how to solve this differential equations with dsolve

x'+2x+y=0

y'+x+2y=0

t=0 => x=1 , y=0

Answer 1

[Star Strider]

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

[Furkan]

Thanks this is what ı want .

best regards

[John BG]

when applying the initial conditions x(0)=1 y(0)=0 the resulting functions vary slightly, have a look to my answer.

when no initial conditions specified dsolve removes C1 and C2 by choosing arbitrary constants.

[Star Strider]

My pleasure!

[Furkan]

Thanks Mr. John BG but first part of answer already enough for me, ı dont need more details at this problem.But both of answers are correct and usefull thanks . Best regards.

[John BG]

it's ok, thanks for reading my answer

Answer 2

[John BG]

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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