MHB Example of systems of the differential linear equations

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The discussion centers on solving a system of differential linear equations using operator notation. The key step involves eliminating the variable x by multiplying the first equation by 4 and applying the operator (D-3) to the second equation. This results in a new equation that combines terms related to y, allowing for simplification. The final expression derived from this process is used to solve for y in the system. Understanding these manipulations is crucial for effectively solving such differential equations.
cbarker1
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Dear Everybody,

I have a question about an example:
"Solve the system:

$x'(t)=3x(t)-4y(t)+1$
$y'(t)=4x(t)-7y(t)+10t$

We write the system using the operator notation:

$(D-3)[x]+4y=1$
$-4x+(D+7)[y]=10t$

We can eliminate x from this system by adding 4 times the first equation to $(D-3)$ applied to the equation. This gives
$(16+(D-3)(D+7))[y]=4*1+(D-3)[10t]=14-30t$"

How does this step work: "We can eliminate x from this system by adding 4 times the first equation to $(D-3)$ applied to the equation."

Thanks
Cbarker1
 
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Cbarker1 said:
Dear Everybody,

I have a question about an example:
"Solve the system:

$x'(t)=3x(t)-4y(t)+1$
$y'(t)=4x(t)-7y(t)+10t$

We write the system using the operator notation:

$(D-3)[x]+4y=1$
$-4x+(D+7)[y]=10t$

We can eliminate x from this system by adding 4 times the first equation to $(D-3)$ applied to the equation. This gives
$(16+(D-3)(D+7))[y]=4*1+(D-3)[10t]=14-30t$"

How does this step work: "We can eliminate x from this system by adding 4 times the first equation to $(D-3)$ applied to the equation."

Thanks
Cbarker1

Hi Cbarker1,

Let's break it up into smaller steps.
We multiply the first equation by 4.
And we apply $(D-3)$ to the second equation:

$4(D-3)[x]+16y=4*1$
$-4(D-3)[x]+(D-3)(D+7)[y]=(D-3)[10t]$

Now we add them together:

$16y+(D-3)(D+7)[y]=4*1+(D-3)[10t]$

And from here we find the given expression.
 

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