $y^{(k)}$,we get the differential equation:
$y''' - 6y'' + 11y' - 6y = 0$.

(C) The equation of the family of parabolas with focus at the origin and $X$-axis as its axis is given by $y^2 = 4a(x+a) = 4ax + 4a^2$.
Differentiating with respect to $x$,we get $2y \frac{dy}{dx} = 4a$,which implies $a = \frac{1}{2} y \frac{dy}{dx}$.
Substituting the value of $a$ into the original equation:
$y^2 = 4 \left( \frac{1}{2} y \frac{dy}{dx} \right) x + 4 \left( \frac{1}{2} y \frac{dy}{dx} \right)^2$
$y^2 = 2xy \frac{dy}{dx} + y^2 \left( \frac{dy}{dx} \right)^2$.
The highest order derivative is $\frac{dy}{dx}$,so the order $m = 1$.
The power of the highest order derivative is $2$,so the degree $n = 2$.
Therefore,$mn - m + n = (1 \times 2) - 1 + 2 = 2 - 1 + 2 = 3$.

(B) Given the family of circles: $(x-a)^2+(y-b)^2=4$.
Differentiating with respect to $x$: $2(x-a)+2(y-b)y'=0 \implies (x-a)+(y-b)y'=0$.
Differentiating again with respect to $x$: $1+(y')^2+(y-b)y''=0 \implies (y-b) = -\frac{1+(y')^2}{y''}$.
Substituting $(y-b)$ back into the first derivative equation: $(x-a) = -y'(y-b) = y' \cdot \frac{1+(y')^2}{y''}$.
Now substitute $(x-a)$ and $(y-b)$ into the original equation: $\left(y' \cdot \frac{1+(y')^2}{y''}\right)^2 + \left(-\frac{1+(y')^2}{y''}\right)^2 = 4$.
This simplifies to: $\frac{(1+(y')^2)^2}{(y'')^2} \cdot ((y')^2+1) = 4$.
Thus,$(1+(y')^2)^3 = 4(y'')^2$,which is $4\left(\frac{d^2 y}{d x^2}\right)^2 = \left[1+\left(\frac{d y}{d x}\right)^2\right]^3$.

(C) Given the family of curves: $y^2=4a(x+a)$
Step $1$: Differentiate with respect to $x$:
$2y \frac{dy}{dx} = 4a$
$\implies a = \frac{y}{2} \frac{dy}{dx}$
Step $2$: Substitute the value of $a$ back into the original equation:
$y^2 = 4 \left( \frac{y}{2} \frac{dy}{dx} \right) \left( x + \frac{y}{2} \frac{dy}{dx} \right)$
$y^2 = 2y \frac{dy}{dx} \left( x + \frac{y}{2} \frac{dy}{dx} \right)$
$y^2 = 2xy \frac{dy}{dx} + y^2 \left( \frac{dy}{dx} \right)^2$
Step $3$: Identify the order and degree:
The highest order derivative is $\frac{dy}{dx}$,so the order $m = 1$.
The highest power of the highest order derivative is $2$,so the degree $n = 2$.
Step $4$: Calculate $m+n^2$:
$m+n^2 = 1 + 2^2 = 1 + 4 = 5$.

(B) Given the family of curves: $(x-2)^2+(y-a)^2=b^2$ $(i)$
Since there are two parameters $a$ and $b$,we differentiate twice.
Differentiating with respect to $x$:
$2(x-2) + 2(y-a)y' = 0$
$(x-2) + (y-a)y' = 0$ (ii)
From (ii),$(y-a) = -\frac{x-2}{y'}$
Substitute this into $(i)$:
$(x-2)^2 + \left(-\frac{x-2}{y'}\right)^2 = b^2$
$(x-2)^2 \left(1 + \frac{1}{(y')^2}\right) = b^2$
$(x-2)^2 \left(\frac{(y')^2+1}{(y')^2}\right) = b^2$
Differentiating again with respect to $x$:
$2(x-2) \left(\frac{(y')^2+1}{(y')^2}\right) + (x-2)^2 \left(\frac{2y'y'' \cdot (y')^2 - ((y')^2+1) \cdot 2y'y''}{(y')^4}\right) = 0$
This differential equation involves the second derivative $y''$,so the order $m = 2$.
The highest power of the highest derivative $y''$ is $1$,so the degree $n = 1$.
Therefore,$m^2+n = (2)^2 + 1 = 4 + 1 = 5$.

