use Laplace transform to solve the following initial value problem x’’+x=8cos7t x(0)=1 x’(0)=0
The Correct Answer and Explanation is:
To solve the differential equation using the Laplace transform, consider: x′′+x=8cos(7t),x(0)=1,x′(0)=0x” + x = 8 \cos(7t), \quad x(0) = 1, \quad x'(0) = 0
Step 1: Take Laplace Transforms
Let X(s)=L{x(t)}X(s) = \mathcal{L}\{x(t)\}
Use the Laplace transforms of derivatives:
- L{x′′(t)}=s2X(s)−sx(0)−x′(0)\mathcal{L}\{x”(t)\} = s^2 X(s) – s x(0) – x'(0)
- L{x(t)}=X(s)\mathcal{L}\{x(t)\} = X(s)
- L{cos(at)}=ss2+a2\mathcal{L}\{\cos(at)\} = \frac{s}{s^2 + a^2}
Apply to both sides: L{x′′+x}=L{8cos(7t)}\mathcal{L}\{x” + x\} = \mathcal{L}\{8 \cos(7t)\} (s2X(s)−s⋅1−0)+X(s)=8⋅ss2+49(s^2 X(s) – s \cdot 1 – 0) + X(s) = 8 \cdot \frac{s}{s^2 + 49} (s2+1)X(s)−s=8ss2+49(s^2 + 1)X(s) – s = \frac{8s}{s^2 + 49}
Step 2: Solve for X(s)X(s)
(s2+1)X(s)=8ss2+49+s(s^2 + 1)X(s) = \frac{8s}{s^2 + 49} + s X(s)=8ss2+49+ss2+1X(s) = \frac{\frac{8s}{s^2 + 49} + s}{s^2 + 1} X(s)=8s(s2+49)(s2+1)+ss2+1X(s) = \frac{8s}{(s^2 + 49)(s^2 + 1)} + \frac{s}{s^2 + 1}
Step 3: Take Inverse Laplace Transform
Use known transforms and partial fractions:
First term: 8s(s2+49)(s2+1)→Inverse is a linear combination ofcos(t),cos(7t)\frac{8s}{(s^2 + 49)(s^2 + 1)} \rightarrow \text{Inverse is a linear combination of} \cos(t), \cos(7t)
Partial fraction decomposition: 8s(s2+1)(s2+49)=As+Bs2+1+Cs+Ds2+49\frac{8s}{(s^2 + 1)(s^2 + 49)} = \frac{As + B}{s^2 + 1} + \frac{Cs + D}{s^2 + 49}
Solving yields: 8s(s2+1)(s2+49)=−1/6s2+1+49/6s2+49\frac{8s}{(s^2 + 1)(s^2 + 49)} = \frac{-1/6}{s^2 + 1} + \frac{49/6}{s^2 + 49}
So, X(s)=(−16+1)ss2+1+496⋅ss2+49X(s) = \left( -\frac{1}{6} + 1 \right)\frac{s}{s^2 + 1} + \frac{49}{6} \cdot \frac{s}{s^2 + 49} X(s)=56⋅ss2+1+496⋅ss2+49X(s) = \frac{5}{6} \cdot \frac{s}{s^2 + 1} + \frac{49}{6} \cdot \frac{s}{s^2 + 49}
Final Answer:
x(t)=56cos(t)+496cos(7t)x(t) = \frac{5}{6} \cos(t) + \frac{49}{6} \cos(7t)
Explanation
The Laplace transform is a powerful tool used to solve linear ordinary differential equations (ODEs), especially those with given initial conditions. This method transforms the ODE from the time domain into the complex frequency domain, where algebraic techniques can be applied. Once the equation is solved in the frequency domain, the inverse Laplace transform is used to return to the time domain.
In the given problem, we are solving the second-order differential equation: x′′+x=8cos(7t),x(0)=1,x′(0)=0x” + x = 8 \cos(7t), \quad x(0) = 1, \quad x'(0) = 0
We begin by taking the Laplace transform of both sides. The transforms of the derivatives incorporate the initial conditions, simplifying the solution process. This reduces the differential equation to an algebraic equation in terms of X(s)X(s), the Laplace transform of x(t)x(t).
After manipulating the equation and combining terms, we solve for X(s)X(s). To simplify this further, we perform partial fraction decomposition. This allows us to break the rational expression into standard forms whose inverse transforms are known, particularly those involving cos(at)\cos(at), which correspond to terms like ss2+a2\frac{s}{s^2 + a^2}.
Taking the inverse Laplace transform of each component results in a linear combination of cosine functions. The final solution, expressed in terms of real-valued trigonometric functions, is: x(t)=56cos(t)+496cos(7t)x(t) = \frac{5}{6} \cos(t) + \frac{49}{6} \cos(7t)
This solution satisfies both the differential equation and the initial conditions. The Laplace transform method is especially effective for nonhomogeneous equations with inputs like sine, cosine, or exponential functions.
