pramrok62

2022-09-30

I need some help solving the differential equation
${y}^{‴}=x+y\phantom{\rule{0ex}{0ex}}y\left(1\right)=3\phantom{\rule{0ex}{0ex}}{y}^{\prime }\left(1\right)=2\phantom{\rule{0ex}{0ex}}{y}^{″}\left(1\right)=1$
and h=0.5 with Euler's method

I don't know how to rewrite the equation to a system of equations of the first order..

Farbwolkenw

Here is how you linearize the system, we are given:
${y}^{‴}\left(x\right)=x+y,y\left(1\right)=3,{y}^{\prime }\left(1\right)=2,{y}^{″}\left(1\right)=1$
We can proceed as follows:
${x}_{1}=y\phantom{\rule{thickmathspace}{0ex}}⟹\phantom{\rule{thickmathspace}{0ex}}{x}_{1}^{\prime }={y}^{\prime }={x}_{2}$
${x}_{2}={y}^{\prime }\phantom{\rule{thickmathspace}{0ex}}⟹\phantom{\rule{thickmathspace}{0ex}}{x}_{2}^{\prime }={y}^{″}={x}_{3}$
${x}_{3}={y}^{″}\phantom{\rule{thickmathspace}{0ex}}⟹\phantom{\rule{thickmathspace}{0ex}}{x}_{3}^{\prime }={y}^{‴}=x+y=x+{x}_{1}$
So, our new system is:
$\begin{array}{rl}{x}_{1}^{\prime }& ={x}_{2}\\ {x}_{2}^{\prime }& ={x}_{3}\\ {x}_{3}^{\prime }& =x+{x}_{1}\end{array}$
Note, the given initial conditions, similarly follow as:
$y\left(1\right)=3\phantom{\rule{thickmathspace}{0ex}}⟹\phantom{\rule{thickmathspace}{0ex}}{x}_{1}\left(1\right)=3$
${y}^{\prime }\left(1\right)=2\phantom{\rule{thickmathspace}{0ex}}⟹\phantom{\rule{thickmathspace}{0ex}}{x}_{2}\left(1\right)=2$
${y}^{″}\left(1\right)=1\phantom{\rule{thickmathspace}{0ex}}⟹\phantom{\rule{thickmathspace}{0ex}}{x}_{3}\left(1\right)=1$
Now, apply Euler's method to this system using $h=0.5$.

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