Numerical Analysis Euler's Method

mayaitagaki
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Please help me with this problem! I have no clue. I know how to deal with only dy/dx. But this includes dy/dx and dz/dx...:cry:
Please see the attachment :)

Step size h = 0.2
range x = 0 to 1
y(0) = 2
z(0) = 4

Thank you,
Maya
 

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What don't you get?
 
Just do two Euler calculations.

Since the d.e. are
\frac{dy}{dx}= -2y+ 4e^{-x}
and
\frac{dz}{dx}= -\frac{yz^2}{3}

use the given initial values x= 0, y= 2, and z= 4 to calculate the two right sides:
\frac{dy}{dx}= -2(2)+ 4e^{0}= -4+ 4= 0
and
\frac{dz}{dx} -\frac{(2)(16){3}= \frac{32/3}= -10.66666...<br /> <br /> Since h= .2, x= 0+ h= 0+ .2= .2, y= 2+ (dy/dx)h= 2 - 0= 2, z= 4+ (dz/dx)h= 4- 2.133333= 1.8666668. <br /> <br /> and repeat.
 
thanks sooooo much! I've got it!
:smile:
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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