Chapter 39: Problem 9
Solve the following equations: \(\frac{\mathrm{d}^{2} y}{\mathrm{~d} x^{2}}-2 \frac{\mathrm{d} y}{\mathrm{~d} x}+3 y=x^{2}-1\)
Short Answer
Expert verified
The general solution to the given second-order, linear differential equation is: \(y(x) = e^x (C_1 \cos{\sqrt{2}x} + C_2 \sin{\sqrt{2}x}) + \frac{1}{3}x^2 + \frac{4}{3}x - \frac{1}{3}\).
Step by step solution
01
Identify the type of differential equation
This is a second-order, linear differential equation with non-constant coefficients due to the presence of the x² term. The given equation is:
\(\frac{d^2y}{dx^2} - 2\frac{dy}{dx} + 3y = x^2 - 1\)
02
Rewrite the equation as a homogeneous and non-homogeneous equation
We will rewrite the differential equation in two parts: the homogeneous part and the non-homogeneous part.
Homogeneous part: \(\frac{d^2y}{dx^2} - 2\frac{dy}{dx} + 3y = 0\)
Non-homogeneous part: \(\frac{d^2y}{dx^2} - 2\frac{dy}{dx} + 3y = x^2 - 1\)
03
Solve the homogeneous equation
To solve the homogeneous part of the differential equation, we will use the \(\textit{Characteristic Equation Method}\). We have:
\[r^2 - 2r + 3 = 0\]
We solve this quadratic equation for r:
The discriminant Δ = b² - 4ac = (-2)² - 4(1)(3) = 4 - 12 = -8
Since the discriminant is negative, the solutions of the characteristic equation will be complex. The complex roots are:
\[r_{1,2} = \frac{-(-2) \pm \sqrt{-8}}{2(1)} = 1 \pm i\sqrt{2}\]
Thus, the general solution of the homogeneous equation is given by:
\(y_h(x) = e^x (C_1 \cos{\sqrt{2}x} + C_2 \sin{\sqrt{2}x}) \)
where \(C_1\) and \(C_2\) are arbitrary constants.
04
Solve the non-homogeneous equation using the Variation of Parameters method
Now, we will solve the non-homogeneous part using the Variation of Parameters method. We assume a particular solution in the form:
\[y_p(x) = Ax^2 + Bx + C\]
To find the coefficients A, B, and C, we will need to differentiate \(y_p(x)\) twice and plug it into the non-homogeneous equation:
\[\frac{dy_p}{dx} = 2Ax + B\]
\[\frac{d^2y_p}{dx^2} = 2A\]
Plug these into the non-homogeneous equation:
\[(2A) - 2(2Ax + B) + 3(Ax^2 + Bx + C) = x^2 - 1\]
Now, compare the coefficients of like terms:
Coefficients of \(x^2\): \(3A = 1 \Rightarrow A = \frac{1}{3}\)
Coefficients of \(x\): \(- 4A + 3B = 0 \Rightarrow B = \frac{4}{3}\)
Constant terms: \(2A - 2B + 3C = -1 \Rightarrow C = -\frac{1}{3}\)
Now, we have the particular solution: \(y_p(x) = \frac{1}{3}x^2 + \frac{4}{3}x - \frac{1}{3}\)
05
Combine the homogeneous and particular solutions
Finally, combine the homogeneous and particular solutions to get the general solution of the given differential equation:
\(y(x) = y_h(x) + y_p(x) = e^x (C_1 \cos{\sqrt{2}x} + C_2 \sin{\sqrt{2}x}) + \frac{1}{3}x^2 + \frac{4}{3}x - \frac{1}{3}\)
This is the general solution to the given second-order, linear differential equation.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
linear homogeneous equation
A linear homogeneous equation in the context of differential equations is a type where every term is a function of the dependent variable and its derivatives, with no standalone constant terms. In a second-order linear homogeneous differential equation, it typically takes the form:
The homogeneous equation portion we derived from the original problem is given by: \[\frac{d^2y}{dx^2} - 2\frac{dy}{dx} + 3y = 0\]
Solving this involves finding the roots of the characteristic equation associated with the differential equation. The characteristic equation results from assuming a solution of the form \( y = e^{rx} \), leading us to solve the polynomial equation for \( r \). Understanding homogeneous equations is vital as they form the basis for analyzing more complicated non-homogeneous equations.
