Page 89 - MATLAB an introduction with applications

74 ——— MATLAB: An Introduction with Applications

Example E1.25: Determine the solution of the following differential equations that satisfies the given initial
conditions.
dy
2
(a) =− 7x ; y (1) = 0.7
dx
dy
2
x
(b) = 5cos y ; y(0) = π/4
dx
dy
y
3x
(c) =− + e ; y(0) = 2
dx
dy
(d) + 5 y = 35 ; y(0) = 4
dt
Solution:
(a) >> dsolve (‘Dy = –7*x^2’,‘y (1) =0.7’)
ans =
–7*x^2*t + 7*x^2 + 7/10
(b) >> dsolve (‘Dy=5*x*cos (y) ^2’,‘y (0) =pi/4’)
ans =
atan (5*t*x + 1)
(c) >> dsolve (‘Dy = –y + exp (3*x)’, ‘y (0) =2’)
ans =
exp (3*x) + exp (– t)*(–exp (3*x) +2)
(d) >> dsolve (‘Dy + 5*y =35’, ‘y (0) =4’)
ans =
7 – 3*exp (–5*t)

Example E1.26: Given the differential equation
2
dx dx
5 = u t ; t ≥ 0
+ 7 + x 8 ( )
dt 2 dt
Using MATLAB program, find
(a) x(t) when all the initial conditions are zero.
(b) x(t) when x (0) = 1 and x = 2.
Solution:
(a) x (t) when all the initial conditions are zero
>> x = dsolve (‘D2x = –7*Dx – 5*x +8’, ‘x (0) = 0’)
x =
8/5+ (–8/5–C2)*exp (1/2*(–7+29^ (1/2))*t) + C2*exp (–1/2*(7+29^
(1/2))*t)

(b) x (t) when x (0) = 1 and x = 2
>> x = dsolve (‘D2x = –7*Dx – 5*x +8’, ‘x (0) = 1’, ‘Dx (0) = 2’)

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