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Chapter 20: Problem 2

Find a particular integral for the equation $$ \frac{\mathrm{d}^{2} y}{\mathrm{~d} x^{2}}+\frac{\mathrm{d} y}{\mathrm{~d} x}+y=1+x $$

Short Answer

Expert verified
What is the particular integral for the given second-order, non-homogeneous linear differential equation with constant coefficients: $$ \frac{d^2y}{dx^2} + \frac{dy}{dx} + y = 1 + x $$

Step by step solution

01

Make an educated guess for the particular integral

In order to find the particular integral of the given equation, we need to determine the form of our guess. The right side of the equation is a polynomial of degree 1, so our guess should also be a polynomial of degree 1. Let's guess: $$ y_p = Ax + B $$ where \(A\) and \(B\) are constants that we will determine.
02

Calculate the derivatives of our guess

Next, we need to compute the first and second derivatives of \(y_p\) in order to substitute them into the given equation. Calculate the first derivative: $$ \frac{dy_p}{dx} = A $$ Now, calculate the second derivative: $$ \frac{d^2y_p}{dx^2} = 0 $$
03

Substitute the guess and its derivatives into the given equation

Now we substitute \(y_p\), \(\frac{dy_p}{dx}\), and \(\frac{d^2y_p}{dx^2}\) into the given equation: $$ \frac{d^2y_p}{dx^2} + \frac{dy_p}{dx} + y_p = 0 + A + Ax + B = 1 + x $$
04

Equate coefficients and solve for \(A\) and \(B\)

In order for our guess to be a valid particular integral, we need the equation to hold for each coefficient. Thus, we can write a system of linear equations by equating coefficients of like powers of \(x\): Coefficients of \(x^1\): $$ A = 1 $$ Coefficients of \(x^0\): $$ A + B = 1 + B = 1 $$ Since \(A = 1\), \(B\) can be found as: $$ B = 0 $$
05

Write down the particular integral

We now have the values of \(A\) and \(B\) which can be used to obtain our particular integral. Substituting \(A = 1\), \(B = 0\) into \(y_p = Ax + B\), we find the particular integral: $$ y_p = x $$ Therefore, the particular integral for the given equation is \(y_p = x\).

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Most popular questions from this chapter

If \(y=A \mathrm{e}^{2 x}\) is the general solution of \(\frac{\mathrm{d} y}{\mathrm{~d} x}=2 y\), find the particular solution satisfying \(y(0)=3\). What is the particular solution satisfying \(\frac{\mathrm{d} y}{\mathrm{~d} x}=2\) when \(x=0 ?\)

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.

Use an integrating factor to obtain the general solution of \(i R+L \frac{\mathrm{d} i}{\mathrm{~d} t}=\sin \omega t\) where \(R, L\) and \(\omega\) are constants.

The general solution of $$ \frac{\mathrm{d}^{2} x}{\mathrm{~d} t^{2}}=-\omega^{2} x $$ is \(x=A \mathrm{e}^{j \omega t}+B \mathrm{e}^{-\mathrm{j} \omega t}\), where \(\mathrm{j}^{2}=-1\). Verify that this is indeed a solution. What is the particular solution satisfying \(x(0)=0\), \(\frac{\mathrm{d} x}{\mathrm{~d} t}(0)=1 ?\) Express the general solution and the particular solution in terms of trigonometrical functions.

Find the general solution of the following: (a) \(\frac{\mathrm{d} x}{\mathrm{~d} t}=x t\) (b) \(\frac{\mathrm{d} y}{\mathrm{~d} x}=\frac{x}{y}\) (c) \(t \frac{\mathrm{d} x}{\mathrm{~d} t}=\tan x\) (d) \(\frac{\mathrm{d} x}{\mathrm{~d} t}=\frac{x^{2}-1}{t}\)
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