Formation of differential equations Questions in English

(D) Given the function $y = \sqrt{\sin x + \sqrt{\sin x + \sqrt{\sin x + \dots \infty}}}$.
Since the series is infinite,we can write it as $y = \sqrt{\sin x + y}$.
Squaring both sides,we get $y^2 = \sin x + y$.
Differentiating both sides with respect to $x$,we get $\frac{d}{dx}(y^2) = \frac{d}{dx}(\sin x + y)$.
Using the chain rule,$2y \frac{dy}{dx} = \cos x + \frac{dy}{dx}$.
Rearranging the terms,we get $2y \frac{dy}{dx} - \frac{dy}{dx} = \cos x$.
Factoring out $\frac{dy}{dx}$,we get $(2y - 1)\frac{dy}{dx} = \cos x$,which can be written as $(2y - 1)\frac{dy}{dx} - \cos x = 0$.

(B) Given the equation $y = a + bx^2$.
First,differentiate with respect to $x$:
$\frac{dy}{dx} = \frac{d}{dx}(a + bx^2) = 0 + 2bx = 2bx$.
Next,differentiate again with respect to $x$ to find the second derivative:
$\frac{d^2y}{dx^2} = \frac{d}{dx}(2bx) = 2b$.
Now,consider the expression $x \frac{d^2y}{dx^2}$:
$x \frac{d^2y}{dx^2} = x(2b) = 2bx$.
Since $\frac{dy}{dx} = 2bx$,we can substitute this into the equation:
$x \frac{d^2y}{dx^2} = \frac{dy}{dx}$.
Thus,option $B$ is correct.

(B) The order of a differential equation representing a family of curves is equal to the number of arbitrary constants present in the equation of the family of curves.
Since the given family of curves contains $4$ arbitrary constants,the order of the corresponding differential equation is $4$.

(A) The general equation of a circle with radius $a$ is given by $(x - h)^2 + (y - k)^2 = a^2$,where $(h, k)$ are the coordinates of the center.
This equation contains two arbitrary constants,$h$ and $k$.
The order of a differential equation is equal to the number of arbitrary constants present in the general solution of the family of curves.
Since there are $2$ arbitrary constants,the order of the differential equation is $2$.

(C) Given the function $y = 4\sin 3x$.
Differentiating with respect to $x$,we get:
$\frac{dy}{dx} = 4 \times 3 \cos 3x = 12 \cos 3x$.
Differentiating again with respect to $x$,we get:
$\frac{d^2y}{dx^2} = 12 \times (-3 \sin 3x) = -36 \sin 3x$.
We can rewrite this as:
$\frac{d^2y}{dx^2} = -9(4 \sin 3x)$.
Since $y = 4 \sin 3x$,we substitute $y$ into the equation:
$\frac{d^2y}{dx^2} = -9y$.
Rearranging the terms,we get:
$\frac{d^2y}{dx^2} + 9y = 0$.
Thus,the correct option is $C$.

(C) The general equation of all lines in the $xy$-plane is given by $y = mx + c$,where $m$ and $c$ are arbitrary constants.
To find the differential equation,we differentiate with respect to $x$:
$\frac{dy}{dx} = m$
Now,differentiate 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 equation is ${x^2} + {y^2} = {a^2}$.
To find the differential equation,we differentiate the given equation with respect to $x$:
$\frac{d}{dx}({x^2} + {y^2}) = \frac{d}{dx}({a^2})$
$2x + 2y\frac{dy}{dx} = 0$
Dividing by $2$,we get:
$x + y\frac{dy}{dx} = 0$
Thus,the differential equation is $x + y\frac{dy}{dx} = 0$.

(B) Given $y = \frac{x}{x + 1}$.
Taking the reciprocal of both sides,we get $\frac{1}{y} = \frac{x + 1}{x} = 1 + \frac{1}{x}$.
Differentiating both sides with respect to $x$:
$-\frac{1}{y^2} \frac{dy}{dx} = 0 - \frac{1}{x^2}$.
Multiplying both sides by $-y^2 x^2$,we get $x^2 \frac{dy}{dx} = y^2$.
Thus,the correct option is $B$.

(A) Given the equation $y = A\sin x + B\cos x$.
Differentiating with respect to $x$ once,we get:
$\frac{dy}{dx} = A\cos x - B\sin x$.
Differentiating with respect to $x$ again,we get:
$\frac{d^2y}{dx^2} = -A\sin x - B\cos x$.
We can rewrite this as:
$\frac{d^2y}{dx^2} = -(A\sin x + B\cos x)$.
Since $y = A\sin x + B\cos x$,we substitute $y$ into the equation:
$\frac{d^2y}{dx^2} = -y$.
Rearranging the terms,we get the required differential equation:
$\frac{d^2y}{dx^2} + y = 0$.

