Find the general solution of y′′ + 6y′ + 9y = f(x) for given conditions

by RoRi

January 31, 2016

Find the general solution of the second-order differential equation

$y'' + 6y' + 9 = f(x),$

where $f(x) = 1$ for $1 \leq x \leq 2$ , and $f(x) = 0$ for all other $x$ .

If $x < 1$ or $x > 2$ , then we have the equation

$y'' + 6y' + 9 = 0.$

This is of the form $y'' + ay' + by = 0$ with $a = 6$ and $b= 9$ . Therefore, $d = a^2 - 4b = 0$ so the solutions are given by

$\begin{align*} y &= e^{-\frac{ax}{2}} (c_1 + c_2 x) \\ &= e^{-3x} (c_1 + c_2 x). \end{align*}$

If $1 \leq x \leq 2$ , then we have the equation

$y'' + 6y' + 9 = 1.$

A particular solution to this equation is given by $y_1 = \frac{1}{9}$ (since $y_1' = y_1'' = 0$ in this case). Therefore, all solutions are of the form

$y = (a+bx)e^{-3x} + \frac{1}{9}.$

Stumbling Robot

Find the general solution of y′′ + 6y′ + 9y = f(x) for given conditions

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