Q.The differential equation representing the family of curve $y = a\sin(x+b)$ is:

Eliminate the single arbitrary slope constant $m$ from $y=mx$ by differentiating once.

The family of lines through the origin is $y = mx$, where $m$ is an arbitrary constant. Differentiating with respect to $x$: $\dfrac{dy}{dx}=m$. Substituting back into $y=mx$: …

Differentiate twice and observe the second derivative is $-y$.

Given $y = A\sin x + B\cos x$.

Step 1: Differentiate once:

$\frac{dy}{dx} = A\cos x - B\sin x$

Step 2: Differentiate again:

$\frac{d^2y}{dx^2} = -A\sin x - B\cos x = -(A\sin x + B\cos x) = -y$

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Two arbitrary constants need two differentiations; eliminating $a,b$ gives $\dfrac{d^2y}{dx^2}+y=0$.

Concept. A family with two arbitrary constants ($a$ and $b$) yields a second-order differential equation, obtained by differentiating twice and eliminating the constants.

Solution. Given $y=a\sin(x+b)$. …

Differentiate twice to eliminate the two arbitrary constants $a,b$.

$y=a\sin(x+b)$

$\dfrac{dy}{dx}=a\cos(x+b)$

$\dfrac{d^2y}{dx^2}=-a\sin(x+b)=-y$

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Differentiate the family and eliminate the arbitrary constant $A$ using the original equation.

$y=Ae^{2x} \Rightarrow \dfrac{dy}{dx}=2Ae^{2x}=2(Ae^{2x})=2y$.

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Writing every non-horizontal line as $x=my+c$ and eliminating the arbitrary constants $m,c$ by differentiating twice gives $\dfrac{d^2x}{dy^2}=0$.

A horizontal line cannot be written as $x=$ (function of $y$) in a one-to-one way, but every non-horizontal line (including vertical ones) can be written as

$x=my+c$

where $m,c$ are arbitrary constants (this family covers all non-horizontal lines in the plane).

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Differentiate once to eliminate the single arbitrary constant $a$; the family $y = ae^{2x} + 5$ yields the differential equation $\frac{dy}{dx} = 2(y - 5)$.

Differentiating $y=c_1e^x+c_2e^{-x}$ twice reproduces $y$ itself, giving $y''-y=0$.

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eliminate the arbitrary constant by differentiating

Given $xy=c^2$. Differentiate both sides with respect to $x$ (the right side is a constant, so its derivative is $0$):

$x\dfrac{dy}{dx}+y\cdot 1=0$

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