Question

Find the solution of the initial-value problem $$ \begin{array}{l} y^{\prime \prime}+2 y^{\prime}+10 y=h(t), \\ y(0)=0, \quad y^{\prime}(0)=0, \end{array} $$ where $$ h(t)=\left\\{\begin{array}{ll} 1, & 0

Short answer

Answer: The general solution is $$y(t) = \left\\{\begin{array}{ll} \frac{1}{10}, & 0<t<1 \\\ 0, & 1<t<2 \end{array}\right.$$

Step-by-step solution

1. Finding the Complementary Solution

To find the complementary solution, we solve the homogeneous equation: $$y'' + 2y' + 10y = 0$$ This is a linear differential equation with constant coefficients. We start by finding the characteristic equation: $$r^2 + 2r + 10 = 0$$ Using the quadratic formula, the roots are: $$r = \frac{-2 \pm \sqrt{(-2)^2 - 4(1)(10)}}{2(1)} = -1 \pm 3i$$ Since we have complex roots, the complementary solution is given by: $$y_c(t) = e^{-t} (c_1 \cos(3t) + c_2 \sin(3t))$$

2. Finding the Particular Solution

Given that h(t) is a piecewise function: $$h(t) = \left\\{\begin{array}{ll} 1, & 0<t<1 \\\ 0, & 1<t<2 \end{array}\right.$$ For \(0<t<1\), the differential equation is: $$y'' + 2y' + 10y = 1$$ We assume a particular solution of the form: $$y_p(t) = A$$ Taking derivatives, $$y_p'(t) = 0$$ $$y_p''(t) = 0$$ Substituting into the differential equation, we get: $$0 + 0 + 10A = 1$$ Solving, we find: $$A = \frac{1}{10}$$ So, for \(0<t<1\), the particular solution is: $$y_p(t) = \frac{1}{10}$$ For \(1<t<2\), the differential equation is: $$y'' + 2y' + 10y = 0$$ As this is the same as the homogeneous equation, the particular solution for this interval is: $$y_p(t) = 0$$

3. Applying Initial Conditions

We can now apply the initial conditions \(y(0)=0\) and \(y'(0)=0\) to the complementary solution: $$y_c(0) = e^0 (c_1 \cos(0) + c_2 \sin(0)) = c_1 = 0$$ $$y_c'(0) = -e^0 (c_1 \cos(3\cdot 0) + c_2 \sin(3\cdot 0)) + 3e^0 (c_2 \cos(3\cdot 0) - c_1 \sin(3\cdot 0)) = 3c_2 = 0$$ $$c_2 = 0$$ This means the complementary solution is simply \(y_c(t) = 0\).

4. General Solution

Combining the complementary and particular solutions, we get the general solution for the given piecewise function: $$y(t) = y_c(t) + y_p(t) = \left\\{\begin{array}{ll} \frac{1}{10}, & 0<t<1 \\\ 0, & 1<t<2 \end{array}\right.$$

Key concepts

Complementary Solution

When solving a differential equation, the complementary solution is crucial. It addresses the homogeneous part of the equation where the right-hand side is set to zero. This part of the solution accounts for the natural behavior of the system without any external inputs. In our problem, we begin with the homogeneous differential equation: \(y'' + 2y' + 10y = 0\). To solve this, we find the characteristic equation by substituting potential solutions of the form \(y = e^{rt}\). This results in \(r^2 + 2r + 10 = 0\). Using the quadratic formula, we solve for \(r\) and find the roots are complex: \(r = -1 \pm 3i\). The complementary solution is thus: \(y_c(t) = e^{-t} (c_1 \cos(3t) + c_2 \sin(3t))\). This represents oscillatory behavior modulated by an exponential decay, which is typical for systems with complex roots. Understanding the complementary solution helps set the stage for finding how the system responds to initial conditions and external forces.

Particular Solution

The particular solution is the component of the solution that directly corresponds to the external forces acting on the system. In our problem, the forcing term is a piecewise function \(h(t)\) that changes its value across different intervals. For \(0 < t < 1\), \(h(t) = 1\), which modifies the given differential equation to \(y'' + 2y' + 10y = 1\). Here, we guess a particular solution of the form \(y_p(t) = A\), where \(A\) is a constant. Substituting this into the differential equation leads to \(10A = 1\), thus \(A = \frac{1}{10}\). For \(1 < t < 2\), since \(h(t) = 0\), the particular solution coincides with the complementary solution and becomes \(y_p(t) = 0\). Understanding the particular solution is essential as it shows how the system specifically responds to unique external inputs on different intervals.

Piecewise Function

Piecewise functions are defined by different expressions depending on the interval of the input. In the problem, \(h(t)\) is a piecewise function defined as: \(\{\{1, & 0 < t < 1\ \;0, & 1 < t < 2\}\}\). This means the function has different constant values over different ranges of \(t\). Piecewise functions are particularly useful in modeling situations where values change abruptly, like a switch turning on or off, which is relevant in real-world applications.When dealing with differential equations, piecewise functions can segment the problem into more manageable portions, allowing you to solve the equation separately for each piece. Then, the overall solution stitches these pieces together, accounting for any condition changes at their boundaries. For our problem, understanding the implementation of piecewise functions simplifies finding solutions within their defined intervals.

Quadratic Formula

The quadratic formula is a tool for solving second-degree polynomial equations of the form \(ax^2 + bx + c = 0\). It provides the roots by:\(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\). In our differential equation, the characteristic equation, \(r^2 + 2r + 10 = 0\), is quadratic. Applying the quadratic formula: \(r = \frac{-2 \pm \sqrt{(-2)^2 - 4 \times 1 \times 10}}{2 \times 1}\), we find the roots are \(-1 \pm 3i\). These complex roots result in a complementary solution involving sinusoidal functions. Deciphering roots using the quadratic formula is crucial for understanding how different types of systems behave, especially when considering oscillatory components decaying exponentially. The quadratic formula is an indispensable tool for tackling quadratic expressions in a straightforward manner.