How to solve $y''+y=x^2$?

Question

I need to solve:

$$y''+y=x^2$$

Taking the Laplace transform (and using the fact that it is a linear operator) on both sides I get:

$$\mathscr{L}(y)=\frac{2}{s^3(s^2+1)}+y(0)\frac{s}{s^2+1}+y'(0)\frac{1}{s^2+1}$$

And hence:

$$y=2G(x)+y(0)\cos x +y'(0)\sin x$$

Where $G(x)$ is the inverse Laplace transform of:

$$\frac{1}{s^3(s^2+1)}$$

My question is how do I find this inverse Laplace transform, I'm used to splitting the fraction into partial fractions but I don't think I'm used to doing a partial fraction like in the above.

Answer 1

As for the Laplace solution you asked for, you can split the fraction like this: $$\frac 1{s^3(s^2+2)}=\frac As+\frac B{s^2}+\frac C{s^3}+\frac{Ds+E}{s^2+1}$$ $$=\frac{As^4+As^2+Bs^3+Bs+Cs^2+2C+Ds^4+Es^3}{s^3(s^2+1)}$$ $$=\frac{(A+D)s^4+(B+E)s^3+(A+C)s^2+Bs+C}{s^3(s^2+1)}$$

By identification, you find $B=0,E=0,C=1,A=-1,D=1$

Answer 2

My question is how do I find this inverse Laplace transform of $\dfrac{1}{s^3(s^2+1)}$?

Hint. If one wants to proceed on your route, by a partial fraction decomposition, one has $$ \frac{1}{s^3(s^2+1)}=-\frac{1}{s}+\frac{1}{s^3}+\frac{s}{1+s^2} $$ giving $$ \mathcal{L}^{-1}\left(\frac{1}{s^3(s^2+1)}\right)(t)=-1+\frac{t^2}2+\cos t $$ using standard properties of the inverse Laplace transform.

Answer 3

Oh, I hate the Laplace transform! I have yet to find a differential equation that cannot be solved more easily using simpler methods. Here, the differential equation is $y''+ y= x^2$. The associated homogeneous equation is $y''+ y= 0$. Its characteristic equation is $r^2+ 1= 0$ which has roots $r= \pm i$ so the general solution to the associated homogeneous equation is $C_1cos(x)+ C_2sin(x)$. The right hand side, $x^2$, is one of the kinds of functions we would expect as a solution to such an equation so we use "undetermined coefficients"- try $y= Ax^2+ Bx+ C$ for constants A, B, C to be determined. Then $y'= 2Ax+ B$ and $y''= 2A$. $y''+ y= 2A+ Ax^2+ Bx+ C= Ax^2+ Bx+ 2A+ C= x^2$. Two polynomials will be equal for all x if and only if "corresponding coefficients" are equal- we must have $A= 1$, $B= 0$, and $2A+ C= 0$. So A= 1, B= 0, and C= -2. The general solution to this differential equation is $y(x)= C_1cos(x)+ C_2sin(x)+ x^2- 2$.

Answer 4

$\newcommand{\angles}[1]{\left\langle\,{#1}\,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{\mathrm{i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\Li}[1]{\,\mathrm{Li}_{#1}} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$

$\ds{{1 \over s^{3}\pars{s^{2} + 1}} :\ ?}$

1. For the $\ds{\color{#f00}{\mbox{pole at}\ s = 0}}$, the residue is given by \begin{align} {1 \over 2!}\,\lim_{s \to 0}\,\totald[2]{}{s}\bracks{% {s^{3} \over s^{3}\pars{s^{2} + 1}}\,\expo{st}} & = {1 \over 2}\,\lim_{s \to 0}\,\totald[2]{}{s}\bracks{% {\pars{1 - s^{2}}\pars{1 + st + \half\,s^{2}t^{2}}}} \\[4mm] & = \color{#f00}{-1 + \half\,t^{2}} \end{align}
2. For the $\ds{\color{#f00}{\mbox{poles at}\ s = \pm\ic}}$, the residue is given by \begin{align} \left.\pars{s \pm \ic}{\expo{st} \over s^{3}\pars{s - \ic}\pars{s + \ic}} \right\vert_{\ s\ \to\ \pm\ic} & = {\expo{\pm\ic t} \over \pm\ic\pars{\mp\ic\ \mp\ \ic}} = \color{#f00}{\half\,\expo{\pm\ic t}} \end{align}

   The final result becomes: $$ \pars{-1 + \half\,t^{2}} + \pars{\half\,\expo{\ic t}} + \pars{\half\,\expo{-\ic t}} = \color{#f00}{-1 + \half\,t^{2} + \cos\pars{t}} $$