用拉普拉斯变换求解含初始条件的二阶常微分方程y''+5y'+6y=0

Let's break this down step by step—no fancy tricks, just clear application of Laplace transform rules to arrive at the solution we already know.

Step 1: Recall Key Laplace Transform Properties #

First, we need the core transform rules for derivatives, which are essential for solving ODEs with this method:

• L{y(x)} = Y(s) (by definition of the Laplace transform)
• L{y'(x)} = sY(s) - y(0) (transform of the first derivative)
• L{y''(x)} = s²Y(s) - s y(0) - y'(0) (transform of the second derivative)

We already have the initial conditions handy: y(0) = 2 and y'(0) = 3, so we’ll plug these in as we go.

Step 2: Apply Laplace Transform to the ODE #

The original equation we’re solving is:

$y'' + 5y' + 6y = 0$

Take the Laplace transform of every term on both sides:
L{y''} + 5L{y'} + 6L{y} = L{0}

Substitute the derivative transforms and initial conditions into the equation:
(s²Y(s) - s*2 - 3) + 5(sY(s) - 2) + 6Y(s) = 0

Now expand and collect like terms to isolate $Y(s)$:

• Expand the left-hand side: $s²Y - 2s - 3 + 5sY - 10 + 6Y = 0$
• Group terms with $Y(s)$: $Y(s)(s² + 5s + 6) - 2s - 13 = 0$
• Rearrange to solve for $Y(s)$: $Y(s) = \frac{2s + 13}{s² + 5s + 6}$

Factor the denominator to prepare for partial fractions: $s² +5s +6 = (s+2)(s+3)$, so:
Y(s) = (2s +13)/[(s+2)(s+3)]

Step 3: Partial Fraction Decomposition #

We need to split this rational function into simpler terms that we can inverse-transform easily. Let’s set up the decomposition:
$\frac{2s +13}{(s+2)(s+3)} = \frac{A}{s+2} + \frac{B}{s+3}$

Multiply both sides by $(s+2)(s+3)$ to eliminate denominators:
$2s +13 = A(s+3) + B(s+2)$

Solve for A and B by plugging in convenient s-values:

• Let $s = -2$: $2(-2) +13 = A(-2+3) → 9 = A*1 → A=9$
• Let $s = -3$: $2(-3) +13 = B(-3+2) →7 = B*(-1) → B=-7$

Our decomposed $Y(s)$ now looks like this:
Y(s) = 9/(s+2) -7/(s+3)

Step 4: Inverse Laplace Transform #

Now use the inverse transform rule we all rely on: $L^{-1}{1/(s+a)} = e^{-ax}$

Apply this to each term:

• $L^{-1}{9/(s+2)} =9e^{-2x}$
• $L^{-1}{-7/(s+3)} = -7e^{-3x}$

Combine these to get the final solution:
$y(x) =9e^{-2x} -7e^{-3x}$

If we factor out $e^{-3x}$, we get exactly the solution provided:
$y(x) = e^{-3x}(-7 +9e^x)$

内容的提问来源于stack exchange，提问作者user517526