Q.Prove that $y = Ax$ is a solution of the differential equation $xy' = y$, $(x \neq 0)$ and $A$ is a constant.

Differentiate twice and add to $y$; the terms cancel to give $0$.

$y=a\cos x+b\sin x$

$y'=-a\sin x+b\cos x$

$y''=-a\cos x-b\sin x = -(a\cos x+b\sin x)=-y$

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Differentiate $y$ and substitute into the differential equation; it should reduce to a true statement.

Working: Given $y = x^2 + 2x + c$.

Differentiate with respect to $x$:

$y' = \frac{dy}{dx} = 2x + 2$

Substitute into the LHS of $y' - 2x - 2 = 0$:

$y' - 2x - 2 = (2x+2) - 2x - 2 = 0$

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Find the first and second derivatives of $y=e^x+1$ and substitute into $y''-y'=0$ to confirm both sides are equal.

Given: $y = e^x + 1$

First derivative:

$y' = \frac{dy}{dx} = e^x$

Second derivative:

$y'' = \frac{d^2y}{dx^2} = e^x$

Substitute into the differential equation $y'' - y' = 0$:

$y'' - y' = e^x - e^x = 0$

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Differentiate $y=Ae^x+B$ twice: both $\dfrac{dy}{dx}$ and $\dfrac{d^2y}{dx^2}$ equal $Ae^x$, so their difference is $0$.

Concept. Eliminating the arbitrary constants of a family gives its differential equation; here we just verify the relation by differentiation. Note $\dfrac{d}{dx}(B)=0$ since $B$ is constant.

First derivative.

$\frac{dy}{dx}=\frac{d}{dx}\!\left(Ae^x+B\right)=Ae^x.$

Second derivative. …

Differentiate $y$ twice and substitute into $y''-y'$; it should simplify to 0.

Given $y = e^x + 1$.

$y' = e^x$

$y'' = e^x$

Substitute into the differential equation:

$y'' - y' = e^x - e^x = 0$

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Compute $y'$ and $y''$ and substitute into the equation.

Given $y = e^x + 1$.

$y' = \dfrac{d}{dx}(e^x+1) = e^x, \qquad y'' = \dfrac{d}{dx}(e^x) = e^x.$

Substitute into $y'' - y'$:

$y'' - y' = e^x - e^x = 0.$

Since the left side equals $0$, $y = e^x + 1$ is a solution of $y'' - y' = 0$.

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Differentiate $y=Ax$, substitute into $xy'=y$ and check both sides agree.

Given $y=Ax$ with $A$ constant. Differentiate:

$y'=\frac{dy}{dx}=A.$

Substitute into the left side of the differential equation $xy'=y$:

$xy'=x\cdot A=Ax.$

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Differentiate $y$ twice and substitute into $y'' - y' = 0$.

Given $y = e^x + 1$. Differentiating,

$y' = \frac{d}{dx}(e^x + 1) = e^x,$

$y'' = \frac{d}{dx}(e^x) = e^x.$

Substitute into the left-hand side of the differential equation:

$y'' - y' = e^x - e^x = 0,$

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Differentiate y twice and compare with y itself.

y = 5e^x + 2e^{-x} + x

dy/dx = 5e^x − 2e^{-x} + 1

d²y/dx² = 5e^x + 2e^{-x}

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y'' = c₁eˣ + c₂e⁻ˣ = y, so the differential equation is y'' − y = 0.

Given the general solution y = c₁eˣ + c₂e⁻ˣ.

Step 1: Differentiate: dy/dx = c₁eˣ − c₂e⁻ˣ.

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