Given Equation is $(D^4-a^4)y = x^4$
Solving the auxiliary equation gives $D = a,-a,+ai,-ai$
So the homogeneous solution is
$y_h = c_1e^{ax} + {c_2e^{-ax}} + c_3sin(ax) + c_4 cos (ax) $
Taking the Laplace transform
$ Y(s) = \frac{24}{s^5(s^4-a^4)}$
Using the incomplete partial fraction method is a lengthy process as $s^5$ is in the denominator.
Finding the particular integral using inverse differential operator seems to be the easier method.
Is there a quicker way to solve this equation using the Laplace transform method??
If we use convolution, since we already know the homogeneous solution, does that help us in some way??
Best Answer
Note that the right hand side of the equation is $x^4$, a polynomial with degree $4$. So a particular solution $y_p$ is a polynomial with degree $4$. Let $$ y_p=a_4x^4+a_3x^3+a_2x^2+a_1x+a_0 $$ be a particular solution. Then putting it in the equation gives $$ 4!a_4-a^4(a_4x^4+a_3x^3+a_2x^2+a_1x+a_0)=x^4$$ Comparing the coefficients gives $$-a^4a_4=1,a_3=a_2=a_1=0,4!a_4-a^4a_0=0 $$ from which one has $$ a_0=-\frac{24}{a^8},a_1=a_2=a_3=0, a_4=-\frac1{a^4}.$$ So $$ y_p=-\frac{x^4}{a^4}-\frac{24}{a^8}. $$ Thus the general solution of the DE is $$ y=y_h+y_p=c_1e^{ax} + {c_2e^{-ax}} + c_3\sin(ax) + c_4 \cos (ax)-\frac1{a^4}x^4-\frac{24}{a^8}.$$
Comment:
If you use Laplace transform to solve this DE, it will take longer time.
You can use the differential operator $D$ to get $y_p$. In fact $$ y_p=\frac{1}{D^4-a^4}x^4=-\frac1{a^4}\frac1{1-(\frac{D}{a})^4}x^4=-\frac1{a^4}\sum_{n=0}^\infty\bigg(\frac{D}{a}\bigg)^nx^4=-\frac1{a^4}\bigg(x^4+\frac{24}{a^4}\bigg)=-\frac{x^4}{a^4}-\frac{24}{a^8}.$$