(B) Given the family of curves $y = a \cos(x + b)$,where $a$ and $b$ are arbitrary constants.
Differentiating with respect to $x$,we get:
$\frac{dy}{dx} = -a \sin(x + b)$
Differentiating again with respect to $x$,we get:
$\frac{d^2y}{dx^2} = -a \cos(x + b)$
Since $y = a \cos(x + b)$,we can substitute this into the second derivative:
$\frac{d^2y}{dx^2} = -y$
Rearranging the terms,we get:
$\frac{d^2y}{dx^2} + y = 0$
Thus,the correct option is $B$.

(C) The equation of a straight line at a unit distance from the origin is given by $x \cos \alpha + y \sin \alpha = 1$ $... (i)$
Differentiating with respect to $x$,we get $\cos \alpha + \frac{dy}{dx} \sin \alpha = 0$ $... (ii)$
From $(ii)$,we have $\cos \alpha = -\frac{dy}{dx} \sin \alpha$. Substituting this into $(i)$:
$x(-\frac{dy}{dx} \sin \alpha) + y \sin \alpha = 1$
$\sin \alpha (y - x \frac{dy}{dx}) = 1$
$y - x \frac{dy}{dx} = \frac{1}{\sin \alpha} = \csc \alpha$ $... (iii)$
From $(ii)$,$\frac{dy}{dx} = -\cot \alpha$,so $(\frac{dy}{dx})^2 = \cot^2 \alpha$.
Using the identity $1 + \cot^2 \alpha = \csc^2 \alpha$,we substitute from $(iii)$ and $(iv)$:
$1 + (\frac{dy}{dx})^2 = \csc^2 \alpha = (y - x \frac{dy}{dx})^2$.

(A) Given the equation: $y = c{e^{{{\sin }^{ - 1}}x}}$.
Differentiating both sides with respect to $x$ using the chain rule:
$\frac{dy}{dx} = c{e^{{{\sin }^{ - 1}}x}} \cdot \frac{d}{dx}(\sin^{-1}x)$
We know that $\frac{d}{dx}(\sin^{-1}x) = \frac{1}{\sqrt{1 - x^2}}$.
Substituting this into the derivative:
$\frac{dy}{dx} = c{e^{{{\sin }^{ - 1}}x}} \cdot \frac{1}{\sqrt{1 - x^2}}$
Since $y = c{e^{{{\sin }^{ - 1}}x}}$,we can substitute $y$ back into the equation:
$\frac{dy}{dx} = \frac{y}{\sqrt{1 - x^2}}$
Thus,the correct option is $A$.

(A) Given the equation of the family of curves: ${x^2}y = a$.
Differentiating both sides with respect to $x$ using the product rule:
$\frac{d}{dx}({x^2}y) = \frac{d}{dx}(a)$
${x^2} \frac{dy}{dx} + y \frac{d}{dx}({x^2}) = 0$
${x^2} \frac{dy}{dx} + y(2x) = 0$
${x^2} \frac{dy}{dx} + 2xy = 0$
Dividing the entire equation by ${x^2}$ (assuming $x \neq 0$):
$\frac{dy}{dx} + \frac{2xy}{x^2} = 0$
$\frac{dy}{dx} + \frac{2y}{x} = 0$.
Thus,the correct option is $A$.

(C) Given the equation $y = e^{mx}$.
Taking the natural logarithm on both sides,we get $\log y = mx$.
From this,we can express the arbitrary constant as $m = \frac{\log y}{x}$.
Now,differentiate $y = e^{mx}$ with respect to $x$:
$\frac{dy}{dx} = m e^{mx}$.
Since $e^{mx} = y$,we substitute this into the derivative:
$\frac{dy}{dx} = m \cdot y$.
Substitute the value of $m = \frac{\log y}{x}$ into the equation:
$\frac{dy}{dx} = \left( \frac{\log y}{x} \right) \cdot y = \left( \frac{y}{x} \right) \log y$.
Thus,the correct option is $C$.

(B) Given the solution is $y = c_1 \cos ax + c_2 \sin ax$.
Differentiating with respect to $x$,we get:
$\frac{dy}{dx} = -c_1 a \sin ax + c_2 a \cos ax$.
Differentiating again with respect to $x$:
$\frac{d^2y}{dx^2} = -c_1 a^2 \cos ax - c_2 a^2 \sin ax$.
Factoring out $-a^2$:
$\frac{d^2y}{dx^2} = -a^2(c_1 \cos ax + c_2 \sin ax)$.
Since $y = c_1 \cos ax + c_2 \sin ax$,we substitute $y$ into the equation:
$\frac{d^2y}{dx^2} = -a^2 y$.
Rearranging the terms,we get:
$\frac{d^2y}{dx^2} + a^2 y = 0$.

(A) Given the equation of the line is $y = mx + c$,where $m$ and $c$ are arbitrary constants.
Step $1$: Differentiate the equation with respect to $x$:
$\frac{dy}{dx} = m$
Step $2$: Differentiate again with respect to $x$ to eliminate the constant $m$:
$\frac{d^2y}{dx^2} = 0$
Thus,the differential equation is $\frac{d^2y}{dx^2} = 0$.

