Solve the initial value problem (IVP) using the Laplace transform method:
$y'' - 2y' + 5y = -8e^{-t}, \quad y(0) = 2, \quad y'(0) = 12.$

Question :

Solve the initial value problem (ivp) using the laplace transform method:
$y'' - 2y' + 5y = -8e^{-t}, \quad y(0) = 2, \quad y'(0) = 12.$

Solution:

Neetesh Kumar | December 7, 2024

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Step-by-step solution:

Step 1: Take the Laplace transform of both sides

The Laplace transform of the equation is: $\mathcal{L}(y'') - 2\mathcal{L}(y') + 5\mathcal{L}(y) = \mathcal{L}(-8e^{-t}).$

Use the Laplace transform rules:

$\mathcal{L}(y'') = s^2Y(s) - sy(0) - y'(0)$ ,
$\mathcal{L}(y') = sY(s) - y(0)$ ,
$\mathcal{L}(y) = Y(s)$ ,
$\mathcal{L}(e^{-t}) = \frac{1}{s+1}$ .

Substitute these into the equation:
$\big[s^2Y(s) - sy(0) - y'(0)\big] - 2\big[sY(s) - y(0)\big] + 5Y(s) = -8 \cdot \frac{1}{s+1}.$

Step 2: Substitute initial conditions

From $y(0) = 2$ and $y'(0) = 12$ , substitute into the equation:
$\big[s^2Y(s) - 2s - 12\big] - 2\big[sY(s) - 2\big] + 5Y(s) = -\frac{8}{s+1}.$

Simplify: $s^2Y(s) - 2s - 12 - 2sY(s) + 4 + 5Y(s) = -\frac{8}{s+1}.$

Combine terms involving $Y(s)$ : $\big[s^2 - 2s + 5\big]Y(s) = \frac{-8}{s+1} + 2s + 8.$

Simplify the right-hand side: $\big[s^2 - 2s + 5\big]Y(s) = \frac{-8}{s+1} + 2s + 8.$

Step 3: Solve for $Y(s)$

$Y(s) = \frac{\frac{-8}{s+1} + 2s + 8}{s^2 - 2s + 5}.$

Combine into a single fraction: $Y(s) = \frac{-8 + (2s + 8)(s+1)}{(s+1)(s^2 - 2s + 5)}.$

Expand the numerator: $-8 + (2s + 8)(s + 1) = -8 + 2s^2 + 2s + 8s + 8 = 2s^2 + 10s.$

Thus: $Y(s) = \frac{2s^2 + 10s}{(s+1)(s^2 - 2s + 5)}.$

Step 4: Partial fraction decomposition

Decompose $Y(s)$ into partial fractions: $Y(s) = \frac{A}{s+1} + \frac{Bs + C}{s^2 - 2s + 5}.$

Multiply through by the denominator: $2s^2 + 10s = A(s^2 - 2s + 5) + (Bs + C)(s+1).$

Expand the terms:

$A(s^2 - 2s + 5) = As^2 - 2As + 5A$ ,
$(Bs + C)(s + 1) = Bs^2 + Bs + Cs + C = Bs^2 + (B + C)s + C$ .

Combine: $2s^2 + 10s - 8 = As^2 - 2As + 5A + Bs^2 + (B + C)s + C.$

Group terms: $(A + B)s^2 + (-2A + B + C)s + (5A + C) = 2s^2 + 10s.$

Compare coefficients:

$A + B = 2$ (for $s^2$ ),
$-2A + B + C = 10$ (for $s$ ),
$5A + C = 0$ (constant).

Step 5: Solve for $A$ , $B$ , and $C$

From $A + B = 2$ , solve for $B$ : $B = 2 - A.$

Substitute $B = 2 - A$ into $-2A + B + C = 10$ :
$-2A + (2 - A) + C = 10.$
$-3A + 2 + C = 10.$
$C = 8 + 3A.$

Substitute $A$ and $C$ into $5A + C = 0$ :
$5A + (8 + 3A) = 0.$
$8A + 8 = 0.$
$8A = -8.$
$A = -1.$

Now substitute $A = -1$ into $B = 2 - A$ : $B = 2 - (-1) = 3.$

Finally, substitute $A = -1$ into $C = 8 + 3A$ : $C = 8 + 3(-1) = 8 - 3 = 5.$

Step 6: Rewrite $Y(s)$

Substitute $A = -1$ , $B = 3$ , and $C = 5$ : $Y(s) = \frac{-1}{s+1} + \frac{3s + 5}{s^2 - 2s + 5}.$

Step 7: Take the inverse Laplace transform

The inverse Laplace of $\frac{-1}{s+1}$ is:
$\mathcal{L}^{-1}\left(\frac{-2}{s+1}\right) = -e^{-t}.$
For $\frac{3s + 5}{s^2 - 2s + 5}$ , rewrite the denominator as $(s - 1)^2 + 4$ .
The term becomes:
$\frac{3s + 5}{(s - 1)^2 + 4} = \frac{3(s - 1) + 8}{(s - 1)^2 + 4}.$

Split into two parts:
$\frac{3(s - 1)}{(s - 1)^2 + 4} + \frac{8}{(s - 1)^2 + 4}.$

The inverse Laplace of $\frac{3(s - 1)}{(s - 1)^2 + 4}$ is $3e^t\cos(2t)$ .
The inverse Laplace of $\frac{8}{(s - 1)^2 + 4}$ is $4e^t\sin(2t)$ .

Final Solution

Combine all terms:
$y(t) = \boxed{-e^{-t} + 3e^t\cos(2t) + 4e^t\sin(2t)}.$

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Solve the initial value problem (IVP) using the Laplace transform method: y′′−2y′+5y=−8e−t,y(0)=2,y′(0)=12.y'' - 2y' + 5y = -8e^{-t}, \quad y(0) = 2, \quad y'(0) = 12.y′′−2y′+5y=−8e−t,y(0)=2,y′(0)=12.