Solve (x^2 d^2+xd)y=0
Answers
Answer:
y=xn.
Replacing this into the original equation, we obtain
n(n−1)xn+nxn−xn=(n2−n+n−1)xn=0.
If the equation is satisfied for every x, then the polynomial in n is zero:
n2−1=0,
or
n=±1.
Therefore, the general solution of the ODE is
y=Ax+Bx,
where A and B are arbitrary constants, to be adjusted from the initial conditions.
Footnotes
[1] Cauchy–Euler equation - Wikipedia
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Mangalam Gopal
, Ph.D Mathematics, University of Michigan (1967)
Answered March 31, 2019
This is an Euler type homogeneous D.E and the standard procedure is to try a solution of the form. y. =. x^n.;
then y' = n x^(n-1) and y'' = n (n-1) x^(n-2).
Substituting into the given equation and simplifying, we get (n^2. - 1) x^n. =. 0.
Thus. n = 1 and n = - 1 are the possibilities so that the general solution is y. = c_1. x. + c_2 . (1/x ).
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Andrew Droffner
, studied Mathematics at Rutgers University (1995)
Answered February 15, 2020
x2d2ydx2+xdydx−y=0
This is an Cauchy–Euler equation or equidimensional differential equation. The power of xr is equal to the derivative dnydxn . That’s how it is equi-dimensional.
Use the equidimensional substitution y=xr . Compute the derivatives and find the roots rn of its characteristic polynomial.
y=xr,dydx=rxr−1,d2ydx2=r(r−1)xr−2
Solve the DE
x2d2ydx2+xdydx−y=0
Substitute the derivatives. The DE reduces to its characteristic polynomial.
x2(r(r−1)xr−2)+x(rxr−1)−xr=0
r(r− … (more)
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Elaine Dawe
, BMath, Mathematics & Computer Science, University of Waterloo (1985)
Answered March 31, 2019
Solve:
x2y′′+xy′−y=0
This is an Euler-Cauchy equation and has solutions of the form: y=xn
y′=xn
y′=nxn−1
y′′=n(n−1)xn−2
Plug back into differential and solve for n :
x2(n(n−1)xn−2)+x(nxn−1)−xn=0
xn(n(n−1)+n−1)=0
n2−n+n−1=0
n2−1=0
n=±1
y1=x,y2=1x
y=C1x+C2x
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