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FrogPad
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Seperation of Variables (double check please :)
I have a final coming up, and I want to make sure I have this method down.
Q: For the second-order wave equation [itex] u_{tt}=u_{xx}[/itex], the substitution of [itex] u=A(x)B(t) [/itex] will give second-order equations for A nd B when the x and t variables are seperated. From [itex] B''/B=A''/A=\omega^2 [/itex], find all solutions of the form [itex] u=A(x)B(t) [/itex]
Assume: [tex] u(x,t)=A(x)B(t) [/tex]
[tex] \frac{A(x)B''(t)}{A(x)B(t)}=\frac{A''(x)B(t)}{A(x)B(t)}[/tex]
[tex] \frac{B''}{B}=\frac{A''}{A}=-\omega^2 [/tex] is a second order ODE of the form:
[tex] B''+\omega^2 B = 0[/tex]
Solving yields:
[tex] B(t)=c_1 \cos \omega t + c_2 \sin \omega t [/tex] with the assumption that [tex] \omega > 0 [/tex]
and:
[tex] A(x)=d_1 \cos \omega x + d_2 \sin \omega x [/tex]
therefore:
[tex]u(x,t) = A(x)B(t)= (c_1 \cos \omega t + c_2 \sin \omega t)(d_1 \cos \omega x + d_2 \sin \omega x)[/tex]
And this is simply all the solutions right? It seems really straightforward, but sometimes when I think it is... it totally isn't. Thanks
I have a final coming up, and I want to make sure I have this method down.
Q: For the second-order wave equation [itex] u_{tt}=u_{xx}[/itex], the substitution of [itex] u=A(x)B(t) [/itex] will give second-order equations for A nd B when the x and t variables are seperated. From [itex] B''/B=A''/A=\omega^2 [/itex], find all solutions of the form [itex] u=A(x)B(t) [/itex]
Assume: [tex] u(x,t)=A(x)B(t) [/tex]
[tex] \frac{A(x)B''(t)}{A(x)B(t)}=\frac{A''(x)B(t)}{A(x)B(t)}[/tex]
[tex] \frac{B''}{B}=\frac{A''}{A}=-\omega^2 [/tex] is a second order ODE of the form:
[tex] B''+\omega^2 B = 0[/tex]
Solving yields:
[tex] B(t)=c_1 \cos \omega t + c_2 \sin \omega t [/tex] with the assumption that [tex] \omega > 0 [/tex]
and:
[tex] A(x)=d_1 \cos \omega x + d_2 \sin \omega x [/tex]
therefore:
[tex]u(x,t) = A(x)B(t)= (c_1 \cos \omega t + c_2 \sin \omega t)(d_1 \cos \omega x + d_2 \sin \omega x)[/tex]
And this is simply all the solutions right? It seems really straightforward, but sometimes when I think it is... it totally isn't. Thanks
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