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

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Isolating the highest order derivative in differential equations simplifies the computational process, particularly in methods like Euler's. This approach allows for the formulation of equations in the form of y'' = f(x, y, y'), facilitating the calculation of derivatives at each step. By making the coefficient of the highest derivative equal to one, it standardizes the equations, making them easier to manipulate and solve. The discussion highlights how knowing the second derivative is crucial for iterative methods, as it enables the calculation of subsequent values of y and its derivatives. Understanding this concept is essential for effectively applying numerical methods to solve differential equations.
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i'm studying differential equations, the book states the following:

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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

\frac{d^2 y}{dx^2} = x*y + \frac{dy}{dx}
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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