Q.Form the differential equation of the family $y = a x^{2} + b$, where $a$ and $b$ are arbitrary constants.

Differentiating $bx+ay=ab$ once gives a constant slope, and differentiating again gives $\frac{d^2y}{dx^2}=0$ — so the statement is true.

The given family is

$bx + ay = ab,$

with $a$ and $b$ as arbitrary constants. Differentiate both sides with respect to $x$:

$b + a\frac{dy}{dx} = 0 \quad\Rightarrow\quad \frac{dy}{dx} = -\frac{b}{a},$

which is a constant. Differentiating once more with respect to $x$:

$a\frac{d^2y}{dx^2} = 0 \quad\Rightarrow\quad \frac{d^2y}{dx^2} = 0 \quad (a \neq 0).$

This matches the stated differential equation.

Differentiating $y = k_1 e^x + k_2 e^{-x}$ twice gives $\dfrac{d^2y}{dx^2} = k_1 e^x + k_2 e^{-x} = y$, i.e. $\dfrac{d^2y}{dx^2} - y = 0$.

Start from

$y = k_1 e^x + k_2 e^{-x}.$

First derivative:

$\frac{dy}{dx} = k_1 e^x - k_2 e^{-x}.$

Second derivative:

$\frac{d^2y}{dx^2} = k_1 e^x + k_2 e^{-x}.$

The right-hand side is exactly $y$ again, so

$\frac{d^2y}{dx^2} = y \;\Rightarrow\; \frac{d^2y}{dx^2} - y = 0.$

(There are two arbitrary constants, matching a second-order equation.)

A solution obtained from the general solution by giving definite values to the arbitrary constants is called a particular solution.

The general solution of a differential equation contains as many arbitrary constants as the order of the equation. When these arbitrary constants are assigned particular (specific) values — usually determined from given initial or boundary conditions — the resulting solution is a single specific member of that family.

This specific solution is called the particular solution.

The form $e^{-2x}(A\cos x+B\sin x)$ means complex roots $-2\pm i$; forming the auxiliary equation gives $y_2+4y_1+5y=0$.

The general solution $y=e^{-2x}(A\cos x+B\sin x)$ corresponds to a linear differential equation with complex conjugate roots of the form $\alpha\pm i\beta$, where $\alpha=-2$ and $\beta=1$. So the roots are:

$m=-2\pm i.$

The auxiliary (characteristic) equation is $(m-(-2+i))(m-(-2-i))=0$, i.e.

$(m+2)^2-(i)^2=0\;\Rightarrow\;(m+2)^2+1=0.$

Expanding:

$m^2+4m+4+1=0\;\Rightarrow\;m^2+4m+5=0.$

Replacing $m^2\to y_2$, $m\to y_1$ and the constant with $y$:

$y_2+4y_1+5y=0.$

Differentiate $y^2 = (x+c)^3$ to get $2y\dfrac{dy}{dx} = 3(x+c)^2$, then eliminate $c$ using the original relation to obtain $8\left(\dfrac{dy}{dx}\right)^3 = 27y$.

The general solution $y^2 = (x+c)^3$ contains one arbitrary constant $c$, so the differential equation is obtained by differentiating once and eliminating $c$.

Differentiate both sides with respect to $x$:

$2y\frac{dy}{dx} = 3(x+c)^2.$

From this, $(x+c)^2 = \dfrac{2y}{3}\dfrac{dy}{dx}$. Cube both sides:

$(x+c)^6 = \left(\frac{2y}{3}\frac{dy}{dx}\right)^3 = \frac{8y^3}{27}\left(\frac{dy}{dx}\right)^3.$

But from the original equation $(x+c)^3 = y^2$, so $(x+c)^6 = \big[(x+c)^3\big]^2 = (y^2)^2 = y^4$. Substituting:

$y^4 = \frac{8y^3}{27}\left(\frac{dy}{dx}\right)^3.$

Dividing both sides by $y^3$ and rearranging:

$27y = 8\left(\frac{dy}{dx}\right)^3.$