Solve the ODE $ \ \ y'''+y= \csc (x) \ $

Question

Solve the ODE $ \ \ y'''+y= \csc (x) \ $

Answer:

The homogeneous equation is

$ y'''+y=0 \ $

The auxiliary equation is

$ m^3+1=0 \\ \Rightarrow (m+1)(m^2-m+1)=0 \\ \Rightarrow m=-1, \ m= \frac{1 \pm \sqrt{3}i}{2}$

Thus the general solution is

$ \large y(x)=Ae^{-x}+ e^{\frac{1}{2}x} [B \cos (\frac{\sqrt{3}}{2}x)+C \sin (\frac{\sqrt{3}}{2}x)] \ $ ,

where $ A,B, C \ $ are arbitrary constants.

Now we have to find out the particular integral:

$ P.I.=\large \frac{1}{D^3+1} \csc (x) =\frac{1}{(D+1)(D^2-D+1)} \csc (x) $

But I am unable to evaluate it.

Help me out with any method

Answer 1

This doesn't answer your main question, but here's another method. From the fundamental solution

$$ y_h(x) = c_1y_1(x) + c_2y_2(x) + c_3y_3(x) $$

where \begin{align} y_1(x) &= e^{-x} \\ y_2(x) &= e^{x/2}\cos \left(\frac{\sqrt{3}}{2}x \right) \\ y_3(x) &= e^{x/2}\sin \left( \frac{\sqrt{3}}{2}x\right) \end{align}

Using variation of parameters, we try to find a particular solution $$ y_p(x) = v_1(x)y_1(x) + v_2(x)y_2(x) + v_3(x)y_3(x) $$

The coefficient functions satisfy \begin{align} {v_1}'{y_1} + {v_2}'{y_2} + {v_3}'{y_3} &= 0 \\ {v_1}'{y_1}' + {v_2}'{y_2}' + {v_3}'{y_3}' &= 0 \\ {v_1}'{y_1}'' + {v_2}'{y_2}'' + {v_3}'{y_3}'' &= \csc x \end{align}

From the characteristic polynomial, we know the basis functions satisfy \begin{align} {y_1}'+ {y_1} = 0 \\ {y_2}'' - {y_2}' + {y_2} = 0 \\ {y_3}'' - {y_3}' + {y_3} = 0 \end{align}

Eliminating $v_2$ and $v_3$ gives \begin{align} {v_1}'({y_1}''-{y_1}'+{y_1}) &= \csc x \\ {v_1}' 3e^{-x} &= \csc x \\ v_1 &= \frac{1}{3}\int e^x \csc x\ dx \end{align}

Eliminating $v_1$ gives \begin{align} {v_2}'({y_2}'+{y_2}) + {v_3}'({y_3}'+{y_3}) &= 0 \\ {v_2}'({y_2}''+{y_2}') + {v_3}'({y_3}''+{y_3}') &= \csc x \end{align}

Since ${y_2}'' = {y_2}'-{y_2}$ and ${y_3}'' = {y_3}'-{y_3}$ \begin{align} {v_2}'({y_2}'+{y_2}) + {v_3}'({y_3}'+{y_3}) &= 0 \\ {v_2}'(2{y_2}'-{y_2}') + {v_3}'(2{y_3}'-{y_3}') &= \csc x \end{align}

\begin{align} {v_2}'\left[\frac{3}{2}\cos\left( \frac{\sqrt{3}}{2}x\right) - \frac{\sqrt{3}}{2}\sin\left( \frac{\sqrt{3}}{2}x\right) \right] + {v_3}'\left[\frac{\sqrt{3}}{2}\cos\left( \frac{\sqrt{3}}{2}x\right) + \frac{3}{2}\sin\left( \frac{\sqrt{3}}{2}x\right) \right] &= 0 \\ {v_2}'\sqrt{3}\sin\left( \frac{\sqrt{3}}{2}x\right) - {v_3}'\sqrt{3}\cos\left( \frac{\sqrt{3}}{2}x\right) &= e^{-x/2}\csc x \end{align}

\begin{align} {v_2}'\cos\left( \frac{\sqrt{3}}{2}x\right) + {v_3}'\sin\left( \frac{\sqrt{3}}{2}x\right) &= \frac{1}{3}e^{-x/2}\csc x \\ {v_2}'\sin \left( \frac{\sqrt{3}}{2}x\right) - {v_3}'\cos \left( \frac{\sqrt{3}}{2}x\right) &= \frac{1}{\sqrt{3}}e^{-x/2}\csc x \end{align}

\begin{align} v_2 &= -\frac{2}{\sqrt{3}}e^{-x/2} + \frac{1}{3}\int e^{-x/2}\cot x \ dx \\ v_3 &= -\frac{2}{3}e^{-x/2} - \frac{1}{\sqrt{3}}\int e^{-x/2}\cot x \ dx \end{align}

The final particular solution is \begin{align} y_p(x) &= \frac{1}{3}e^{-x}\int e^x \csc x\ dx \\ &\quad + \frac{1}{3}e^{x/2}\cos\left( \frac{\sqrt{3}}{2}x\right)\int e^{-x/2}\cot x \ dx \\ &\quad - \frac{1}{\sqrt{3}}e^{x/2}\sin\left( \frac{\sqrt{3}}{2}x\right)\int e^{-x/2}\cot x \ dx \\ &\quad - \frac{2}{\sqrt{3}}\cos\left( \frac{\sqrt{3}}{2}x\right) - \frac{2}{3}\sin\left( \frac{\sqrt{3}}{2}x\right) \end{align}

The integrals above are non-elementary

Answer 2

HINT :

Since you have at least one solution $e^{-x}$ of the homogeneous ODE, change of function :

$$y(x)=e^{-x}u(x)$$

$y'=e^{-x}(u'-u)$

$y''=e^{-x}(u''-2u'+u)$

$y'''=e^{-x}(u'''-3u''+3u'-u)$

$y'''+y=e^{-x}(u'''-3u''+3u'-u+u)=e^{-x}(u'''-3u''+3u')=\csc(x)$ $$u'''-3u''+3u'=e^x\csc(x)$$ Let $v(x)=u'(x)$ $$v''-3v'+3v=e^x\csc(x)$$ The third order ODE is reduced to a second order ODE.

Solve it for $v(x)$ ( If you like, you can use the same method to reduce it to a first order ODE).