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Problem # 1 [30 Points] Superposition-The Backbone of Linear Systems: For an engineer who wishes to find the response u to a
and BCs. Ju(0, t) = 0 0<t< oo P3 u(1, t)0 u(z,0)=sin(2mz), IC: These two problems (i.e., P2 and P3) can be solved individuall
Problem # 1 [30 Points] Superposition-The Backbone of Linear Systems: For an engineer who wishes to find the response u to a linear system from an input f, a common approach is (a) Break finto elementary parts, f-L (b) Find the system response u to fk (c) Add (superimpose) the simple responses uto get u-k It turns out if the system is linear, then the sum u is the response we get if the function f were inputed directly; this is the Principle of Superposition (see Figure 1 below). Could be a linear is a differential operator f is the input and the output Figure 1: Basic idea of Superposition Using this basic idea, solve the following initial-boundary problem (ie, a non-homogeneous heat equation) u(0,t)0 IC: u(,0)sin(2x), by breaking the problem into subproblems. Note that the method of solve the above problem. Hence, one possible approach would be to con- sider two problems separation of variables is not a viable method to u(0,t) 0 IC: u(r,0)0,
and BCs. Ju(0, t) = 0 0
0 0
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Cs -t -0 Then we find Dn; Hente vtnt) o 惌.

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