Chapter 20: Problem 2
By integrating twice find the general solution of \(y^{\prime \prime}=12 x^{2}\)
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 20: Problem 2
By integrating twice find the general solution of \(y^{\prime \prime}=12 x^{2}\)
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
Use Euler's method to find a numerical solution of \(\frac{\mathrm{dy}}{\mathrm{d} x}=\frac{x y}{x^{2}+2}\) subject to \(y(1)=3\). Take \(h=0.1\) and hence approximate \(y(1.5)\). Obtain the true solution using the method of separation of variables. Work throughout to six decimal places.
Give one example each of a first-order linear equation, first-order non-linear equation, second-order linear equation, second-order non-linear equation.
The charge, \(q\), on a capacitor in an \(L C R\) series circuit satisfies the second-order differential equation $$ L \frac{\mathrm{d}^{2} q}{\mathrm{~d} t^{2}}+R \frac{\mathrm{d} q}{\mathrm{~d} t}+\frac{1}{C} q=E $$ where \(L, R, C\) and \(E\) are constants. Show that if \(2 L=C R^{2}\) the general solution of this equation is $$ \begin{aligned} &q= \\ &\mathrm{e}^{-t /(C R)}\left(A \cos \frac{1}{C R} t+B \sin \frac{1}{C R} t\right)+C E \end{aligned} $$ If \(i=\frac{\mathrm{d} q}{\mathrm{~d} t}=0\) and \(q=0\) when \(t=0\) show that the current in the circuit is $$ i=\frac{2 E}{R} \mathrm{e}^{-t /(C R)} \sin \frac{1}{C R} t $$
Obtain the general solutions, that is the complementary functions, of the following homogeneous equations: (a) \(\frac{\mathrm{d}^{2} y}{\mathrm{~d} x^{2}}-2 \frac{\mathrm{d} y}{\mathrm{~d} x}+y=0\) (b) \(\frac{\mathrm{d}^{2} y}{\mathrm{~d} t^{2}}+\frac{\mathrm{d} y}{\mathrm{~d} t}+5 y=0\) (c) \(\frac{\mathrm{d}^{2} y}{\mathrm{~d} x^{2}}+\frac{\mathrm{d} y}{\mathrm{~d} x}-2 y=0\) (d) \(\frac{\mathrm{d}^{2} y}{\mathrm{~d} x^{2}}+9 y=0\) (e) \(\frac{\mathrm{d}^{2} y}{\mathrm{~d} x^{2}}-2 \frac{\mathrm{d} y}{\mathrm{~d} x}=0\) (f) \(\frac{\mathrm{d}^{2} x}{\mathrm{~d} t^{2}}-16 x=0\)
Find a particular integral for the equation $$ \frac{\mathrm{d}^{2} y}{\mathrm{~d} x^{2}}+\frac{\mathrm{d} y}{\mathrm{~d} x}+y=1+x $$
What do you think about this solution?
We value your feedback to improve our textbook solutions.