(D) The equation of a straight line passing through the point $(1, -1)$ with slope $m$ is given by:
$y - y_1 = m(x - x_1)$
Substituting $(x_1, y_1) = (1, -1)$,we get:
$y - (-1) = m(x - 1)$
$y + 1 = m(x - 1)$
Since $m$ represents the slope of the line,we can write $m = \frac{dy}{dx}$.
Substituting this into the equation,we get:
$y + 1 = \frac{dy}{dx}(x - 1)$
Rearranging the terms to solve for $y$:
$y = (x - 1)\frac{dy}{dx} - 1$
Thus,the correct option is $D$.

(B) Given the family of curves: ${y^2} = 4a(x + a) \quad (i)$
Differentiating both sides with respect to $x$:
$2y \frac{{dy}}{{dx}} = 4a$
$a = \frac{y}{2} \frac{{dy}}{{dx}} \quad (ii)$
Substitute the value of $a$ from equation $(ii)$ into equation $(i)$:
${y^2} = 4 \left( \frac{y}{2} \frac{{dy}}{{dx}} \right) \left( x + \frac{y}{2} \frac{{dy}}{{dx}} \right)$
${y^2} = 2y \frac{{dy}}{{dx}} \left( x + \frac{y}{2} \frac{{dy}}{{dx}} \right)$
Divide by $y$ (assuming $y \neq 0$):
$y = 2 \frac{{dy}}{{dx}} \left( x + \frac{y}{2} \frac{{dy}}{{dx}} \right)$
$y = 2x \frac{{dy}}{{dx}} + y {\left( \frac{{dy}}{{dx}} \right)^2}$
Rearranging the terms:
$y - y {\left( \frac{{dy}}{{dx}} \right)^2} = 2x \frac{{dy}}{{dx}}$
$y \left[ {1 - {{\left( {\frac{{dy}}{{dx}}} \right)}^2}} \right] = 2x \frac{{dy}}{{dx}}$
Thus,the correct option is $B$.

(C) Given the equation $v = \frac{A}{r} + B$.
Differentiating with respect to $r$,we get:
$\frac{dv}{dr} = -\frac{A}{r^2}$.
Differentiating again with respect to $r$,we get:
$\frac{d^2v}{dr^2} = \frac{2A}{r^3}$.
From the first derivative,we have $A = -r^2 \frac{dv}{dr}$.
Substituting this into the second derivative:
$\frac{d^2v}{dr^2} = \frac{2(-r^2 \frac{dv}{dr})}{r^3} = -\frac{2}{r} \frac{dv}{dr}$.
Rearranging the terms,we get:
$\frac{d^2v}{dr^2} + \frac{2}{r} \frac{dv}{dr} = 0$.

(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.
Step $1$: Differentiate with respect to $x$:
$\frac{dy}{dx} = 2Ax + B$
Step $2$: Differentiate again with respect to $x$:
$\frac{d^2y}{dx^2} = 2A$
Step $3$: Differentiate a third time with respect to $x$:
$\frac{d^3y}{dx^3} = 0$
Since there are $3$ arbitrary constants,we differentiate $3$ times to eliminate them. Thus,the required differential equation is $\frac{d^3y}{dx^3} = 0$.

(D) Given equation: $y = (x + K)e^{-x}$
Differentiating both sides with respect to $x$ using the product rule:
$\frac{dy}{dx} = \frac{d}{dx}(x + K) \cdot e^{-x} + (x + K) \cdot \frac{d}{dx}(e^{-x})$
$\frac{dy}{dx} = 1 \cdot e^{-x} + (x + K) \cdot (-e^{-x})$
$\frac{dy}{dx} = e^{-x} - (x + K)e^{-x}$
Since $y = (x + K)e^{-x}$,we substitute $y$ into the equation:
$\frac{dy}{dx} = e^{-x} - y$
Rearranging the terms gives:
$\frac{dy}{dx} + y = e^{-x}$

(B) Given the equation $y = cx + c - c^3$.
To find the differential equation,we differentiate the given equation with respect to $x$:
$\frac{dy}{dx} = c$.
Now,we substitute $c = \frac{dy}{dx}$ back into the original equation $y = cx + c - c^3$:
$y = x\left(\frac{dy}{dx}\right) + \frac{dy}{dx} - \left(\frac{dy}{dx}\right)^3$.
Thus,the required differential equation is $y = x\frac{dy}{dx} + \frac{dy}{dx} - \left(\frac{dy}{dx}\right)^3$.

(B) Given the equation $y = e^x(A\cos x + B\sin x)$.
Differentiating with respect to $x$ using the product rule:
$\frac{dy}{dx} = e^x(A\cos x + B\sin x) + e^x(-A\sin x + B\cos x)$
$\frac{dy}{dx} = e^x((A+B)\cos x + (B-A)\sin x) = y + e^x(-A\sin x + B\cos x)$
Differentiating again with respect to $x$:
$\frac{d^2y}{dx^2} = \frac{dy}{dx} + e^x(-A\sin x + B\cos x) + e^x(-A\cos x - B\sin x)$
$\frac{d^2y}{dx^2} = \frac{dy}{dx} + (\frac{dy}{dx} - y) - y$
$\frac{d^2y}{dx^2} = 2\frac{dy}{dx} - 2y$.

