Q.Show that each of the following expressions is a solution of the corresponding given differential equation.

Differentiating twice reproduces $y$ itself, since $e^x$ and $e^{-x}$ are both fixed (up to sign) by two derivatives.

1. $y=Ae^x+Be^{-x}$. First derivative: $y'=Ae^x-Be^{-x}$.
2. Second derivative: $y''=Ae^x+Be^{-x}$.
3. Comparing, $y''=Ae^x+Be^{-x}=y$ exactly — the two arbitrary constants $A,B$ have been eliminated.
4. So the differential equation of the family is $y''-y=0$, i.e. $\dfrac{d^2y}{dx^2}-y=0$.

Eliminating the arbitrary constant $c$ from $y=cx-c^2$ gives the differential equation $(y')^2-xy'+y=0$.

1. The family of curves is $y=cx-c^2$, with $c$ an arbitrary constant.
2. Differentiate with respect to $x$: $y'=c$ (since $c$ is constant along each member of the family).
3. Substitute $c=y'$ back into the original equation: $y=(y')x-(y')^2=xy'-(y')^2$.
4. Rearranging: $(y')^2-xy'+y=0$.
5. This is the differential equation whose general solution is the given family $y=cx-c^2$ (in fact this family is the general solution of this Clairaut-type equation).

Eliminating the arbitrary radius $r$ from $x^2+y^2=r^2$ by differentiation gives the differential equation $x\,dx+y\,dy=0$.

1. The family of all circles centred at the origin is $x^2+y^2=r^2$, where $r$ is an arbitrary constant (one parameter, so a first-order differential equation is expected).
2. Differentiate both sides with respect to $x$: $2x + 2y\dfrac{dy}{dx} = 0$.
3. Divide by $2$: $x + y\dfrac{dy}{dx} = 0$.
4. Multiply through by $dx$: $x\,dx + y\,dy = 0$. The constant $r$ has been eliminated, as required for the differential equation of the whole family.

Differentiate the given family once with respect to $x$ and substitute back $ke^{\lambda x}=y$ to eliminate the single arbitrary constant $k$, giving a first-order ODE.

1. Given: $y = ke^{\lambda x}$, with $k$ arbitrary and $\lambda$ a fixed constant (not to be eliminated).
2. Since there is exactly one arbitrary constant ($k$), one differentiation suffices to eliminate it.
3. Differentiate with respect to $x$: $\dfrac{dy}{dx} = k\lambda e^{\lambda x}$.
4. Recognise $ke^{\lambda x}=y$ from the original equation, so $\dfrac{dy}{dx} = \lambda\left(ke^{\lambda x}\right) = \lambda y$.
5. This is a first-order linear ODE with no arbitrary constant remaining.
6. This matches option (a).

Eliminating the two arbitrary constants $m$ and $c$ from $y=mx+c$ by differentiating twice yields $y''=0$.

1. The general equation of a straight line not parallel to the $y$-axis is $y=mx+c$, containing two independent arbitrary constants $m$ (slope) and $c$ (intercept).
2. To form the differential equation representing all such lines, we must eliminate both constants, which (since there are two constants) requires differentiating twice.
3. Differentiate once: $\dfrac{dy}{dx}=m$. This still contains the constant $m$ (it is not yet free of arbitrary constants).
4. Differentiate again (with respect to $x$): since $m$ is a constant, $\dfrac{d}{dx}(m)=0$, giving $\dfrac{d^2y}{dx^2}=0$.
5. This final equation contains no arbitrary constants and is satisfied by every line $y=mx+c$ for any choice of $m,c$ — exactly the family of all non-vertical straight lines.
6. Distractors: (a) $\dfrac{dy}{dx}=$ a constant is true for one particular line (fixed $m$), not the whole family (this isn't even a proper differential equation, since it still has the arbitrary constant $m$ in it); (c) and (d) introduce a dependence on $y$ itself, which is not implied by a straight line's equation.