how to solve this differential equations with dsolve
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x'+2x+y=0
y'+x+2y=0
t=0 => x=1 , y=0
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Star Strider
le 23 Déc 2016
It is straightforward to incorporate the initial conditions in the dsolve call:
syms x(t) y(t)
Dx = diff(x);
Dy = diff(y);
[x,y] = dsolve(Dx + 2*x + y == 0, Dy + x + 2*y == 0, x(0) == 1, y(0) == 0)
x =
exp(-t)/2 + exp(-3*t)/2
y =
exp(-3*t)/2 - exp(-t)/2
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John BG
le 23 Déc 2016
Modifié(e) : John BG
le 23 Déc 2016
1.
solving the system
syms x(t) y(t)
z=dsolve(diff(x)==-y-2*y,diff(y)==-x-2*y)
z.x
=
C2*exp(-3*t) - 3*C1*exp(t)
z.y
=
C1*exp(t) + C2*exp(-3*t)
2.
applying initial conditions, A(t=0):
A=[1 -3;1 1]
b=[1;0]
s=A\b
=
0.250000000000000
-0.250000000000000
C1=s(1)
C1 =
0.250000000000000
C2=s(2)
C2 =
-0.250000000000000
3. Build real functions
fx=matlabFunction(z.x)
fx =
@(C1,C2,t)C1.*exp(t).*-3.0+C2.*exp(t.*-3.0)
fy=matlabFunction(z.y)
fy =
@(C1,C2,t)C1.*exp(t)+C2.*exp(t.*-3.0)
t=[10:.1:10]
fx(C1,C2,t)
=
-1.651984934610504e+04
fy(C1,C2,t)
=
5.506616448701680e+03
if you find these lines useful would you please mark my answer as Accepted Answer?
To any other reader, please if you find this answer of any help, click on the thumbs-up vote link,
thanks in advance for time and attention
John BG
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