pramrok62

2022-09-30

I need some help solving the differential equation

${y}^{\u2034}=x+y\phantom{\rule{0ex}{0ex}}y(1)=3\phantom{\rule{0ex}{0ex}}{y}^{\prime}(1)=2\phantom{\rule{0ex}{0ex}}{y}^{\u2033}(1)=1$

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..

${y}^{\u2034}=x+y\phantom{\rule{0ex}{0ex}}y(1)=3\phantom{\rule{0ex}{0ex}}{y}^{\prime}(1)=2\phantom{\rule{0ex}{0ex}}{y}^{\u2033}(1)=1$

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

Beginner2022-10-01Added 6 answers

Here is how you linearize the system, we are given:

${y}^{\u2034}(x)=x+y,y(1)=3,{y}^{\prime}(1)=2,{y}^{\u2033}(1)=1$

We can proceed as follows:

${x}_{1}=y\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}{x}_{1}^{\prime}={y}^{\prime}={x}_{2}$

${x}_{2}={y}^{\prime}\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}{x}_{2}^{\prime}={y}^{\u2033}={x}_{3}$

${x}_{3}={y}^{\u2033}\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}{x}_{3}^{\prime}={y}^{\u2034}=x+y=x+{x}_{1}$

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:

$y(1)=3\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}{x}_{1}(1)=3$

${y}^{\prime}(1)=2\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}{x}_{2}(1)=2$

${y}^{\u2033}(1)=1\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}{x}_{3}(1)=1$

Now, apply Euler's method to this system using $h=0.5$.

${y}^{\u2034}(x)=x+y,y(1)=3,{y}^{\prime}(1)=2,{y}^{\u2033}(1)=1$

We can proceed as follows:

${x}_{1}=y\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}{x}_{1}^{\prime}={y}^{\prime}={x}_{2}$

${x}_{2}={y}^{\prime}\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}{x}_{2}^{\prime}={y}^{\u2033}={x}_{3}$

${x}_{3}={y}^{\u2033}\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}{x}_{3}^{\prime}={y}^{\u2034}=x+y=x+{x}_{1}$

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:

$y(1)=3\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}{x}_{1}(1)=3$

${y}^{\prime}(1)=2\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}{x}_{2}(1)=2$

${y}^{\u2033}(1)=1\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}{x}_{3}(1)=1$

Now, apply Euler's method to this system using $h=0.5$.

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