- \( a\frac{d^2 y}{dx^2} + b\frac{dy}{dx} + cy = 0 \)
The homogeneous equation portion we derived from the original problem is given by: \[\frac{d^2y}{dx^2} - 2\frac{dy}{dx} + 3y = 0\]
Solving this involves finding the roots of the characteristic equation associated with the differential equation. The characteristic equation results from assuming a solution of the form \( y = e^{rx} \), leading us to solve the polynomial equation for \( r \). Understanding homogeneous equations is vital as they form the basis for analyzing more complicated non-homogeneous equations.
variation of parameters
The variation of parameters is a powerful technique used to solve non-homogeneous differential equations. Consider a situation where the equation is of the form:
To implement this method, first solve the homogeneous version of the equation to get the fundamental set of solutions, which are then used to construct a particular solution to the non-homogeneous equation. The particular solution comes in the form of a new set of functions, \(v_1(x)\) and \(v_2(x)\), modifying the homogeneous solution:
\[y_h(x) = C_1 y_1(x) + C_2 y_2(x)\]
The key idea is to allow the constants \(C_1\) and \(C_2\) to vary with \(x\), which turns them into functions that can absorb \(f(x)\). Finally, using the original solutions \(y_1(x)\) and \(y_2(x)\), and inserting them into a set of equations derived using integration, you find \(v_1(x)\) and \(v_2(x)\), thus discovering the particular solution \(y_p(x)\). By combining \(y_h(x)\) and \(y_p(x)\), you get the complete solution.
- \(\frac{d^2y}{dx^2} - 2\frac{dy}{dx} + 3y = f(x) \)
To implement this method, first solve the homogeneous version of the equation to get the fundamental set of solutions, which are then used to construct a particular solution to the non-homogeneous equation. The particular solution comes in the form of a new set of functions, \(v_1(x)\) and \(v_2(x)\), modifying the homogeneous solution:
\[y_h(x) = C_1 y_1(x) + C_2 y_2(x)\]
The key idea is to allow the constants \(C_1\) and \(C_2\) to vary with \(x\), which turns them into functions that can absorb \(f(x)\). Finally, using the original solutions \(y_1(x)\) and \(y_2(x)\), and inserting them into a set of equations derived using integration, you find \(v_1(x)\) and \(v_2(x)\), thus discovering the particular solution \(y_p(x)\). By combining \(y_h(x)\) and \(y_p(x)\), you get the complete solution.
complex roots
When solving linear homogeneous equations, you may encounter complex roots while dealing with characteristic equations of the form \( ax^2 + bx + c = 0 \). A complex root arises when the discriminant \( b^2 - 4ac \) is negative. In such cases, the roots are not real numbers but instead take on the form \( m \pm ni \), where \( i \) is the imaginary unit satisfying \( i^2 = -1 \).
Complex roots generate solutions that involve exponential terms mixed with sine and cosine functions. Specifically, if you have roots \( r = \. \pm bi \), the solutions can be expressed as:
\[y_h(x) = e^x (C_1 \cos{\sqrt{2}x} + C_2 \sin{\sqrt{2}x})\]
The appearance of complex roots indicates an oscillatory behavior in the solution, essential for understanding phenomena that exhibit wave-like properties. Understanding this behavior helps in applying differential equations to real-world physical systems that involve periodicity and damping.
Complex roots generate solutions that involve exponential terms mixed with sine and cosine functions. Specifically, if you have roots \( r = \. \pm bi \), the solutions can be expressed as:
- \( y(x) = e^{mx} \left( C_1 \cos{bx} + C_2 \sin{bx} \right) \)
\[y_h(x) = e^x (C_1 \cos{\sqrt{2}x} + C_2 \sin{\sqrt{2}x})\]
The appearance of complex roots indicates an oscillatory behavior in the solution, essential for understanding phenomena that exhibit wave-like properties. Understanding this behavior helps in applying differential equations to real-world physical systems that involve periodicity and damping.