Chapter 5: Problem 10
In Exercises \(1-12\) find the general solution. $$ y^{\prime \prime}+6 y^{\prime}+13 y=0 $$
Short Answer
Step by step solution
Key Concepts
These are the key concepts you need to understand to accurately answer the question.
/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none}
Learning Materials
Features
Discover
Chapter 5: Problem 10
In Exercises \(1-12\) find the general solution. $$ y^{\prime \prime}+6 y^{\prime}+13 y=0 $$
These are the key concepts you need to understand to accurately answer the question.
All the tools & learning materials you need for study success - in one app.
Get started for free
Find the general solution, given that \(y_{1}\) satisfies the complementary equation. As a byproduct, find a fundamental set of solutions of the complementary equation. \((2 x+1) y^{\prime \prime}-2 y^{\prime}-(2 x+3) y=(2 x+1)^{2} ; \quad y_{1}=e^{-x}\)
(a) Verify that if $$ y_{p}=A(x) \cos \omega x+B(x) \sin \omega x $$ where \(A\) and \(B\) are twice differentiable, then $$ \begin{aligned} y_{p}^{\prime} &=\left(A^{\prime}+\omega B\right) \cos \omega x+\left(B^{\prime}-\omega A\right) \sin \omega x \text { and } \\ y_{p}^{\prime \prime} &=\left(A^{\prime \prime}+2 \omega B^{\prime}-\omega^{2} A\right) \cos \omega x+\left(B^{\prime \prime}-2 \omega A^{\prime}-\omega^{2} B\right) \sin \omega x \end{aligned} $$ (b) Use the results of (a) to verify that $$ \begin{aligned} a y_{p}^{\prime \prime}+b y_{p}^{\prime}+c y_{p}=&\left[\left(c-a \omega^{2}\right) A+b \omega B+2 a \omega B^{\prime}+b A^{\prime}+a A^{\prime \prime}\right] \cos \omega x+\\\ &\left[-b \omega A+\left(c-a \omega^{2}\right) B-2 a \omega A^{\prime}+b B^{\prime}+a B^{\prime \prime}\right] \sin \omega x \end{aligned} $$ (c) Use the results of (a) to verify that $$ y_{p}^{\prime \prime}+\omega^{2} y_{p}=\left(A^{\prime \prime}+2 \omega B^{\prime}\right) \cos \omega x+\left(B^{\prime \prime}-2 \omega A^{\prime}\right) \sin \omega x . $$ (d) Prove Theorem 5.5.2.
Suppose \(p_{1}\) and \(p_{2}\) are continuous on \((a, b) .\) Let \(y_{1}\) be a solution of $$y^{\prime \prime}+p_{1}(x) y^{\prime}+p_{2}(x) y=0$$ that has no zeros on \((a, b),\) and let \(x_{0}\) be in \((a, b) .\) Use reduction of order to show that \(y_{1}\) and $$y_{2}(x)=y_{1}(x) \int_{x_{0}}^{x} \frac{1}{y_{1}^{2}(t)} \exp \left(-\int_{x_{0}}^{t} p_{1}(s) d s\right) d t$$ form a fundamental set of solutions of \((\mathrm{A})\) on \((a, b) .\) (NOTE: This exercise is related to Exercise \(9 .)\)
Find a particular solution. $$ y^{\prime \prime}+3 y^{\prime}-2 y=-e^{2 x}(5 \cos 2 x+9 \sin 2 x) $$
In Exercises \(24-29\) use the principle of superposition to find a particular solution. $$ y^{\prime \prime}+y=e^{-x}\left(2-4 x+2 x^{2}\right)+e^{3 x}\left(8-12 x-10 x^{2}\right) $$
What do you think about this solution?
We value your feedback to improve our textbook solutions.