Question

(a) Verify that the given function, \(y_{P}(t)\), is a particular solution of the differential equation. (b) Determine the complementary solution, \(y_{C}(t)\). (c) Form the general solution and impose the initial conditions to obtain the unique solution of the initial value problem. $$y^{\prime \prime}-2 y^{\prime}+y=e^{t}, \quad y(0)=-2, \quad y^{\prime}(0)=2, \quad y_{P}(t)=\frac{1}{2} t^{2} e^{t}$$

Short answer

Question: Determine the unique solution of the second-order linear differential equation \(y'' - 2y' + y = e^t\) satisfying the initial conditions \(y(0) = -2\) and \(y'(0) = 2\). Given that \(y_{P}(t) = \frac{1}{2}t^{2}e^{t}\), verify the particular solution and find the complementary solution. Answer: The unique solution of the given second-order linear differential equation with the given initial conditions is \(y(t) = -2 e^{t} + 4t e^{t} + \frac{1}{2}t^2 e^{t}\).

Step-by-step solution

Verify the particular solution

Let's check if \(y_{P}(t)\) is a valid particular solution for the given differential equation. We need to calculate \(y^{\prime \prime}_P(t)-2 y^{\prime}_P(t)+y_P(t)\) and see if it equals \(e^{t}\). The given function is \(y_{P}(t)=\frac{1}{2}t^{2}e^{t}\). First, differentiate \(y_{P}(t)\) to get \(y^{\prime}_P(t)\): $$y^{\prime}_P(t) =\frac{d}{dt}(\frac{1}{2}t^{2}e^{t})=t e^{t} + \frac{1}{2}t^{2}e^{t} $$ Next, differentiate \(y^{\prime}_P(t)\) to get \(y^{\prime \prime}_P(t)\): $$y^{\prime \prime}_P(t) =\frac{d}{dt}(t e^{t} + \frac{1}{2}t^{2}e^{t})= (2t+1)e^{t}+\frac{1}{2}t^{2}e^{t}$$ Now, plug these values into the differential equation to verify if we get \(e^{t}\): $$y^{\prime \prime}_P(t)- 2 y^{\prime}_P(t)+y_P(t)=(2t+1)e^{t}+\frac{1}{2}t^{2}e^{t} - 2(te^{t}+\frac{1}{2}t^2 e^{t}) + \frac{1}{2}t^{2}e^{t} = e^{t}$$ Since the expression is equal to \(e^t\), \(y_{P}(t)\) is indeed a particular solution.

Find the complementary solution

We need to find the complementary solution, \(y_{C}(t)\), by solving the homogeneous equation of the given differential equation: $$y^{\prime \prime}-2 y^{\prime}+y=0$$ This is a linear homogeneous differential equation with constant coefficients, so we can solve it using the characteristic equation: $$r^2-2r+1=0$$ Solve this quadratic equation to find the roots: The roots are \(r=1\) (same root, with multiplicity 2). So the complementary solution is in the form: $$y_{C}(t) = C_1 e^{t} + C_2 t e^{t} $$

Obtain the general solution

Now that we have \(y_{C}(t)\) and \(y_{P}(t)\), add these two to form the general solution \(y(t)\): $$y(t) = y_C(t) + y_P(t) = C_1 e^{t} + C_2 t e^{t} + \frac{1}{2}t^2 e^{t} $$

Apply the initial conditions

To find the unique solution, apply the given initial conditions: \(y(0) = -2\) and \(y^{\prime}(0) = 2\): 1. For \(y(0)=-2\): $$-2 = C_1 e^{0} + C_2 (0) e^{0} + \frac{1}{2}(0)^2 e^{0} \Rightarrow -2 = C_1$$ 2. For \(y^{\prime}(0)=2\): First, let's find the derivative of \(y(t)\): $$y^{\prime}(t) = \frac{d}{dt}(C_1 e^{t} + C_2 t e^{t} + \frac{1}{2}t^2 e^{t})= C_1 e^{t} + C_2 (t+1)e^{t} + t^2e^{t}$$ Now, apply the initial condition \(y^{\prime}(0) = 2\): $$2 = C_1 e^{0} + C_2 (0+1)e^{0} + (0)^2 e^{0} \Rightarrow 2 = C_1 +C_2$$ Substitute C_1=-2: $$2 = -2 + C_2 \Rightarrow C_2 = 4$$ So the unique solution is: $$y(t) = -2 e^{t} + 4t e^{t} + \frac{1}{2}t^2 e^{t}$$

Key concepts

Particular Solution Verification

When solving a differential equation, particularly, a non-homogeneous differential equation, it's crucial to find a solution that incorporates the specific structure of the non-homogeneity. This is known as the particular solution, denoted typically as \(y_P(t)\).

Verifying that a function is the particular solution involves checking if it satisfies the original differential equation when substituted in place of the dependent variable. Here’s how this verification works: you must first calculate its derivatives, plug them into the equation, and see if it simplifies to the non-homogeneous term. In our given problem, the non-homogeneous term is \(e^t\), and the proposed particular solution \(y_P(t) = \frac{1}{2} t^2 e^t\) must, after plugging in its derivatives, simplify to that term. Successful verification suggests that the chosen \(y_P(t)\) is indeed part of the family of functions solving the differential equation.

Complementary Solution

Another essential aspect of solving non-homogeneous differential equations is finding the complementary solution, often denoted as \(y_C(t)\). It is associated with the homogeneous part of the equation—when the non-homogeneous term is set to zero. The complementary solution represents the general solution of this homogeneous equation.

To find \(y_C(t)\), one must solve the characteristic polynomial arising from the homogeneous equation. The roots of this polynomial dictate the form of \(y_C(t)\). In the case of real, repeated roots—as we have from \(r^2 - 2r + 1 = 0\) resulting in a double root at \(r = 1\)—the complementary solution is a combination of exponentials multiplied by the powers of \(t\), thus \(y_C(t) = C_1 e^t + C_2 t e^t\). These exponential terms reflect the behavior of the system's response when no external force (non-homogeneity) is applied.

General Solution

The general solution of a non-homogeneous differential equation is the sum of the particular and complementary solutions. It incorporates both the steady state achieved under the influence of the non-homogeneous part and the natural response described by the homogeneous equation.

The general formula for this solution is given by \(y(t) = y_C(t) + y_P(t)\). In our exercise, after obtaining the particular solution \(y_P(t)\) and the complementary solution \(y_C(t)\), we combined them to get the complete picture of the system's behavior. The resulting general solution furnishes a family of functions, and it's complete in the sense of capturing all possible scenarios described by the differential equation.

Applying Initial Conditions

To pinpoint a specific solution among the various possibilities presented by the general solution, we use initial conditions. Initial conditions provide values for the function and often its derivatives at a particular point, usually \(t = 0\). Applying these conditions allows us to solve for the arbitrary constants in the general solution.

For the exercise at hand, the initial conditions are \(y(0) = -2\) and \(y'(0) = 2\). We use these conditions to find the values of \(C_1\) and \(C_2\) in the general solution. By plugging these constants back into the general solution, we arrive at the unique solution that not only satisfies the differential equation but also passes through the specified points at \(t = 0\). This unique solution represents the specific state of the system at the outset of the problem, tailoring the general formula to a particular scenario.