Solve the IVP using the method of Laplace transform: y” + 6y = 1 _ 1; y(0) = 0_ y’ (0) = -1

Solve the IVP using the method of Laplace transform: y” + 6y = 1 _ 1; y(0) = 0_ y’ (0) = -1

The Correct Answer and Explanation is:

We are given the initial value problem (IVP):y′′+6y=1;y(0)=0,y′(0)=−1y” + 6y = 1 \quad ; \quad y(0) = 0, \quad y'(0) = -1y′′+6y=1;y(0)=0,y′(0)=−1

Step 1: Take the Laplace Transform of both sides

Recall:

• L{y′′(t)}=s2Y(s)−sy(0)−y′(0)\mathcal{L}\{y”(t)\} = s^2Y(s) – sy(0) – y'(0)L{y′′(t)}=s2Y(s)−sy(0)−y′(0)
• L{y(t)}=Y(s)\mathcal{L}\{y(t)\} = Y(s)L{y(t)}=Y(s)
• L{1}=1s\mathcal{L}\{1\} = \frac{1}{s}L{1}=s1​

Apply the Laplace transform:L{y′′+6y}=L{1}\mathcal{L}\{y” + 6y\} = \mathcal{L}\{1\}L{y′′+6y}=L{1}

Substitute:s2Y(s)−sy(0)−y′(0)+6Y(s)=1ss^2Y(s) – sy(0) – y'(0) + 6Y(s) = \frac{1}{s}s2Y(s)−sy(0)−y′(0)+6Y(s)=s1​

Using the initial conditions y(0)=0y(0) = 0y(0)=0, y′(0)=−1y'(0) = -1y′(0)=−1:s2Y(s)−0−(−1)+6Y(s)=1ss^2Y(s) – 0 – (-1) + 6Y(s) = \frac{1}{s}s2Y(s)−0−(−1)+6Y(s)=s1​s2Y(s)+1+6Y(s)=1ss^2Y(s) + 1 + 6Y(s) = \frac{1}{s}s2Y(s)+1+6Y(s)=s1​

Group Y(s)Y(s)Y(s):(s2+6)Y(s)=1s−1(s^2 + 6)Y(s) = \frac{1}{s} – 1(s2+6)Y(s)=s1​−1Y(s)=1s−1s2+6Y(s) = \frac{\frac{1}{s} – 1}{s^2 + 6}Y(s)=s2+6s1​−1​Y(s)=1−ss(s2+6)Y(s) = \frac{1 – s}{s(s^2 + 6)}Y(s)=s(s2+6)1−s​

Step 2: Partial Fraction Decomposition

Break down:1−ss(s2+6)=As+Bs+Cs2+6\frac{1 – s}{s(s^2 + 6)} = \frac{A}{s} + \frac{Bs + C}{s^2 + 6}s(s2+6)1−s​=sA​+s2+6Bs+C​

Multiply both sides by the denominator:1−s=A(s2+6)+(Bs+C)s1 – s = A(s^2 + 6) + (Bs + C)s1−s=A(s2+6)+(Bs+C)s

Expand:1−s=A(s2+6)+Bs2+Cs=As2+6A+Bs2+Cs=(A+B)s2+Cs+6A1 – s = A(s^2 + 6) + Bs^2 + Cs = As^2 + 6A + Bs^2 + Cs = (A + B)s^2 + Cs + 6A1−s=A(s2+6)+Bs2+Cs=As2+6A+Bs2+Cs=(A+B)s2+Cs+6A

Match coefficients:

• s2:A+B=0s^2: A + B = 0s2:A+B=0
• s1:C=−1s^1: C = -1s1:C=−1
• Constant: 6A=16A = 16A=1

Solve:

• A=16A = \frac{1}{6}A=61​
• B=−16B = -\frac{1}{6}B=−61​
• C=−1C = -1C=−1

Now:Y(s)=16s−16⋅ss2+6−1s2+6Y(s) = \frac{1}{6s} – \frac{1}{6} \cdot \frac{s}{s^2 + 6} – \frac{1}{s^2 + 6}Y(s)=6s1​−61​⋅s2+6s​−s2+61​

Step 3: Inverse Laplace Transform

Use known transforms:

• L−1{1s}=1\mathcal{L}^{-1}\left\{ \frac{1}{s} \right\} = 1L−1{s1​}=1
• L−1{ss2+a2}=cos⁡(at)\mathcal{L}^{-1}\left\{ \frac{s}{s^2 + a^2} \right\} = \cos(at)L−1{s2+a2s​}=cos(at)
• L−1{1s2+a2}=1asin⁡(at)\mathcal{L}^{-1}\left\{ \frac{1}{s^2 + a^2} \right\} = \frac{1}{a} \sin(at)L−1{s2+a21​}=a1​sin(at)

Here, a2=6⇒a=6a^2 = 6 \Rightarrow a = \sqrt{6}a2=6⇒a=6​

Apply inverse transforms:y(t)=16−16cos⁡(6t)−16sin⁡(6t)y(t) = \frac{1}{6} – \frac{1}{6} \cos(\sqrt{6}t) – \frac{1}{\sqrt{6}} \sin(\sqrt{6}t)y(t)=61​−61​cos(6​t)−6​1​sin(6​t)

Final Answer:

y(t)=16−16cos⁡(6t)−16sin⁡(6t)\boxed{y(t) = \frac{1}{6} – \frac{1}{6} \cos(\sqrt{6}t) – \frac{1}{\sqrt{6}} \sin(\sqrt{6}t)}y(t)=61​−61​cos(6​t)−6​1​sin(6​t)​

Explanation (300 words):

To solve this second-order initial value problem using Laplace transforms, we start by transforming the differential equation into an algebraic equation. The Laplace transform turns derivatives into polynomial expressions involving sss, allowing the equation to be solved in the frequency domain. We apply known Laplace formulas, especially for second derivatives and constants, and incorporate the given initial conditions y(0)=0y(0) = 0y(0)=0 and y′(0)=−1y'(0) = -1y′(0)=−1. This results in a transformed equation that can be manipulated algebraically.

We isolate Y(s)Y(s)Y(s), the Laplace transform of y(t)y(t)y(t), and simplify the expression into a rational function. To make it suitable for inverse Laplace transformation, we decompose the rational function using partial fractions. This step requires algebraic expansion and matching coefficients to identify constants A, B, and C.

Once the expression is decomposed, we use inverse Laplace formulas. These include transforms for terms like 1s\frac{1}{s}s1​, ss2+a2\frac{s}{s^2 + a^2}s2+a2s​, and 1s2+a2\frac{1}{s^2 + a^2}s2+a21​. Recognizing that a=6a = \sqrt{6}a=6​ allows us to apply these directly.

Finally, the solution is constructed by summing the inverse transforms, resulting in a combination of a constant term, a cosine function, and a sine function. This final expression satisfies the original differential equation and initial conditions, confirming that the Laplace transform method provides an accurate and systematic approach to solving linear differential equations with given initial values.