1. ## DIFFERENTIAL equation

hi

y''-y= $x^n$

using method of separation of variables...

the roots come out to be 1,-1

so when i find variable parameters A(X) and B(x)
i get stuck

2. If we write the DE in terms of LT we have...

$s^{2}\cdot Y(s) -s\cdot y(0) -y^{'}(0) - Y(s) = \frac{n!}{s^{n+1}}$ (1)

... and from it we obtain...

$Y(s)= \frac{s\cdot y(0)+y^{'}(0)}{s^{2}-1} + \frac{n!}{s^{n+1}\cdot (s^{2}-1)}$ (2)

The solution of the DE can be [formally...] obtained computing the inverse LT of (2) and is...

$y(t)= y(0)\cdot \cosh x + y^{'}(0)\cdot \sinh x + \int_{0}^{x} (x-u)^{n}\cdot \sinh u\cdot du$ (3)

Kind regards

$\chi$ $\sigma$

3. ## about the solution above

Many thanks for the reply ...
but instead of using the Laplace Method...we are supposed to do it via variation of parameters...
and when i solve for the parameters i get the integral

$integralx^n*e^x dx$

4. Variation of parameters:
The general solution to the associated homogeneous equation, y"- y= 0, is, as you say, $Ce^x+ De^{-x}$.

In "variation of parameters" we look for a solution of the form $y(x)= u(x)e^x+ v(x)e^{-x}$. Note that there are an infinite number of such solutions. In fact, given any solution, Y(x), we could just take u(x) identically 0 and then solve for v(x) or, conversely, take v(x) identically 0 and solve for u(x).

In any case, if $y(x)= u(x)e^x+ v(x)e^{-x}$ then $y'= u'e^x+ ue^x+ v'e^{-x}- ve^{-x}$. Since there are an infinite number of "u" and "v" that will work, we can "narrow the search" and simplify by requiring, also, that $u'e^x+ v'e^{-x}= 0$.

With that we have $y'= ue^x- ve^{-x}$ and, differentiating again, $y"= u'e^x+ ue^x- v'e^{-x}+ ve^{-x}$. Putting that into the differential equation, $y"- y= u'e^x+ ue^x- v'e^{-x}+ ve^{-x}- (ue^x+ ve^{-x})= u'e^x- v'e^{-x}= x^n$.

Our requirement that " $u'e^x+ v'e^{-x}= 0$ means that we will have no u" or v" and the fact that $e^x$ and $e^{-x}$ satisfy the homogeneous equation means there is no u or v. We have only u' and v' in the equation.

Now we can solve the two equations $u'e^x+ v'e^{-x}= 0$ and $u'e^x- v'e^{-x}= x^n$ "algebraically" for u' and v'. Adding the two equations gives $2u'e^x= x^n$ so $u'= \frac{1}{2}x^n e^{-x}$. Subtracting the second equation from the first gives $2v'e^{-x}= -x^n$ so $v'= -\frac{1}{2}x^ne^x$.

Integrate to find u(x) and v(x). You can do those integrals by integration by parts, taking $u= x^n$ and $dv= e^xdx$ in the one case and $dv= e^{-x}dx$ in the other. That will reduce to an integral in $x^{n-1}$. Repeat until you have completely reduced that to $x^0$. After a few example, you should be able to find a formula for the general case.

I will confess that I would have done this a third way, using "undetermined coefficients". Try a specific solution of the form $y(x)= A_nx^n+ A_{n-1}x^{n-1}+ \cdot \cdot \cdot + A_1 x+ A_0$, put that into the differential equation and solve for the coefficients.

5. ## Reduction formula

wow!!!

thanks a lot HallsofIvy for the post...and certainly the part solved is almost same as i did except the last part...and yes i am trying to do the integral...its like series of n n-1 n-2 n-3 and in the end for the power of $x^0$ is n!
so t sum up all the series to find a general solution ....