(C) Given equation is $y = A + Bx + C{e^{ - x}} \dots (i)$
Differentiating with respect to $x$:
$\frac{dy}{dx} = B - C{e^{ - x}} \dots (ii)$
Differentiating again with respect to $x$:
$\frac{d^2y}{dx^2} = C{e^{ - x}} \dots (iii)$
Differentiating a third time with respect to $x$:
$\frac{d^3y}{dx^3} = -C{e^{ - x}} \dots (iv)$
From equation $(iii)$,we have $C{e^{ - x}} = \frac{d^2y}{dx^2}$.
Substituting this into equation $(iv)$:
$\frac{d^3y}{dx^3} = -\frac{d^2y}{dx^2}$
Rearranging the terms,we get:
$\frac{d^3y}{dx^3} + \frac{d^2y}{dx^2} = 0$,which is $y''' + y'' = 0$.

(A) Given the equation: $y = A \cos \omega t + B \sin \omega t$
Differentiating with respect to $t$:
$y' = -A \omega \sin \omega t + B \omega \cos \omega t$
Differentiating again with respect to $t$:
$y'' = -A \omega^2 \cos \omega t - B \omega^2 \sin \omega t$
Factoring out $-\omega^2$:
$y'' = -\omega^2 (A \cos \omega t + B \sin \omega t)$
Since $y = A \cos \omega t + B \sin \omega t$,we substitute $y$ into the equation:
$y'' = -\omega^2 y$
Thus,the differential equation is $y'' + \omega^2 y = 0$.

(C) The equation of all straight lines passing through the origin is given by $y = mx$,where $m$ is an arbitrary constant.
Differentiating both sides with respect to $x$,we get:
$\frac{dy}{dx} = m$.
From the original equation $y = mx$,we can write $m = \frac{y}{x}$.
Substituting the value of $m$ into the differentiated equation,we get:
$\frac{dy}{dx} = \frac{y}{x}$.

(D) Given the equation: $y = a{e^{mx}} + b{e^{-mx}}$
Differentiating with respect to $x$,we get:
$\frac{dy}{dx} = ma{e^{mx}} - mb{e^{-mx}}$
Differentiating again with respect to $x$,we get:
$\frac{d^2y}{dx^2} = m^2a{e^{mx}} + m^2b{e^{-mx}}$
Factoring out $m^2$:
$\frac{d^2y}{dx^2} = m^2(a{e^{mx}} + b{e^{-mx}})$
Since $y = a{e^{mx}} + b{e^{-mx}}$,we can substitute $y$ into the equation:
$\frac{d^2y}{dx^2} = m^2y$
Rearranging the terms gives:
$\frac{d^2y}{dx^2} - m^2y = 0$
Thus,the correct option is $D$.

(C) Given the equation: $y = \sec(\tan^{-1}x)$.
Differentiating both sides with respect to $x$ using the chain rule:
$\frac{dy}{dx} = \frac{d}{dx}[\sec(\tan^{-1}x)]$
Using the derivative formula $\frac{d}{dx}[\sec(u)] = \sec(u)\tan(u)\frac{du}{dx}$ and $\frac{d}{dx}[\tan^{-1}x] = \frac{1}{1+x^2}$:
$\frac{dy}{dx} = \sec(\tan^{-1}x) \cdot \tan(\tan^{-1}x) \cdot \frac{1}{1+x^2}$
Since $\tan(\tan^{-1}x) = x$ and $y = \sec(\tan^{-1}x)$:
$\frac{dy}{dx} = y \cdot x \cdot \frac{1}{1+x^2}$
Rearranging the terms:
$(1 + x^2)\frac{dy}{dx} = xy$.

(B) Given the family of curves: $y = ax \cos \left( \frac{1}{x} + b \right) \dots (i)$
Differentiating with respect to $x$ using the product rule:
$y_1 = a \cos \left( \frac{1}{x} + b \right) + ax \left( -\sin \left( \frac{1}{x} + b \right) \right) \left( -\frac{1}{x^2} \right)$
$y_1 = a \cos \left( \frac{1}{x} + b \right) + \frac{a}{x} \sin \left( \frac{1}{x} + b \right) \dots (ii)$
Differentiating again with respect to $x$:
$y_2 = a \left( -\sin \left( \frac{1}{x} + b \right) \right) \left( -\frac{1}{x^2} \right) + a \left( -\frac{1}{x^2} \right) \sin \left( \frac{1}{x} + b \right) + \frac{a}{x} \cos \left( \frac{1}{x} + b \right) \left( -\frac{1}{x^2} \right)$
$y_2 = \frac{a}{x^2} \sin \left( \frac{1}{x} + b \right) - \frac{a}{x^2} \sin \left( \frac{1}{x} + b \right) - \frac{a}{x^3} \cos \left( \frac{1}{x} + b \right)$
$y_2 = -\frac{a}{x^3} \cos \left( \frac{1}{x} + b \right)$
Multiply both sides by $x^4$:
$x^4 y_2 = -ax \cos \left( \frac{1}{x} + b \right)$
Since $y = ax \cos \left( \frac{1}{x} + b \right)$,we have:
$x^4 y_2 = -y$
$x^4 y_2 + y = 0$

