({{ Q1/12    ,    Q2   ,    Q3   ,    Q4   , Q5, Q6, Q7, Q8,    Total   }, {      ,       ,       ,       ,      ,      ,      ,      ,       }})

Struthers: MA3520 Final  
Name_________________

Show polynomials matrices etc. you enter in calculators, no points for dsolve, no imaginary answers.
Put extra work on the back of pages.

Solve 
({{X', =, (({{2, -1}, {-1, 2}})).X+}})(({{2+6 t}, {1}}))  with x(0)=(({{5}, {8}}))

Eigensystem[(({{2, -1}, {-1, 2}}))]
xp[t_]= {a1,a2} t + {b1,b2};
SolveAlways[ xp'[t]=={{2,-1},{-1,2}}.xp[t]+{2+6 t,1},t]

LinearSolve[ {{-1,1},{1,1}}, {5,8}+{5,4}]

Solve 
({{X', =, (({{3, 5}, {-5, 3}})).X+cos(2t)}})(({{1}, {1}})).

({{Score/12}, {      }})

Solve the Ordinary Differential Equation 
({{X', =, (({{-24, -24, 32}, {-14, -8, 16}, {-27, -26, 36}})).X+(e^(2t))}})(({{6}, {2}, {7}})). Mathematica gives eigenvalues as {-4,4,4} and eigenvectors as {{2,1,2},{4,2,5},{0,0,0}}.  
Fill in the matrix you row-reduce to satisfy x[0]=(({{7}, {13}, {-2}})).
({{ ,  ,  ,     }, {      ,       ,       ,        }, {      ,       ,       ,        }})

({{Score/13}, {      }})

Solve x y' - 2 (x^2) y = 6(e^(x^2))  with y[1]==2

y[x]/.DSolve[ {x y'[x]- 2 (x^2)y[x]== 6 (E^(x^2)),y[1]==2}, y[x],x][[1]]

({{Score/12}, {      }})

Solve (3(y^2)+ 2 x y cos(x (y^2)))(dy/dx)=-(2x + (y^2)cos(x (y^2))) with y(1)=0.

D[ Sin[x (y^2)]+(x^2)+(y^3),{{x,y}}]

DSolve[ (t^2)y''[t]+ 5 t y'[t]+4 y[t]==50(t^3), y[t],t]

({{Score/12}, {      }})

Use Euler method to obtain a 4 decimal approximation of y(1.2) for 
y'=-2y+(e^x) with y(1)=1.2. 
First use h=0.2 and then h=0.1. Find an explict solution for the initial value problem and fill in the tables.

({{Score/13}, {      }})

Clear[y,x,ySol]
f[x_,y_]:= -2 y +  (E^x)
ySol[x_]=Expand[ y[x]/.DSolve[ {y'[x]==f[x,y[x]],y[-1]==0.2},y[x],x][[1]]]
h=0.2;
{x,y}={1,1.2};
TableForm[
Join[ {{x,y,0.3,0.0,0.0}},
Table[
{x,y}={x,y}+h {1, f[x,y]};
(*{x,y,ySol[x],Abs[y-ySol[x]], 100Abs[y-ySol[x]]/Abs[ySol[x]]},*)
{x,"","","",""},
{1}]
],
TableHeadings->{Automatic,{"(Subscript[x, n])", "(Subscript[y, n])", "Actual Value", "Abs. Error", "%Relative Error"}}
]
h=0.1;
{x,y}={1,1.2};
TableForm[
Join[ {{x,y,0.3,0.0,0.0}},
Table[
{x,y}={x,y}+h {1, f[x,y]};
(*{x,y,ySol[x],Abs[y-ySol[x]], 100Abs[y-ySol[x]]/Abs[ySol[x]]},*)
{x,"","","",""},
{2}]
],
TableHeadings->{Automatic,{"(Subscript[x, n])", "(Subscript[y, n])", "Actual Value", "Abs. Error", "%Relative Error"}}
]

Solve y''+2y'+y=t+(e^(-t))

Clear[y,t]
DSolve[ 2 (t^2)y''[t]+5 t y'[t] +y[t]== Log[t], y[t],t] 
{{y[t]->-3+((C[1])/(Sqrt[t]))+((C[2])/t)+Log[t]}}
(2m+1)(m+1)//Expand
1+3 m+2 (m^2)

({{Score/13}, {      }})

Show that (Subscript[y, 1])=(t^(-1)) solves 2(t^2)y''+5 t y'+y=0.
Solve 2(t^2)y''+5 t y'+y=ln(t) with y(1)=1 and y'(1)=2 given that
(Subscript[y, c])= (Subscript[c, 1])(Subscript[y, 1])+(Subscript[c, 2])(t^(-0.5)) is the complementary solution. Show equations and explain process.

Clear[y,t]
DSolve[ 2 (t^2)y''[t]+5 t y'[t] +y[t]== Log[t], y[t],t] 
{{y[t]->-3+((C[1])/(Sqrt[t]))+((C[2])/t)+Log[t]}}
(2m+1)(m+1)//Expand
1+3 m+2 (m^2)

({{Score/13}, {      }})