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## windsylph 3 years ago PDE's: For the partial differential equation u_tt = 9u_xx, a general solution is u(x,t) = F(x+3t)+G(x-3t). Find a particular solution with initial conditions u(x,0)=0, u_t(x,0)=sin(pi*x), u(0,t)=u(1,t)=0.

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1. windsylph

I'm having trouble especially using/interpreting the last initial condition.

2. myininaya

before we look at initial conditions do you know how to do this part: solve for particular solution $u_{t t}=9u_{xx}$ ? is it separation of variables?

3. windsylph

If the problem asked me to solve it, yeah I would use separation of variables since that's the only method I know so far. but the problem only stated that the given solution is indeed a solution. actually, the first part of this problem was to show that the given solution is indeed a solution, the second part is this.

4. myininaya

ok

5. myininaya

$u(x,0)=F(x+3(0))+G(x-3(0))=F(x)+G(x)=0$ f(x)+g(x)=0 $u_t=3f(x+3t)-3g(x-3t)$ $u_t(x,0)=3f(x)-3g(x)=\sin(\pi x)$ $u(0,t)=u(1,t)=0$ $u(0,t)=F(0+3t)+G(0-3t)=F(3t)+G(-3t)=0$ $u(1,t)=F(1+3t)+G(1-3t)=0$

6. myininaya

so we have $F(x)+G(x)=0$ $f(x)+g(x)=0$ $3f(x)-3g(x)=\sin(\pi x)$ $F(3t)+G(-3t)=0$ $F(1+3t)+G(1-3t)=0$

7. myininaya

$f(x)+g(x)=0 => f(x)=-g(x) =>3f(x)+3f(x)=\sin(\pi x)$

8. myininaya

$f(x)=\frac{1}{6} \sin(\pi x)$

9. myininaya

F(x)=$\int\limits_{}^{}f dx=\int\limits_{}^{}\frac{1}{6} \sin(\pi x) dx=-\frac{1}{6} \cos(\pi x) +c_1$

10. myininaya

$3f(3t)-3g(-3t)=0 => f(3t)-g(-3t)=0 => f(3t)=g(-3t)$ $3f(1+3t)-3g(1-3t)=0 => f(1+3t)=g(1-3t)$ i hope some of this will help i have to leave

11. windsylph

thanks still! ill look at it

12. myininaya

you might want to check what i wrote for mistakes gl

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