Answer to Question #151680 in Differential Equations for Yasser Pangcoga

Question

(D3 - 1)y = 4 sin x.

Answer 1

Given differential equation $(D^3-1)y=sinx$

Auxiliary equation corresponding to the given homogeneous differential equation is

$m^3-1=0\Rightarrow (m-1)(m^2+m+1)=0$

$\Rightarrow m=1,m=\frac{-1\pm \sqrt{1-4}}{2}=\frac{-1\pm \sqrt{3}i}{2}$

Solution to the homogeneous differential equation is

$y_{h}(x)=c_{1}e^{x}+e^{\frac{-x}{2}}\left [ c_{2}cos(\frac{\sqrt{3}x}{2})+c_{3}sin(\frac{\sqrt{3}x}{2}) \right ]$

Now let us find the particular solution using the method of undetermined coefficeints.

A particular solution using the method of undetermined coefficients is

$y_{p}(x)=Asinx+Bcosx$

$\Rightarrow y'_{p}(x)=Acosx-Bsinx, y''_{p}(x)=-Asinx-Bcosx,$

$y'''_{p}(x)=-Acosx+Bsinx$

Now $y'''_{p}(x)-y_{p}(x)=sinx$

$\Rightarrow -Acosx+Bsinx-Acosx+Bsinx=sinx$

$\Rightarrow -2Acosx+2Bsinx=sinx$

$\Rightarrow A=0, B=\frac{1}{2}$

So, the particular solution is $y_{p}(x)=\frac{1}{2}cosx$

Therefore, general solution to the given differential equation is

$y(x)=y_{h}(x)+y_{p}(x)$

That is

$y(x)=c_{1}e^{x}+e^{\frac{x}{2}}\left [ c_{2}cos(\frac{\sqrt{3}x}{2})+c_{3}sin(\frac{\sqrt{3}x}{2}) \right ]$ $+\frac{1}{2}cosx$