How to solve this differential equation? y''-\frac{y'}{x}=4x^{2}y

Judith McQueen 2022-01-16 Answered
How to solve this differential equation? yyx=4x2y
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Debbie Moore
Answered 2022-01-17 Author has 43 answers
Note that we have xyy=4x3y. The LHS in some sense is dimensional consistent and looks something similar to a quotient rule provided we divide by x2. So dividing by x2 and doing algebraic manipulations we get (1xy)=4xy. Now this looks familiar to some extent.
Rewriting, we get (y2x)=2xy
Now let y2x=z(x). Plug this in and simplify to get z=yzy
(Replace 2x by yz)
So we have z2=y2+c
Thus we have now converted a second order differential equation in terms of first order differential equation, viz,
12xdydx=±y2+c where c is a constant.
(You could plug this in and check that this satisfies the second order differential equation.)
We now need other conditions (boundary/ initial conditions) to completely solve the problem i.e. to eliminate c and other constant which will arise after solving the first order differential equation to get y(x)=exp(x2)
(Note that taking c=0 we get a simple ode and the solution to which is y(x)=y(0)exp(±x2)).
EDIT:
The first order ODE can be solved as follows:
CASE 1:
Let c>0, then c=a2
Rearranging, we get
dyy2+a2=±d(x2)
y=atan(θ), we get dy=asec2(θ)dθ
Hence, the ode now becomes,
sec(θ)dθ=±d(x2)
d(log(sec(θ)+tan(θ)))=±d(x2)
log(sec(θ)+tan(θ)))=±(x2+k)
sec(θ)+tan(θ)=exp(±(x2+k))
Substitute for θ in terms of y to get,
ya±1+(ya)2=Kexp(±x2)
CASE 2:
Let c>0, then let c=a2
Rearranging, we get
Cassandra Ramirez
Answered 2022-01-18 Author has 30 answers
y=fy
y=fy+f2y
y=ffy+f2y
So y=1xy+4x2yf=±2x
and y=±2xyy=ce±x2

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alenahelenash
Answered 2022-01-24 Author has 368 answers

yyx=4x2y
xd2ydx2dydx4x3y=0
From there, we get the hints of left t=xn,
Then dydx=dydtdtdx=nxn1dydt
d2dx2=ddx(nxn1dydt)=nxn1ddx(dydt)+n(n1)xn2dydt=nxn1ddt(dydt)dtdx+n(n1)xn2dydt=nxn1d2ydt2nxn1+n(n1)xn2dydt=n2x2n2d2ydt2+n(n1)xn2dydt
x(n2x2n1d2ydt2+n(n1)xn2dydt)nxn1dydt4x3y=0
n2x2n1d2ydt2+n(n1)xn1dydtnxn1dydt4x3y=0
n2x2n1d2ydt2+n(n2)xn1dydt4x3y=0
n2x2n4d2ydt2+n(n2)xn4dydt4y=0
n2t2n4nd2ydt2+n(n2)tn4ndydt4y=0
The ODE has the simplest form when we choose n=2
The ODE becomes
4d2ydt24y=0
d2ydt2y=0
The auxiliary equation is
λ21=0
λ=±1
y=C1et+C2et

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