Question

Question regarding one step in this problem

Answer 1

\[x^2 y'' - 7xy' + 16; y_1 = x^4\]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?

Answer 2

divide by x^2

Answer 3

why is it that we have to divide it by x^2 after changing it to standard form?

Answer 4

you dicide to change it to standard form

Answer 5

That looks like it, unless the equation is inhomogeneous and you have to figure out the inhomogeneous part by the given solution.

Answer 6

divide*, you cannot have any f(x) multiypling your highest degree

Answer 7

\(x^2 y'' - 7xy' + 16=0\) is not in standard form because there is \(x^2\) in front of \(y''\) where in standard form there is nothing.

Answer 8

ohh I see!

Answer 9

first solve the homogenous equation then use variation of parameters to get general solution to your non homogenous equation

Answer 10

so to set it to standard form whatever is in front of y'' has to be divided to just have y'' alone

Answer 11

\[
x^2 y'' - 7xy' + 16\\
y_1 = x^4
\]
But plugging \(y_1\) into the above expression I get \(-16x^4+16\).

Answer 12

oh no you have to do reduction of sides

Answer 13

Explain?

Answer 14

\[y_2 = y_1 \int\limits_{}^{}\frac{ e ^{-\int\limits_{}^{}P(x)dx} }{y_1^2 }\]

Answer 15

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?

Answer 16

Or is \(y_1\) not a solution?

Answer 17

The final answer is \[Y=C_1x^4 + C_2 x^4\ln(x)\]or simplified \[Y = x^4(C_1 + C_2 \ln(x))\]

Answer 18

using reduction of sides you find y_2

Answer 19

In other words, \(y_1=x^4\) should satisfy the equation \(x^2 y'' - 7xy' + 16=0\)?
\[
x^2(x^4)''-7x(x^4)'+16\\
12x^4-28x^4+16\\
-16x^4+16\neq0
\]

Answer 20

\[y'' \neq y\]

Answer 21

try to search for solution of this type:
\[y = k{x^s},\quad s \in \mathbb{Z}\]

Answer 22

\[\Large y = k{x^s},\quad s \in \mathbb{Z}\]

Answer 23

after the substitution, you should get a quadratic equation for \(s\)

Answer 24

\[
x^2 y'' - 7xy' + 16=0 \text{ not }x^2 y'' - 7xy' + 16y=0
\]
I got\[
r(r-1)x^r-7rx^r+16=0
\]
after assuming \(y=x^r\).

Answer 25

\(+16x^r\)

Answer 26

16 does not have y behind it in the differential equation. Not sure whether it is a typo or not.

Answer 27

from first reply I see 16y/x^2

Answer 28

This is horrible!

Answer 29

if we ahve only \(16\) and not \(16y\) then we have to apply my substitution above to the homogeneous equation only

Answer 30

have*

Answer 31

Exactly the same equation on this website.

http://tutorial.math.lamar.edu/Classes/DE/EulerEquations.aspx

Answer 32

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.

Answer 33

oh yeahI ate the y after the 16, sorry. I do undeerstand this though, just had doubt in the beginning

Answer 34

Y U ATE THE Y? YYYYY?

Sorry.....