Mathematical Algorithms Solutions 2022

Q1
a. ,
Assuming an interval of 0≤ x ≤ 1 and h=0.2
dy
dx =x− y +1
For y(0.2)
f 0=f ( 0 ,0.2 ) = ( 0.2 ) − ( 0 ) +1
= 1.2
y(0.2)= y0 +h f 0
= 1 + 0.2(1.2)
= 1.24
For y(0.4)
f 0=f ( 0,0. 4 ) = ( 0. 4 )− ( 0 ) +1
= 1.4
y(0.2)= y0 +h f 0
= 1 + 0.2(1.4)
= 1.28
For y(0.6)
f 0=f ( 0,0.6 ) = ( 0. 6 ) − ( 0 ) +1

= 1.6
y(0.2)= y0 +h f 0
= 1 + 0.2(1.2)
= 1.32
b. ,
The solution would be given as
U = Ae λt+ Ae− λt
Inserting the bounderly conditions
X (0) = A + B = 0 , therefore A=-B
X(1) = A eλ +B e−λ = 0 , and A(e2 λ−1) =0
Since |λ | > 0 thereby having A=B=
u =0 this is the case with λ< 0
Now trying the same for λ> 0.
X(t) = A cos (√ λ x) + B sin ((√ λ x)
When applying boundary conditions.
0 = X(0) = A and B sin √ λ =0 since B≠ 0
Sin √ λ=0
√ λ=nπ where n = 1, 2, 3, . . .
Now solving for U
U' ( t ) = −n2 π2 T (t )
The solution is
T N =cn e−n2 π2 t and n= 1,2,3,4...