(B) Given equation is $\sin^{-1} x + \sin^{-1} y = c$ ... $(i)$
Differentiating both sides with respect to $x$,we get:
$\frac{d}{dx}(\sin^{-1} x) + \frac{d}{dx}(\sin^{-1} y) = \frac{d}{dx}(c)$
$\frac{1}{\sqrt{1 - x^2}} + \frac{1}{\sqrt{1 - y^2}} \cdot \frac{dy}{dx} = 0$
Multiplying by $\sqrt{1 - x^2} \sqrt{1 - y^2}$,we get:
$\sqrt{1 - y^2} + \sqrt{1 - x^2} \frac{dy}{dx} = 0$
Rearranging the terms,we get:
$\sqrt{1 - x^2} \, dy + \sqrt{1 - y^2} \, dx = 0$.

(B) Given the family of curves: $y = A{e^{3x}} + B{e^{5x}}$
Step $1$: Differentiate with respect to $x$:
$\frac{{dy}}{{dx}} = 3A{e^{3x}} + 5B{e^{5x}}$
Step $2$: Differentiate again with respect to $x$:
$\frac{{{d^2}y}}{{d{x^2}}} = 9A{e^{3x}} + 25B{e^{5x}}$
Step $3$: We eliminate the arbitrary constants $A$ and $B$. The characteristic equation for the roots $m_1 = 3$ and $m_2 = 5$ is $(m - 3)(m - 5) = 0$,which simplifies to $m^2 - 8m + 15 = 0$.
Step $4$: Replacing $m^k$ with $\frac{{d^k}y}{{dx^k}}$,we get the differential equation:
$\frac{{{d^2}y}}{{d{x^2}}} - 8\frac{{dy}}{{dx}} + 15y = 0$
Thus,the correct option is $B$.

(B) The length of the normal to a curve is given by the formula $L = |y|\sqrt{1 + (\frac{dy}{dx})^2}$.
Given that the length of the normal is a constant $k$,we have $|y|\sqrt{1 + (\frac{dy}{dx})^2} = k$.
Squaring both sides,we get $y^2(1 + (\frac{dy}{dx})^2) = k^2$.
Expanding the equation,we get $y^2 + y^2(\frac{dy}{dx})^2 = k^2$.
Rearranging the terms,we obtain $y^2(\frac{dy}{dx})^2 = k^2 - y^2$,which can be written as $(y\frac{dy}{dx})^2 = k^2 - y^2$.

(A) Given $\frac{d^2y}{dx^2} = 0$.
Integrating with respect to $x$,we get $\frac{dy}{dx} = a$,where $a$ is an arbitrary constant.
Integrating again with respect to $x$,we get $\int \frac{dy}{dx} dx = \int a dx + b$,where $b$ is another arbitrary constant.
Therefore,$y = ax + b$.

(D) Given equation is $Ax^2 + By^2 = 1 \quad \dots(1)$
Differentiating with respect to $x$:
$2Ax + 2By \frac{dy}{dx} = 0 \implies Ax + By \frac{dy}{dx} = 0 \quad \dots(2)$
Differentiating again with respect to $x$:
$A + B \left( y \frac{d^2y}{dx^2} + \left( \frac{dy}{dx} \right)^2 \right) = 0 \quad \dots(3)$
From $(2)$,$A = -\frac{By}{x} \frac{dy}{dx}$. Substituting this into $(3)$:
$-\frac{By}{x} \frac{dy}{dx} + By \frac{d^2y}{dx^2} + B \left( \frac{dy}{dx} \right)^2 = 0$
Dividing by $B$ (assuming $B \neq 0$):
$-\frac{y}{x} \frac{dy}{dx} + y \frac{d^2y}{dx^2} + \left( \frac{dy}{dx} \right)^2 = 0$
Multiplying by $x$:
$xy \frac{d^2y}{dx^2} + x \left( \frac{dy}{dx} \right)^2 - y \frac{dy}{dx} = 0$
The highest order derivative is $\frac{d^2y}{dx^2}$,so the order is $2$. The power of the highest order derivative is $1$,so the degree is $1$.

