Why is it useful to isolate the highest order derivative when solving DEs?

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SUMMARY

The discussion centers on the importance of isolating the highest order derivative in differential equations (DEs) for both theoretical and computational purposes. Specifically, the example provided illustrates how to apply Euler's method to a second-order differential equation, d²y/dx² = xy + dy/dx, with initial conditions y(0)=1 and y'(0)=1. The process of calculating successive values of y and its derivatives relies on knowing the second derivative, emphasizing the necessity of expressing the DE in the form y'' = f(x, y, y') for effective numerical solutions.

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cocopops12
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i'm studying differential equations, the book states the following:

11v4f2u.png


i want to know what are some of the "theoretical and computational purposes" behind it?

i always notice the book likes to make the coefficient of the highest derivative equal to 1
but why is that?!
 
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If you've done Euler's method or separation of variables, you saw in the first order case how you have to solve for the derivative as a function of x and y. This extends to higher orders... suppose we have a second order differential equation

[tex]\frac{d^2 y}{dx^2} = x*y + \frac{dy}{dx}[/tex]
and y(0)=1, y'(0)=1

I want to do an Euler's method kind of calculation. This requires knowing the derivative at every step. Unfortunately I don't have a formula for the derivative, so I have to use the second derivative to re-calculate the derivative each time

y(.1)=y(0)+.1*y'(0) = 1.1
y'(.1)=y'(0)+.1*y''(0)
I know what y''(0) is from the differential equation: y''(0)=0*1+1=1
y'(.1)=1+.1*1=1.1

Now if I want to update to get y(.2) and y'(.2), I need to know what y''(.1) is. So I use the differential equation
y''(.1) = xy+y' = .1*1.1+1.1 = 1.21

Now I can go to x=.2
y(.2)=y(.1)+.1*y'(.1)=1.21
y'(.2)=y'(.1)+.1*y''(.1)=1.221

Now I need to know what y''(.2) is because I want to be able to calculate y and y' at x=.3, etc. So

y''(.2)=x*y+y' = .2*1.21+1.221 = 1.463

Now I can keep going, estimating y(x) for as large a value of x as I want. The computation only required being able to solve for the second derivative of y at each step, since I already knew all the lower derivatives, so it was useful to have the equation in the form y''=f(x,y,y')
 
thank you sir!, haven't studied Euler's Method yet, but i do get the idea.

anyone got more info or examples? :biggrin:
 

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