MATH528Due date: 10/4/18

Homework 05

1Exercises

Exercise. PS4.3.1

Solution. The general solution of the system 𝒚'=𝑨⁡𝒚 with

𝑨=( 1 1 3 -1 )

is 𝒚(t)=c1eλ1t⁡𝒙1+c2eλ2t⁡𝒙2, with λ1,λ2 eigenvalues of 𝑨, and 𝒙1,𝒙2 the corresponding eigenvectors, and 𝒄=𝑿-1𝒚(0).

In[58]:=

A={{1,1},{3,-1}}; MatrixForm[A]

( 1 1 3 -1 )

In[59]:=

lambda = Eigenvalues[A]; Lambda = DiagonalMatrix[lambda]; MatrixForm[Lambda]

( -2 0 0 2 )

In[62]:=

X = Transpose[Eigenvectors[A]]; MatrixForm[X]

( -1 1 3 1 )

The general solution is

𝒚=c1e-2⁡t( -1 3 )+c2e2⁡t( 1 1 ).

Here are some additional verifications not required in your homework solution, but given to show that the above result is confirmed by the Mathematica DSolve function.

In[74]:=

y[t_]={y1[t],y2[t]};

sol[t_]=Expand[DSolveValue[ {y'[t] == A . y[t], y1[0]==y10, y2[0]==y20},y[t],t]]

{12⁢e-4⁢ty10+12⁢e-2⁢ty10-12⁢e-4⁢ty20+12⁢e-2⁢ty20,-12⁢e-4⁢ty10+12⁢e-2⁢ty10+12⁢e-4⁢ty20+12⁢e-2⁢ty20}

In[49]:=

sol[0]

{y10,y20}

In[75]:=

c = Inverse[X]. {y10,y20};

ans[t_]=Expand[c[[1]] Exp[lambda[[1]] t] X[[1]] + c[[2]] Exp[lambda[[2]] t] X[[2]]]

{12⁢e-4⁢ty10+12⁢e-2⁢ty10-12⁢e-4⁢ty20+12⁢e-2⁢ty20,-12⁢e-4⁢ty10+12⁢e-2⁢ty10+12⁢e-4⁢ty20+12⁢e-2⁢ty20}

In[78]:=

Simplify[ans[t] == sol[t]]

True

Exercise. PS1.1.6

Solution.

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Exercise. PS1.1.7

Solution.

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Exercise. PS1.1.8

Solution.

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Exercise. PS1.2.2

Solution.

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Exercise. PS1.2.3

Solution.

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Exercise. PS1.2.4

Solution.

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Exercise. PS1.3.5

Solution.

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2Problems

Problem. PS1.1.16

Solution.

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Problem. PS1.1.17

Solution.

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Problem. PS1.1.18

Solution.

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Problem. PS1.3.22

Solution.

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3Projects

3.1PS2.1.16