hi
can someone please help me with the following equation
y''-y=$\displaystyle x^n$
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
please guide me.
hi
can someone please help me with the following equation
y''-y=$\displaystyle x^n$
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
please guide me.
If we write the DE in terms of LT we have...
$\displaystyle s^{2}\cdot Y(s) -s\cdot y(0) -y^{'}(0) - Y(s) = \frac{n!}{s^{n+1}}$ (1)
... and from it we obtain...
$\displaystyle Y(s)= \frac{s\cdot y(0)+y^{'}(0)}{s^{2}-1} + \frac{n!}{s^{n+1}\cdot (s^{2}-1)}$ (2)
The solution of the DE can be [formally...] obtained computing the inverse LT of (2) and is...
$\displaystyle y(t)= y(0)\cdot \cosh x + y^{'}(0)\cdot \sinh x + \int_{0}^{x} (x-u)^{n}\cdot \sinh u\cdot du$ (3)
Kind regards
$\displaystyle \chi$ $\displaystyle \sigma$
Variation of parameters:
The general solution to the associated homogeneous equation, y"- y= 0, is, as you say, $\displaystyle Ce^x+ De^{-x}$.
In "variation of parameters" we look for a solution of the form $\displaystyle y(x)= u(x)e^x+ v(x)e^{-x}$. 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 $\displaystyle y(x)= u(x)e^x+ v(x)e^{-x}$ then $\displaystyle y'= u'e^x+ ue^x+ v'e^{-x}- ve^{-x}$. Since there are an infinite number of "u" and "v" that will work, we can "narrow the search" and simplify by requiring, also, that $\displaystyle u'e^x+ v'e^{-x}= 0$.
With that we have $\displaystyle y'= ue^x- ve^{-x}$ and, differentiating again, $\displaystyle y"= u'e^x+ ue^x- v'e^{-x}+ ve^{-x}$. Putting that into the differential equation, $\displaystyle y"- y= u'e^x+ ue^x- v'e^{-x}+ ve^{-x}- (ue^x+ ve^{-x})= u'e^x- v'e^{-x}= x^n$.
Our requirement that "$\displaystyle u'e^x+ v'e^{-x}= 0$ means that we will have no u" or v" and the fact that $\displaystyle e^x$ and $\displaystyle e^{-x}$ 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 $\displaystyle u'e^x+ v'e^{-x}= 0$ and $\displaystyle u'e^x- v'e^{-x}= x^n$ "algebraically" for u' and v'. Adding the two equations gives $\displaystyle 2u'e^x= x^n$ so $\displaystyle u'= \frac{1}{2}x^n e^{-x}$. Subtracting the second equation from the first gives $\displaystyle 2v'e^{-x}= -x^n$ so $\displaystyle v'= -\frac{1}{2}x^ne^x$.
Integrate to find u(x) and v(x). You can do those integrals by integration by parts, taking $\displaystyle u= x^n$ and $\displaystyle dv= e^xdx$ in the one case and $\displaystyle dv= e^{-x}dx$ in the other. That will reduce to an integral in $\displaystyle x^{n-1}$. Repeat until you have completely reduced that to $\displaystyle x^0$. 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 $\displaystyle y(x)= A_nx^n+ A_{n-1}x^{n-1}+ \cdot \cdot \cdot + A_1 x+ A_0$, put that into the differential equation and solve for the coefficients.
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 $\displaystyle x^0$ is n!
so t sum up all the series to find a general solution ....
will upload the result