Question

Find the \(\gamma_{p}\) and the \(y_{c \prime}\) the general solution, and the definite solution of each of the following: $$2 y^{\prime \prime}(t)-12 y^{\prime}(t)+20 y=40 ; y(0)=4, y^{\prime}(0)=5$$

Short answer

The definite solution is \(y(t) = 2e^{3t} - t e^{3t} + 2\).

Step-by-step solution

Identify the Problem Type

The given equation is a second-order linear homogeneous differential equation with constant coefficients and a forcing function (non-homogeneous part) on the right-hand side: \(2y''(t) - 12y'(t) + 20y = 40\). We are also given initial conditions \(y(0) = 4\) and \(y'(0) = 5\).

Find the Complementary Solution

Start by solving the homogeneous part of the equation, \(2y''(t) - 12y'(t) + 20y = 0\). This involves finding the characteristic equation: \(2r^2 - 12r + 20 = 0\). Solve this quadratic equation for \(r\).

Step 2.1: Solve the Characteristic Equation

Using the quadratic formula \(r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 2\), \(b = -12\), and \(c = 20\), calculate \(r\): \[r = \frac{12 \pm \sqrt{(-12)^2 - 4 \times 2 \times 20}}{4}\] Solve to find \(r = 3\). This is a repeated root.

Write the Complementary Solution

Since there is a repeated root \(r = 3\), the complementary solution \(y_c(t)\) is:\[y_c(t) = c_1 e^{3t} + c_2 t e^{3t}\]

Find the Particular Solution

Assume a particular solution \(y_p(t)\) for the equation \(2y'' - 12y' + 20y = 40\). Since the non-homogeneous term is a constant (40), assume \(y_p(t) = A\) where \(A\) is a constant and substitute back into the differential equation and solve for \(A\).

Step 4.1: Substitute and Solve for A

Substitute \(y_p(t) = A\) into the equation, so that:\[2(0) - 12(0) + 20A = 40\]\[20A = 40 \rightarrow A = 2\]Thus, \(y_p(t) = 2\).

Write the General Solution

Combine the complementary solution \(y_c(t)\) and the particular solution \(y_p(t)\):\[y(t) = c_1 e^{3t} + c_2 t e^{3t} + 2\]

Apply Initial Conditions

Use the initial conditions \(y(0) = 4\) and \(y'(0) = 5\) to find \(c_1\) and \(c_2\). Substitute \(t = 0\) into the general solution and its derivative, then solve for the constants.

Step 6.1: Apply \(y(0) = 4\)

Substitute into \(y(t)\):\[y(0) = c_1 e^{0} + c_2 \cdot 0 \cdot e^{0} + 2 = 4\]\[c_1 + 2 = 4 \quad \Rightarrow \quad c_1 = 2\]

Step 6.2: Derivative and Apply \(y'(0) = 5\)

Find the derivative of \(y(t)\), \(y'(t) = 3c_1 e^{3t} + c_2 e^{3t} + 3c_2 t e^{3t}\).Substitute into \(y'(0) = 5\):\[y'(0) = 3c_1 + c_2 = 5\]\[3(2) + c_2 = 5\]\[6 + c_2 = 5 \quad \Rightarrow \quad c_2 = -1\]

Write the Definite Solution

Substitute \(c_1\) and \(c_2\) into the general solution to find the definite solution:\[y(t) = 2e^{3t} - t e^{3t} + 2\]

Key concepts

Homogeneous Equations

Homogeneous equations are a type of differential equation where the function equals zero. They usually take the form of a linear differential equation such as \(ay'' + by' + cy = 0\), where the constants \(a\), \(b\), and \(c\) will create specific conditions for the solution.

In our example, the homogeneous part of the given equation is \(2y''(t) - 12y'(t) + 20y = 0\). This is a second-order polynomial equation with constant coefficients. Solving such equations typically involves finding the characteristic equation. For our problem, this step converted it into a quadratic equation \(2r^2 - 12r + 20 = 0\), which provides roots indicating the nature of the solution.

• If the roots are real and distinct, the solutions are two separate exponentials.
• If the roots are real and repeated, as in our case where \(r=3\), the solution contains a term with \(t\) multiplying the exponential.
• If the roots are complex, the solution will involve sine and cosine terms.

Finding the solution to the homogeneous equation provides the 'framework' or 'model' around which other solutions, like particular solutions, are built.

Particular Solution

The particular solution is an individual solution to the non-homogeneous differential equation. It specifically satisfies the entire equation, including the non-zero "forcing" term.

In our example, the expression \(2y'' - 12y' + 20y = 40\) features the constant 40 as the forcing term, indicating that this is a non-homogeneous differential equation. Unlike the homogeneous counterpart where the equation equals zero, here we need a solution that accounts for the additional constant term.

• For equations with a constant forcing term, a common assumption for the particular solution is that it is also a constant, say \(A\).
• Upon substitution into the differential equation, if we assume \(y_p(t) = A\), it leads to solving a simple algebraic equation.

For our problem, it turned out that \(A = 2\), resulting in the particular solution \(y_p(t) = 2\). This addresses the specific needs of the forcing term, shifting the solution away from the 'net zero' of the homogeneous case.

Complementary Solution

The complementary solution, represented as \(y_c(t)\), forms a crucial part of the total solution to the differential equation. This solution comes directly from solving the homogeneous equation and forms the basis of any variation due to non-homogeneity.

In the given equation \(2y''(t) - 12y'(t) + 20y = 0\), we solve for the roots of the characteristic equation, \(2r^2 - 12r + 20 = 0\). A repeated root \(r = 3\) signifies that the exponential term will combine with a polynomial multiplicand (usually \(t\)) to give

• \(y_c(t) = c_1 e^{3t} + c_2 t e^{3t}\)

This approach accounts for all solutions of the homogeneous counterpart, as it describes all possible behaviors modeled by the differential equation without any external influences. The idea is to capture the intrinsic nature of the system's responses.

Initial Conditions

Initial conditions are specific values given at the beginning of the scenario we’re observing, often before the variable has had a chance to change significantly. They are crucial for determining particular solutions that satisfy not just the differential equation but also any real-world constraints or starting points.

In the example, the initial conditions are \(y(0) = 4\) and \(y'(0) = 5\). These are applied to the general solution composed of the complementary and particular solutions. By substituting \(t = 0\) into \(y(t) = c_1 e^{3t} + c_2 t e^{3t} + 2\) along with its derivative, we solve for unknowns \(c_1\) and \(c_2\).

• Using \(y(0) = 4\), solve: \(c_1 + 2 = 4\) leading to \(c_1 = 2\).
• Using \(y'(0) = 5\), solve the derivative’s equation: \(3c_1 + c_2 = 5\), finding \(c_2 = -1\).

What the initial conditions give us is specificity. By applying them, we tailor a general form into the exact solution needed to match the system's precise state or behavior at the commencement of observation.