(A) All circles passing through the origin and having their center on the $x$-axis at $(a, 0)$ will have a radius of $a$.
The general equation of such circles is given by:
$(x - a)^2 + y^2 = a^2$
$x^2 - 2ax + a^2 + y^2 = a^2$
$x^2 + y^2 - 2ax = 0$ --- $(i)$
This equation has only one arbitrary constant,$a$. Therefore,we differentiate it once with respect to $x$:
$2x + 2y\frac{dy}{dx} - 2a = 0$
$x + y\frac{dy}{dx} = a$
Substituting the value of $a$ into equation $(i)$:
$x^2 + y^2 - 2x(x + y\frac{dy}{dx}) = 0$
$x^2 + y^2 - 2x^2 - 2xy\frac{dy}{dx} = 0$
$y^2 - x^2 - 2xy\frac{dy}{dx} = 0$
$y^2 = x^2 + 2xy\frac{dy}{dx}$

(C) Given the family of curves: $y = c_1 e^{c_2 x} \dots (i)$
Differentiating with respect to $x$,we get:
$y' = c_1 c_2 e^{c_2 x} = c_2 y \dots (ii)$
Again,differentiating with respect to $x$,we get:
$y'' = c_2 y' \dots (iii)$
From equation $(ii)$,we have $c_2 = \frac{y'}{y}$.
Substituting this value of $c_2$ into equation $(iii)$:
$y'' = \left( \frac{y'}{y} \right) y'$
Multiplying both sides by $y$:
$y y'' = (y')^2$
This is the required differential equation.

(D) The general equation of a circle passing through the origin with its center on the $x$-axis is given by $(x - h)^2 + y^2 = h^2$,where $h$ is the $x$-coordinate of the center.
Expanding this,we get $x^2 - 2hx + h^2 + y^2 = h^2$,which simplifies to $x^2 + y^2 - 2hx = 0$ ..... $(i)$.
Differentiating equation $(i)$ with respect to $x$,we get $2x + 2y \frac{dy}{dx} - 2h = 0$.
Dividing by $2$,we get $x + y \frac{dy}{dx} - h = 0$,so $h = x + y \frac{dy}{dx}$.
Substituting the value of $h$ into equation $(i)$,we get $x^2 + y^2 - 2x(x + y \frac{dy}{dx}) = 0$.
$x^2 + y^2 - 2x^2 - 2xy \frac{dy}{dx} = 0$.
$y^2 - x^2 - 2xy \frac{dy}{dx} = 0$.
Rearranging the terms,we get $2xy \frac{dy}{dx} = y^2 - x^2$,which gives $\frac{dy}{dx} = \frac{y^2 - x^2}{2xy}$.

(C) Given the family of curves is ${x^2} + {y^2} - 2ay = 0$ ..... $(i)$
Differentiating both sides with respect to $x$,we get:
$2x + 2yy' - 2ay' = 0$
$2ay' = 2x + 2yy'$
$2a = \frac{2x + 2yy'}{y'} = \frac{2x}{y'} + 2y$ ..... $(ii)$
Substituting the value of $2a$ from equation $(ii)$ into equation $(i)$:
${x^2} + {y^2} - (\frac{2x}{y'} + 2y)y = 0$
${x^2} + {y^2} - \frac{2xy}{y'} - 2{y^2} = 0$
${x^2} - {y^2} - \frac{2xy}{y'} = 0$
Multiplying by $y'$,we get:
$({x^2} - {y^2})y' - 2xy = 0$
$({x^2} - {y^2})y' = 2xy$.

(C) The equation of the family of parabolas with latus rectum $4a$ and axis parallel to the $x$-axis is $(y - k)^2 = 4a(x - h)$,where $(h, k)$ are arbitrary constants.
Differentiating with respect to $x$:
$2(y - k) \frac{dy}{dx} = 4a$
$\Rightarrow (y - k) \frac{dy}{dx} = 2a$
Differentiating again with respect to $x$:
$(y - k) \frac{d^2y}{dx^2} + \left( \frac{dy}{dx} \right)^2 = 0$
From the first derivative,we have $(y - k) = \frac{2a}{dy/dx}$. Substituting this into the second derivative equation:
$\left( \frac{2a}{dy/dx} \right) \frac{d^2y}{dx^2} + \left( \frac{dy}{dx} \right)^2 = 0$
Multiplying by $\frac{dy}{dx}$:
$2a \frac{d^2y}{dx^2} + \left( \frac{dy}{dx} \right)^3 = 0$
In this differential equation,the highest order derivative is $\frac{d^2y}{dx^2}$,so the order is $2$. The power of the highest order derivative is $1$,so the degree is $1$.

