Chapter 16: Problem 6
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
Expert verified
The definite solution is \(y(t) = 2e^{3t} - t e^{3t} + 2\).
Step by step solution
01
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 \(y'(0) = 5\).
02
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 \(r\).
03
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.
04
Write the Complementary Solution
Since there is a repeated root \(r = 3\), the complementary solution \(y_c(t)\) is:\[y_c(t) = c_1 e^{3t} + c_2 t e^{3t}\]
05
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 \(A\).
06
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, \(y_p(t) = 2\).
07
Write the General Solution
Combine the complementary solution \(y_c(t)\) and the particular solution \(y_p(t)\):\[y(t) = c_1 e^{3t} + c_2 t e^{3t} + 2\]
08
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.
09
Step 6.1: Apply \(y(0) = 4\)
Substitute into \(y(t)\):\[y(0) = c_1 e^{0} + c_2 \cdot 0 \cdot e^{0} + 2 = 4\]\[c_1 + 2 = 4 \quad \Rightarrow \quad c_1 = 2\]
10
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 \(y'(0) = 5\):\[y'(0) = 3c_1 + c_2 = 5\]\[3(2) + c_2 = 5\]\[6 + c_2 = 5 \quad \Rightarrow \quad c_2 = -1\]
11
Write the Definite Solution
Substitute \(c_1\) and \(c_2\) into the general solution to find the definite solution:\[y(t) = 2e^{3t} - t e^{3t} + 2\]
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
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.
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.
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.
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.
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
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
- \(y_c(t) = c_1 e^{3t} + c_2 t e^{3t}\)
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 \(c_2\).
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 \(c_2 = -1\).