Test is coming

Test is coming

Postby Guest » Tue May 07, 2019 10:49 am

Please check this (easy) diff. equation for me.
I have made the laplace transform and got [tex]\frac{1}{s^2+1}\cdot\frac{1}{s}\cdot\tanh(s*\pi/2)[/tex]
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Re: Test is coming

Postby Guest » Tue May 14, 2019 7:43 am

What, exactly, was the question? To find the Laplace transform of y? It's simpler to solve the differential equation without using the Laplace transform.
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Re: Test is coming

Postby Guest » Mon Jun 10, 2019 8:34 am

The characteristic equation is [tex]r^2= 1[/tex] which has solutions r= i and r= -I. So the general solution to the differential equation is [tex]y(t)= C_1cos(t)+ C_2sin(t)[/tex]. To find a solution to the entire equation using "variation of parameters", we seek a solution of the form [tex]y(t)= u(t)cos(t)+ v(t)sin(t)[/tex]. Then [tex[y'(t)= u'(t)cos(t)- u(t)sin(t)+ v'(t)sin(t)+ v(t)cos(t)[/tex]. There are, in fact, infinitely many solutions of that form so to "narrow the search" and simplify, we assume that [tex]u'(t)cos(t)+ v'(t)sin(t)= 0[/tex]. In that case, [tex]y'(t)= -u(t)sin(t)+ v(t)cos(t)[/tex]. Differentiating again, [tex]y''(t)= -u''(t)sin(t)- u(t)cos(t)+ v'(t)cos(t)- v(t)sin(t)[/tex] so that [tex]y''+ y= -u'(t)sin(t)- u(t)cos(t)+ v'(t)cos(t)- v(t)sin(t)+ u(t)cos(t)+ v(t)sin(t)= -u'(t)sin(t)+ v'(t)cos(t)= f(t)[/tex].

We have the two equations [tex]u'(t)cos(t)+ v'(t)sin(t)= 0[/tex] and [tex]-u'(tt)sin(t)+ v'(t)cos(t)= f(t)[/tex]. Treat those as two linear equations to solve for u' and v'. Multiply the first equation by sin(t) to get [tex]u'(t)cos(t)sin(t)+ v'(t)sin^2(t)= 0[/tex], multiply the second equation by cos(t) to get [tex]-u(t)sin(t)cos(t)+ v'(t)cos^2(t)= f(t)cos(t).[/tex] Adding eliminates u'(t): [tex]v'(t)(sin^2(t)+ cos^2(t))= v'(t)= f(t)cos(t)[/tex]. Integrate f(t)cos(t) to get v(t). Similarly, multiplying the first equation by cos(t) gives [tex]u'cos^2(t)+ v'sin(t)cos(t)= 0[/tex] and the multiplying the second equation by sin(t) gives [tex]-u'(t)sin^2(t)+ v'(t)sin(t)cos(t)= f(t)sin(t)[/tex] Subtracting the second of those equations from the first, [tex]u'(t)= -f(t) sin(t)[/tex]. Integrate that to get u(t).

To do those integrals, since f(t) is defined "piecewise" you will have to integrate by "pieces". For example for t between 0 and [tex]\pi[/tex], f(t)= -1 so [tex]u'(t)= sin(t)[/tex] and [tex]u(t)= -cos(t)[/tex] while [tex]v'(t)= -cos(t)[/tex] so [tex]v(t)= sin(t)[/tex] (since we just want one out of infinitely many solutions, we can take the "constants of integration" to be 0). So for x between 0 and [tex]\pi[/tex], [tex]y(t)= C_1cos(t)+ C_2sin(t)- cos^2(t)+ sin^2(t)[/tex]. Also [tex]y'(t)= -C_1sin(t)+ C_2 cos(t)+ 4cos(t)sin(t).[/tex] Using the initial conditions, [tex]y(0)= C_1- 1= 0\ so\ y(0)= 1\ and\ y'(0)= C_2= 0.[/tex] For t between 0 and [tex]\pi[/tex], [tex]y(t)=cos(t)- cos^2(t)+ sin^2(t)[/tex].

Now do the same for t between [tex]\pi[/tex] and [tex]2\pi[/tex] with f(x)= 1. Use the values from the first solution at [tex]x= \pi[/tex] as initial conditions.
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