Solve the initial value problem

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Homework Statement

Solve the initial value problem

y₁'=-13y₁+4y₂
y₂'=-24y₁+7y₂

y₁(0)=5, y₂(0)=2

Homework Equations

The Attempt at a Solution

Here is what I have:
y'=[-13 4, -24 7]y

I change it to A=[-13 4, -24 7]
My eigenvalues are λ=-1 and λ=-5.
My basis are [1/3 1] and [1/2 1].

Now I have to find u'=Du, which is:
[u₁', u₂'] = [-1 0, 0 -5][u₁, u₂]

Which then yields:
u₁'=-1u₁
u₂'=-5u₂

With these solutions:
u₁=C₁e^x
u₂=C₂e^2x

Now I obtain y=Pu, which is:
y=[1/3 1/2, 1 1][C₁e^x, C₂e^2x]
y=[(C₁e^x/3) + (C₂e^2x/2), C₁e^x + C₂e^2x]

Now I have:
y₁=(C₁e^x/3) + (C₂e^2x/2)
y₂=C₁e^x + C₂e^2x

Apply initial conditions to obtain:
y₁=(C₁/3) + (C₂/2)=5
y₂= C₁ + C₂=2

I solved and obtained:
C₁=28
C₂=-26

y₁=(28e^x/3) - (26e^2x/2)
y₂=28e^x - 26e^2x

Would this be right or did I make a mistake?

Thank you.

 

[Ray Vickson]

Homework Statement

Solve the initial value problem

y₁'=-13y₁+4y₂
y₂'=-24y₁+7y₂

y₁(0)=5, y₂(0)=2

Homework Equations

The Attempt at a Solution

Here is what I have:
y'=[-13 4, -24 7]y

I change it to A=[-13 4, -24 7]
My eigenvalues are λ=-1 and λ=-5.
My basis are [1/3 1] and [1/2 1].

Now I have to find u'=Du, which is:
[u₁', u₂'] = [-1 0, 0 -5][u₁, u₂]

Which then yields:
u₁'=-1u₁
u₂'=-5u₂

With these solutions:
u₁=C₁e^x
u₂=C₂e^2x

Now I obtain y=Pu, which is:
y=[1/3 1/2, 1 1][C₁e^x, C₂e^2x]
y=[(C₁e^x/3) + (C₂e^2x/2), C₁e^x + C₂e^2x]

Now I have:
y₁=(C₁e^x/3) + (C₂e^2x/2)
y₂=C₁e^x + C₂e^2x

Apply initial conditions to obtain:
y₁=(C₁/3) + (C₂/2)=5
y₂= C₁ + C₂=2

I solved and obtained:
C₁=28
C₂=-26

y₁=(28e^x/3) - (26e^2x/2)
y₂=28e^x - 26e^2x

Would this be right or did I make a mistake?

Thank you.

You can check it for yourself: just plug your solution into the DE system to see if it works. Checking your solution that way is something you should get in the habit of doing always, every single time.