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Question regarding one step in this problem

Mathematics
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\[x^2 y'' - 7xy' + 16; y_1 = x^4\]I have to set it in standard form following this rule\[y''+P(x)y'+Q(x)=0\]which results in\[y''-\frac{ 7 }{ x } y' + \frac{ 16 }{ x^2 } y = 0\] How did the x and x^2 moved on the denominator?
divide by x^2
why is it that we have to divide it by x^2 after changing it to standard form?

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Other answers:

you dicide to change it to standard form
That looks like it, unless the equation is inhomogeneous and you have to figure out the inhomogeneous part by the given solution.
divide*, you cannot have any f(x) multiypling your highest degree
\(x^2 y'' - 7xy' + 16=0\) is not in standard form because there is \(x^2\) in front of \(y''\) where in standard form there is nothing.
ohh I see!
first solve the homogenous equation then use variation of parameters to get general solution to your non homogenous equation
so to set it to standard form whatever is in front of y'' has to be divided to just have y'' alone
\[ x^2 y'' - 7xy' + 16\\ y_1 = x^4 \] But plugging \(y_1\) into the above expression I get \(-16x^4+16\).
oh no you have to do reduction of sides
Explain?
\[y_2 = y_1 \int\limits_{}^{}\frac{ e ^{-\int\limits_{}^{}P(x)dx} }{y_1^2 }\]
I am assuming \(y_1=x^4\) is a solution to the incomplete differential equation \(x^2 y'' - 7xy' + 16\). What does this equation equal to?
Or is \(y_1\) not a solution?
The final answer is \[Y=C_1x^4 + C_2 x^4\ln(x)\]or simplified \[Y = x^4(C_1 + C_2 \ln(x))\]
using reduction of sides you find y_2
In other words, \(y_1=x^4\) should satisfy the equation \(x^2 y'' - 7xy' + 16=0\)? \[ x^2(x^4)''-7x(x^4)'+16\\ 12x^4-28x^4+16\\ -16x^4+16\neq0 \]
\[y'' \neq y\]
try to search for solution of this type: \[y = k{x^s},\quad s \in \mathbb{Z}\]
\[\Large y = k{x^s},\quad s \in \mathbb{Z}\]
after the substitution, you should get a quadratic equation for \(s\)
\[ x^2 y'' - 7xy' + 16=0 \text{ not }x^2 y'' - 7xy' + 16y=0 \] I got\[ r(r-1)x^r-7rx^r+16=0 \] after assuming \(y=x^r\).
\(+16x^r\)
16 does not have y behind it in the differential equation. Not sure whether it is a typo or not.
from first reply I see 16y/x^2
This is horrible!
if we ahve only \(16\) and not \(16y\) then we have to apply my substitution above to the homogeneous equation only
have*
Exactly the same equation on this website. http://tutorial.math.lamar.edu/Classes/DE/EulerEquations.aspx
If the equation is \[ x^2y''-7xy'+16y=0 \] Divide the whole thing by \(x^2\) to get the standard form. Or just read the website.
oh yeahI ate the y after the 16, sorry. I do undeerstand this though, just had doubt in the beginning
Y U ATE THE Y? YYYYY? Sorry.....

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