anonymous
  • anonymous
Question regarding one step in this problem
Mathematics
  • Stacey Warren - Expert brainly.com
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schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
\[x^2 y'' - 7xy' + 16; y_1 = x^4\]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?
dan815
  • dan815
divide by x^2
anonymous
  • anonymous
why is it that we have to divide it by x^2 after changing it to standard form?

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dan815
  • dan815
you dicide to change it to standard form
thomas5267
  • thomas5267
That looks like it, unless the equation is inhomogeneous and you have to figure out the inhomogeneous part by the given solution.
dan815
  • dan815
divide*, you cannot have any f(x) multiypling your highest degree
thomas5267
  • thomas5267
\(x^2 y'' - 7xy' + 16=0\) is not in standard form because there is \(x^2\) in front of \(y''\) where in standard form there is nothing.
anonymous
  • anonymous
ohh I see!
dan815
  • dan815
first solve the homogenous equation then use variation of parameters to get general solution to your non homogenous equation
anonymous
  • anonymous
so to set it to standard form whatever is in front of y'' has to be divided to just have y'' alone
thomas5267
  • thomas5267
\[ x^2 y'' - 7xy' + 16\\ y_1 = x^4 \] But plugging \(y_1\) into the above expression I get \(-16x^4+16\).
anonymous
  • anonymous
oh no you have to do reduction of sides
thomas5267
  • thomas5267
Explain?
anonymous
  • anonymous
\[y_2 = y_1 \int\limits_{}^{}\frac{ e ^{-\int\limits_{}^{}P(x)dx} }{y_1^2 }\]
thomas5267
  • thomas5267
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?
thomas5267
  • thomas5267
Or is \(y_1\) not a solution?
anonymous
  • anonymous
The final answer is \[Y=C_1x^4 + C_2 x^4\ln(x)\]or simplified \[Y = x^4(C_1 + C_2 \ln(x))\]
anonymous
  • anonymous
using reduction of sides you find y_2
thomas5267
  • thomas5267
In other words, \(y_1=x^4\) should satisfy the equation \(x^2 y'' - 7xy' + 16=0\)? \[ x^2(x^4)''-7x(x^4)'+16\\ 12x^4-28x^4+16\\ -16x^4+16\neq0 \]
anonymous
  • anonymous
\[y'' \neq y\]
Michele_Laino
  • Michele_Laino
try to search for solution of this type: \[y = k{x^s},\quad s \in \mathbb{Z}\]
Michele_Laino
  • Michele_Laino
\[\Large y = k{x^s},\quad s \in \mathbb{Z}\]
Michele_Laino
  • Michele_Laino
after the substitution, you should get a quadratic equation for \(s\)
thomas5267
  • thomas5267
\[ x^2 y'' - 7xy' + 16=0 \text{ not }x^2 y'' - 7xy' + 16y=0 \] I got\[ r(r-1)x^r-7rx^r+16=0 \] after assuming \(y=x^r\).
Michele_Laino
  • Michele_Laino
\(+16x^r\)
thomas5267
  • thomas5267
16 does not have y behind it in the differential equation. Not sure whether it is a typo or not.
Michele_Laino
  • Michele_Laino
from first reply I see 16y/x^2
thomas5267
  • thomas5267
This is horrible!
Michele_Laino
  • Michele_Laino
if we ahve only \(16\) and not \(16y\) then we have to apply my substitution above to the homogeneous equation only
Michele_Laino
  • Michele_Laino
have*
thomas5267
  • thomas5267
Exactly the same equation on this website. http://tutorial.math.lamar.edu/Classes/DE/EulerEquations.aspx
thomas5267
  • thomas5267
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.
anonymous
  • anonymous
oh yeahI ate the y after the 16, sorry. I do undeerstand this though, just had doubt in the beginning
thomas5267
  • thomas5267
Y U ATE THE Y? YYYYY? Sorry.....

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