If $y=3e^{2x}+2e^{3x}$, prove that $\dfrac{d^2y}{dx^2}-5\dfrac{dy}{dx}+6y=0$.

Question

Accepted Answer

> [!TLDR]
> Differentiate twice and substitute: $y''-5y'+6y$ collapses to $0$ because the coefficients of $e^{2x}$ and $e^{3x}$ each cancel.

**Step 1 — First derivative.**
$$y=3e^{2x}+2e^{3x}\;\Rightarrow\;\frac{dy}{dx}=3(2e^{2x})+2(3e^{3x})=6e^{2x}+6e^{3x}.$$

**Step 2 — Second derivative.**
$$\frac{d^2y}{dx^2}=6(2e^{2x})+6(3e^{3x})=12e^{2x}+18e^{3x}.$$

**Step 3 — Substitute into the expression.**
$$\frac{d^2y}{dx^2}-5\frac{dy}{dx}+6y=\big(12e^{2x}+18e^{3x}\big)-5\big(6e^{2x}+6e^{3x}\big)+6\big(3e^{2x}+2e^{3x}\big).$$ …

Order	Leibniz	Lagrange	Newton
1st	$\frac{dy}{dx}$	$f'(x)$	$\dot{y}$
2nd	$\frac{d^2y}{dx^2}$	$f''(x)$	$\ddot{y}$
$n$ th	$\frac{d^ny}{dx^n}$	$f^{(n)}(x)$	—

Q.If $y=3e^{2x}+2e^{3x}$ , prove that $\dfrac{d^2y}{dx^2}-5\dfrac{dy}{dx}+6y=0$ .

Concept understanding — Successive Differentiation

Successive Differentiation — Repeated Slopes

The definition and notation

Seeing the pattern

Q.If y=3e2x+2e3xy=3e^{2x}+2e^{3x}y=3e2x+2e3x, prove that d2ydx2−5dydx+6y=0\dfrac{d^2y}{dx^2}-5\dfrac{dy}{dx}+6y=0dx2d2y​−5dxdy​+6y=0.

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Q.If $y=3e^{2x}+2e^{3x}$ , prove that $\dfrac{d^2y}{dx^2}-5\dfrac{dy}{dx}+6y=0$ .

Unlock everything free for 14 days