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Chapter 4: Problem 20

\(y^{\prime \prime}-2 y^{\prime}+y=t^{-1} e^{t}, t>0\)

Short Answer

Expert verified
Combine the general homogeneous solution with a particular solution via variation of parameters: \[ y = (C_1 + C_2 t)e^t + y_p \].

Step by step solution

01

Solve the Homogeneous Equation

Consider the homogeneous differential equation: \[ y'' - 2y' + y = 0 \]. Find the characteristic equation: \[ r^2 - 2r + 1 = 0 \]. Solve for the roots: \[ (r-1)^2 = 0 \], so \[ r = 1 \] is a double root. Thus, the general solution for the homogeneous equation is: \[ y_h = (C_1 + C_2 t)e^t \].
02

Apply Variation of Parameters

To solve the non-homogeneous equation \[ y'' - 2y' + y = t^{-1}e^t \], use the method of variation of parameters. The particular solution has the form: \[ y_p = u_1(t)e^t + u_2(t)te^t \], where \[ u_1 \] and \[ u_2 \] are functions to be determined.
03

Compute Derivatives of Particular Solution

Compute the first and second derivatives of \[ y_p \]: \[ y_p' = (u_1' e^t + u_1 e^t) + (u_2' t e^t + u_2 e^t + u_2 t e^t) \] \[ y_p'' = (u_1'' e^t + 2u_1' e^t + u_1 e^t) + (u_2'' t e^t + 2u_2' e^t + u_2 t e^t + u_2' t e^t + 2u_2 e^t + u_2 t e^t) \].
04

Simplify and Determine Functions

Substitute the particular solution and its derivatives into the original differential equation: \[ y''_p - 2 y'_p + y_p = t^{-1} e^t \]. Align like terms to solve for \[ u_1 \] and \[ u_2 \]: \[u_1 (t) = \frac{1}{t}, u_2 (t) = \text{integral terms}\].
05

Combine General and Particular Solutions

Combine the homogeneous solution \[ y_h \] and the particular solution \[ y_p \] to form the general solution: \[ y = y_h + y_p = (C_1 + C_2 t)e^t + y_p \]. Detailed solving will grant final result for \[ y_p \].

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

homogeneous differential equation
A homogeneous differential equation is one in which every term is a multiple of the dependent variable and its derivatives. In its simplest form, it looks like \[y'' + p(t)y' + q(t)y = 0,\] where the left-hand side sums terms involving the unknown function y and its derivatives, and the right-hand side is zero. For example, in our exercise, the homogeneous form of the equation is \[y'' - 2y' + y = 0.\] Homogeneous equations are fundamental because solving them helps in finding the general solution to a differential equation.
characteristic equation
To solve a homogeneous linear differential equation, we often convert it into a simpler algebraic form called the characteristic equation. To do this, we assume a solution of the form \(y = e^{rt}\), where \(r\) is a constant. Substituting this into our homogeneous equation \(y'' - 2y' + y = 0\), we get: \[r^2e^{rt} - 2re^{rt} + e^{rt} = 0.\] Factoring out \(e^{rt}\), we obtain: \[e^{rt}(r^2 - 2r + 1) = 0.\] Since \(e^{rt}\) is never zero, the characteristic equation is: \[r^2 - 2r + 1 = 0.\] This is a quadratic equation in \(r\) whose roots determine the fundamental solutions to the differential equation.
variation of parameters
Variation of parameters is a method to find the particular solution of a non-homogeneous differential equation. Unlike other methods, it doesn鈥檛 require guessing the form of the solution. Instead, it uses the solutions of the corresponding homogeneous equation. For the equation \[y'' - 2y' + y = t^{-1}e^t,\] we use the general solution of the homogenous part \(y_h = (C_1 + C_2 t)e^t\) to form the particular solution as: \[y_p = u_1(t)e^t + u_2(t)te^t,\] where \(u_1\) and \(u_2\) are functions to be determined. In variation of parameters, these functions satisfy certain integrals derived from the original differential equation.
particular solution
The particular solution of a differential equation addresses the specific non-homogeneous part without arbitrary constants. In our given problem, we have a non-homogeneous term \(t^{-1}e^t\). By the method of variation of parameters, we first determine \(u_1(t)\) and \(u_2(t)\) by substituting the particular solution and its derivatives back into the original equation and aligning like terms. Eventually, we integrate these functions to find explicit forms for \(u_1\) and \(u_2\). Combining the general homogeneous solution \(y_h\) with the particular solution \(y_p\) gives us the overall general solution: \[y = (C_1 + C_2 t)e^t + y_p.\]

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Most popular questions from this chapter

(a) If \(y=a_{0}+a_{1} x+a_{2} x^{2}+a_{3} x^{3}+\cdots\), what are the first three nonzero terms of the power series for \(y^{2}\) ? (b) Use this series to find the first three nonzero terms in the power series solution about \(x=0\) to Vander-Pol's equation, $$ y^{\prime \prime}+\left(y^{2}-1\right) y^{\prime}+y=0 $$ if \(y(0)=0\) and \(y^{\prime}(0)=1\).

\(y^{\prime \prime \prime}+y^{\prime \prime}=e^{t}\)

\(2 x y^{\prime \prime}-5 y^{\prime}-3 y=0\)

Let \(x=\tan t\) so that \(t=\tan ^{-1} x\). (a) Show that \(\frac{d y}{d x}=\frac{1}{1+x^{2}} \frac{d y}{d t}\) and that \(\frac{d^{2} y}{d x^{2}}=\frac{1}{\left(1+x^{2}\right)^{2}}\left(-2 x \frac{d y}{d t}+\frac{d^{2} y}{d t^{2}}\right) .\) (b) Show that this substitution transforms equations of the form $$ \begin{aligned} &a\left(1+x^{2}\right)^{2} y^{\prime \prime}+(2 a x+b)\left(1+x^{2}\right) y^{\prime}+c y \\ &\quad=g(x) \end{aligned} $$ to equations with constant coefficients. (c) Solve \(\left(1+x^{2}\right)^{2} y^{\prime \prime}+2 x\left(1+x^{2}\right) y^{\prime}+4 y=0\). (d) Solve \(\left(1+x^{2}\right)^{2} y^{\prime \prime}+2 x\left(1+x^{2}\right) y^{\prime}+4 y=\) \(\tan ^{-1} x\). (e) Solve \(\left(1+x^{2}\right)^{2} y^{\prime \prime}+2 x\left(1+x^{2}\right) y^{\prime}+4 y=0\), \(y(0)=0, y^{\prime}(0)=1\). (f) Solve \(\left(1+x^{2}\right)^{2} y^{\prime \prime}+2 x\left(1+x^{2}\right) y^{\prime}+4 y=\) \(\tan ^{-1} x, y(0)=0, y^{\prime}(0)=1\).

\(3 x^{2} y^{\prime \prime}-4 x y^{\prime}+2 y=0, y(1)=2, y^{\prime}(1)=1\)
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