(C) The order of the differential equation $\frac{d^2 y}{d x^2}-2\left(\frac{d y}{d x}\right)^3+\sin \left(\frac{d y}{d x}\right)+y=0$ is $l=2$.
To find the degree of $\left(1+\frac{d^2 y}{d x^2}\right)^{\frac{2}{3}}=\left[2-\left(\frac{d y}{d x}\right)^3\right]^{\frac{3}{2}}$,we cube both sides: $\left(1+\frac{d^2 y}{d x^2}\right)^2 = \left[2-\left(\frac{d y}{d x}\right)^3\right]^{\frac{9}{2}}$. Squaring again to remove fractional powers,the highest derivative $\frac{d^2 y}{d x^2}$ will have power $2 \times 2 = 4$. Thus,$m=4$.
Given $y=A x^2+B e^{4 x}$.
$y^{\prime}=2 A x+4 B e^{4 x}$
$y^{\prime \prime}=2 A+16 B e^{4 x}$
Eliminating $A$ and $B$:
$y^{\prime \prime}-4 y^{\prime} = 2A + 16Be^{4x} - 8Ax - 16Be^{4x} = 2A - 8Ax$.
$y^{\prime}-2Ax = 4Be^{4x} \implies y^{\prime \prime}-4y^{\prime} = 2A - 8Ax$.
Solving for constants leads to the differential equation $\left(2 x^2-x\right) y^{\prime \prime}-\left(8 x^2-1\right) y^{\prime}+(16 x-4) y=0$.

(C) The equation of the family of parabolas with focus at the origin $(0,0)$ and the $X$-axis as its axis is given by $(y-0)^2 = -4a(x-a)$,where $a$ is the parameter.
$y^2 = -4ax + 4a^2$ $(i)$
Differentiating with respect to $x$:
$2y y' = -4a$
$a = -\frac{y y'}{2}$
Substituting the value of $a$ into equation $(i)$:
$y^2 = -4\left(-\frac{y y'}{2}\right)x + 4\left(-\frac{y y'}{2}\right)^2$
$y^2 = 2x y y' + 4\left(\frac{y^2 y'^2}{4}\right)$
$y^2 = 2x y y' + y^2 y'^2$
Dividing by $y$ (assuming $y \neq 0$):
$y = 2x y' + y y'^2$
$y\left(\frac{dy}{dx}\right)^2 + 2x\left(\frac{dy}{dx}\right) - y = 0$
The order $m$ of this differential equation is $1$ and the degree $n$ is $2$.
Therefore,$mn - m + n = (1)(2) - 1 + 2 = 2 - 1 + 2 = 3$.

(C) To find the differential equation for each option,we differentiate with respect to $x$ and eliminate the arbitrary constant.
$(a)$ $x^2+y^2+2gx+4y+2=0$. Differentiating,$2x+2y\frac{dy}{dx}+2g+4\frac{dy}{dx}=0$. This leads to a first-order,first-degree equation.
$(b)$ $x^2=a^2(1+y^2)$. Differentiating,$2x=a^2(2y\frac{dy}{dx})$. Substituting $a^2$ back,we get a first-order,first-degree equation.
$(c)$ $y^2=2c(x+\sqrt{c})$. Differentiating,$2y\frac{dy}{dx}=2c$,so $c=y\frac{dy}{dx}$. Substituting $c$ into the original equation: $y^2=2(y\frac{dy}{dx})(x+\sqrt{y\frac{dy}{dx}})$. Rearranging: $y^2-2xy\frac{dy}{dx}=2y\frac{dy}{dx}\sqrt{y\frac{dy}{dx}}$. Squaring both sides: $(y^2-2xy\frac{dy}{dx})^2 = 4y^2(\frac{dy}{dx})^2(y\frac{dy}{dx}) = 4y^3(\frac{dy}{dx})^3$. This equation has order $1$ and degree $3$.
$(d)$ $y^2=4ax$. Differentiating,$2y\frac{dy}{dx}=4a$. Substituting $a$ gives $y^2=2xy\frac{dy}{dx}$,which is first-order,first-degree.
Thus,option $(c)$ is the correct answer.