(B) Given the general solution: $y = c_1 \cos(x + c_2) - c_3 e^{-x + c_4} + c_5 \sin x$
Using trigonometric identity $\cos(A+B) = \cos A \cos B - \sin A \sin B$ and property of exponents $e^{A+B} = e^A e^B$:
$y = c_1(\cos x \cos c_2 - \sin x \sin c_2) - (c_3 e^{c_4}) e^{-x} + c_5 \sin x$
Let $A = c_1 \cos c_2$,$B = (c_1 \sin c_2 - c_5)$,and $C = c_3 e^{c_4}$.
Then $y = A \cos x - B \sin x - C e^{-x} \dots (i)$
Differentiating with respect to $x$:
$y' = -A \sin x - B \cos x + C e^{-x} \dots (ii)$
$y'' = -A \cos x + B \sin x - C e^{-x} \dots (iii)$
$y''' = A \sin x + B \cos x + C e^{-x} \dots (iv)$
Adding $(i)$ and $(iii)$: $y'' + y = -2C e^{-x} \dots (v)$
Adding $(ii)$ and $(iv)$: $y''' + y' = 2C e^{-x} \dots (vi)$
Adding $(v)$ and $(vi)$: $y''' + y'' + y' + y = 0$
Thus,the differential equation is $\frac{d^3y}{dx^3} + \frac{d^2y}{dx^2} + \frac{dy}{dx} + y = 0$.

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

(C) The given equation is ${y^2} = \sqrt{c}(x + 2c)$.
Since there is only one arbitrary constant $c$,the order of the differential equation is $1$.
To find the degree,we differentiate with respect to $x$:
$2y \frac{dy}{dx} = \sqrt{c}$.
Squaring both sides,we get $4y^2 \left(\frac{dy}{dx}\right)^2 = c$.
Substitute $\sqrt{c} = 2y \frac{dy}{dx}$ into the original equation:
${y^2} = 2y \frac{dy}{dx} (x + 2(4y^2 \left(\frac{dy}{dx}\right)^2))$.
${y^2} = 2xy \frac{dy}{dx} + 16y^3 \left(\frac{dy}{dx}\right)^3$.
Dividing by $y$ (assuming $y \neq 0$):
$y = 2x \frac{dy}{dx} + 16y^2 \left(\frac{dy}{dx}\right)^3$.
The highest power of the derivative $\frac{dy}{dx}$ is $3$,so the degree is $3$.

(A) Given the equation: $y = ke^{\sin^{-1} x} + 3$.
Subtracting $3$ from both sides,we get: $y - 3 = ke^{\sin^{-1} x}$.
Differentiating both sides with respect to $x$:
$\frac{d}{dx}(y - 3) = \frac{d}{dx}(ke^{\sin^{-1} x})$.
$\frac{dy}{dx} = k \cdot e^{\sin^{-1} x} \cdot \frac{1}{\sqrt{1 - x^2}}$.
Since $ke^{\sin^{-1} x} = y - 3$,we substitute this into the derivative:
$\frac{dy}{dx} = (y - 3) \cdot \frac{1}{\sqrt{1 - x^2}}$.
Multiplying both sides by $\sqrt{1 - x^2}$,we obtain:
$\sqrt{1 - x^2} \frac{dy}{dx} = y - 3$.

(A) Given the equation of the family of ellipses: $\frac{x^2}{a^2} + \frac{y^2}{b^2} = c$.
Differentiating with respect to $x$: $\frac{2x}{a^2} + \frac{2yy'}{b^2} = 0 \Rightarrow \frac{x}{a^2} + \frac{yy'}{b^2} = 0$.
Differentiating again with respect to $x$: $\frac{1}{a^2} + \frac{y y'' + (y')^2}{b^2} = 0$.
From the first derivative,$\frac{1}{a^2} = -\frac{yy'}{xb^2}$.
Substituting this into the second derivative equation: $-\frac{yy'}{xb^2} + \frac{yy'' + (y')^2}{b^2} = 0$.
Multiplying by $\frac{b^2}{y'}$ (assuming $y' \neq 0$): $-\frac{y}{x} + \frac{yy'' + (y')^2}{y'} = 0$.
Rearranging the terms: $\frac{yy'' + (y')^2}{y'} = \frac{y}{x} \Rightarrow \frac{yy''}{y'} + y' = \frac{y}{x}$.
Dividing by $y$: $\frac{y''}{y'} + \frac{y'}{y} = \frac{1}{x} \Rightarrow \frac{y''}{y'} + \frac{y'}{y} - \frac{1}{x} = 0$.

(D) Given the equation: $y = Ae^{2x} + Be^{-2x}$
Differentiating with respect to $x$ once:
$\frac{dy}{dx} = 2Ae^{2x} - 2Be^{-2x}$
Differentiating with respect to $x$ again:
$\frac{d^2y}{dx^2} = 4Ae^{2x} + 4Be^{-2x}$
Factoring out $4$:
$\frac{d^2y}{dx^2} = 4(Ae^{2x} + Be^{-2x})$
Since $y = Ae^{2x} + Be^{-2x}$,we substitute $y$ into the equation:
$\frac{d^2y}{dx^2} = 4y$

(C) The equation of a circle with center $(h, 0)$ on the $x$-axis and radius $h$ (since it touches the $y$-axis at the origin) is $(x - h)^2 + y^2 = h^2$.
Expanding this,we get $x^2 - 2xh + h^2 + y^2 = h^2$,which simplifies to $x^2 + y^2 = 2xh$.
To eliminate the arbitrary constant $h$,we differentiate with respect to $x$:
$2x + 2y \frac{dy}{dx} = 2h$.
Substituting $h = x + y \frac{dy}{dx}$ into the original equation $x^2 + y^2 = 2x(h)$:
$x^2 + y^2 = 2x(x + y \frac{dy}{dx})$.
$x^2 + y^2 = 2x^2 + 2xy \frac{dy}{dx}$.
$y^2 - x^2 = 2xy \frac{dy}{dx}$.
Thus,$\frac{dy}{dx} = \frac{y^2 - x^2}{2xy}$.

