Find the general solution of the differential equation by using the Indefinite Coefficients Method.

Question

$$''' + y' = 2^2 + 4\sin(x)$$

Find the general solution of the differential equation by using the Indefinite Coefficients Method.

Answer 1

The method of undetermined coefficients is described here.

Trying a solution of the form $e^{mx}$ for the homogeneous equation, we find that the characteristic equation (or auxiliary equation) is $m^{3}+m=m(m^{2}+1)=0,$ so the homogenous solution is $y_{c}(x)=A+B\sin(x)+C\cos(x).$

Then for the particular integral we try $y_{p}(x)=ax^{3}+bx^{2}+cx+d+hx\sin(x)$ and substituting into the equation you will find that $a=\frac{2}{3}$, $b=0$, $c=-4$ and $h=-2$ (we already know that $d$ can be combined in the constant solution from $y_{c}(x)$).

Thus the general solution is: $$y(x)=y_{c}(x)+y_{p}(x)=C_{1}+C_{2}\sin(x)+C_{3}\cos(x)+\frac{2}{3}x^{3}-4x-2x\sin(x).$$

Answer 2

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ Set $\ds{z \equiv y'}$ such that we are left with $\ds{z'' + z = 2^{2} + 4\sin\pars{x}}$. Lets $\ds{\xi = z' + \ic z}$ such that $\ds{z = \Im\pars{\xi}}$. Then, \begin{align} z'' + z = \xi'- \xi\,\ic & = 2^{2} + 4\sin\pars{x} \\[5mm] \totald{\pars{\expo{-\ic x}\xi}}{x} & = \expo{-\ic x}\bracks{2^{2} + 4\sin\pars{x}} \\[5mm] \expo{-\ic x}\xi & = \int\expo{-\ic x}\bracks{2^{2} + 4\sin\pars{x}}\dd x + C\,,\quad \pars{~C:\ \mbox{constant}~} \\[5mm] \xi & = -4\ic - \expo{-\ic x} -2\ic x\expo{\ic x} + 4x + 2\ic x^{2} + C\expo{\ic x} \\[5mm] z & = -4 + \sin\pars{x} -2x\cos{x} + 2 x^{2} + \Im\pars{C\expo{\ic x}} \end{align} Now, you are in the right path.