## anonymous one year ago Question regarding one step in this problem

1. 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?

2. dan815

divide by x^2

3. anonymous

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

4. dan815

you dicide to change it to standard form

5. thomas5267

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

6. dan815

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

7. 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.

8. anonymous

ohh I see!

9. dan815

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

10. anonymous

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

11. thomas5267

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

12. anonymous

oh no you have to do reduction of sides

13. thomas5267

Explain?

14. anonymous

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

15. 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?

16. thomas5267

Or is $$y_1$$ not a solution?

17. anonymous

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

18. anonymous

using reduction of sides you find y_2

19. thomas5267

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

20. anonymous

$y'' \neq y$

21. Michele_Laino

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

22. Michele_Laino

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

23. Michele_Laino

after the substitution, you should get a quadratic equation for $$s$$

24. 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 $$y=x^r$$.

25. Michele_Laino

$$+16x^r$$

26. thomas5267

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

27. Michele_Laino

from first reply I see 16y/x^2

28. thomas5267

This is horrible!

29. Michele_Laino

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

30. Michele_Laino

have*

31. thomas5267

Exactly the same equation on this website. http://tutorial.math.lamar.edu/Classes/DE/EulerEquations.aspx

32. 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.

33. anonymous

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

34. thomas5267

Y U ATE THE Y? YYYYY? Sorry.....