(D) The equation of an ellipse centered at the origin is $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$.
Since the ellipse passes through $(0,3)$,we have $\frac{0^2}{a^2} + \frac{3^2}{b^2} = 1$,which gives $b^2 = 9$.
Thus,the equation of the ellipse is $\frac{x^2}{a^2} + \frac{y^2}{9} = 1$.
Differentiating with respect to $x$,we get $\frac{2x}{a^2} + \frac{2y}{9} y' = 0$,which simplifies to $\frac{x}{a^2} + \frac{y y'}{9} = 0$,or $\frac{1}{a^2} = -\frac{y y'}{9x}$.
Substituting $\frac{1}{a^2}$ back into the equation $\frac{x^2}{a^2} + \frac{y^2}{9} = 1$,we get $x^2(-\frac{y y'}{9x}) + \frac{y^2}{9} = 1$.
Multiplying by $9$,we get $-x y y' + y^2 = 9$,or $x y y' - y^2 + 9 = 0$.

(C) The equation of the family of all circles touching the $x-$axis at the origin $(0, 0)$ with center $(0, a)$ and radius $|a|$ is given by:
$x^2 + (y - a)^2 = a^2$
$x^2 + y^2 - 2ay + a^2 = a^2$
$x^2 + y^2 - 2ay = 0$ ... $(1)$
Differentiating both sides with respect to $x$:
$2x + 2y\frac{dy}{dx} - 2a\frac{dy}{dx} = 0$
$x + y\frac{dy}{dx} = a\frac{dy}{dx}$
$a = \frac{x + y\frac{dy}{dx}}{\frac{dy}{dx}}$
Substituting the value of $a$ into equation $(1)$:
$x^2 + y^2 - 2y \left( \frac{x + y\frac{dy}{dx}}{\frac{dy}{dx}} \right) = 0$
$(x^2 + y^2)\frac{dy}{dx} - 2y(x + y\frac{dy}{dx}) = 0$
$(x^2 + y^2)\frac{dy}{dx} - 2xy - 2y^2\frac{dy}{dx} = 0$
$(x^2 - y^2)\frac{dy}{dx} = 2xy$
Comparing this with the given equation $(x^2 - y^2)\frac{dy}{dx} = g(x)y$,we get:
$g(x)y = 2xy$
$g(x) = 2x$

(A) Given the equation of the family of curves: $x^{2} = 4b(y+b)$.
Differentiating both sides with respect to $x$,we get:
$2x = 4b y^{\prime}$
$b = \frac{2x}{4y^{\prime}} = \frac{x}{2y^{\prime}}$.
Substitute the value of $b$ back into the original equation:
$x^{2} = 4 \left( \frac{x}{2y^{\prime}} \right) \left( y + \frac{x}{2y^{\prime}} \right)$.
Simplify the expression:
$x^{2} = \frac{2x}{y^{\prime}} \left( \frac{2yy^{\prime} + x}{2y^{\prime}} \right)$.
$x^{2} = \frac{2x(2yy^{\prime} + x)}{2(y^{\prime})^{2}}$.
$x^{2} = \frac{x(2yy^{\prime} + x)}{(y^{\prime})^{2}}$.
Dividing by $x$ (assuming $x \neq 0$):
$x(y^{\prime})^{2} = 2yy^{\prime} + x$.

(N/A) Given function is $y=e^{-3x}$.
Differentiating both sides with respect to $x$,we get:
$\frac{dy}{dx} = -3e^{-3x}$ ... $(1)$
Now,differentiating $(1)$ with respect to $x$ again,we get:
$\frac{d^{2}y}{dx^{2}} = 9e^{-3x}$
Substituting the values of $\frac{d^{2}y}{dx^{2}}$,$\frac{dy}{dx}$,and $y$ into the left-hand side $(L.H.S.)$ of the given differential equation:
$L.H.S. = \frac{d^{2}y}{dx^{2}} + \frac{dy}{dx} - 6y$
$L.H.S. = 9e^{-3x} + (-3e^{-3x}) - 6(e^{-3x})$
$L.H.S. = 9e^{-3x} - 3e^{-3x} - 6e^{-3x}$
$L.H.S. = 9e^{-3x} - 9e^{-3x} = 0$
Since $L.H.S. = R.H.S. = 0$,the function $y=e^{-3x}$ is a solution of the given differential equation.