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I am stuck on the following problem, hoping that someone will be able to help me.

I have a following second order differential equation: $$y''=-\sqrt{y}+0.5y'$$ with the following initial conditions: $$y(0)=0, y'(0)=0$$

I am familiar with Euler's method for first order differential equation when $y'=f(x,y)$, but In the above case I get stuck. When I start from the initial point $(x_0, y_0)$, which based on the initial condition would be $(0, 0)$ and then moving to the point $x_1=1$ and using the second initial condition I get $(x_1, y_1) = (1, 0)$. And now, how do I get the value of $y'(1)$ to plot the next point, let's say $(2, y_2)$?

I tried to reduce the order of the initial equation with the following substitution $y'=u$ and then: $$y'' = u' \implies u'=-\sqrt{y}+0.5u$$ but I run into a similar problem trying to plot $u(y)$.

If I try to write it as $$\frac{du}{dy}u=-\sqrt{y}+0.5u$$ and express $\frac{du}{dy}$ as $f(y,u)$ I run into a problem at point $(0, 0)$ where $u=0$.

I will appreciate if someone could give me some tips on how to tackle this, or what I am doing wrong here.

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  • $\begingroup$ I meant $y''$. Thanks for catching it. $\endgroup$ Commented Nov 30 at 8:53
  • $\begingroup$ The more I think about this question it leads me to a conclusion that with given initial conditions the solution would be just a straight line $y=0$. Can someone confirm that this is correct? $\endgroup$ Commented Nov 30 at 9:00
  • $\begingroup$ @najek81. This is correct (at least to me). $\endgroup$ Commented Nov 30 at 9:43
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    $\begingroup$ You say you are stuck on a problem but you don't state the problem. Euler will not move you away from the given initial conditions since (0,0) is an equilibrium point (the vector field is zero there). The fact that y=0 is a solution of the given equation is easily checked by substitution. So what is the precise question here? $\endgroup$ Commented Nov 30 at 19:23
  • $\begingroup$ Letting $y=v^{2}$ and studying the resulting ODE in the case $v'(0)=0$ with various values of $v(0) \gt 0$, I obtain numerically a curve $v(x)$ which is close to $v(x)=v_0 \sqrt{1-(x/a)^2}$. Hence the solution to the ODE looks qualitatively close to an ellipse in the first quadrant. $\endgroup$ Commented Nov 30 at 23:06

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