Question

Let an electrical circuit be governed by the following system of differential equation where i1,i2,i3 are the current in each branch of electrical circuit.find the current i1,i2,i3 internship branch of the electrical circuit by using diagonalisationmethod i1'=2i1+2i2+i3;i2'=i1+3i2+i3; i3'=i1+2i2+2i3

Answer 1

Solution

Let $x(t)=\left(\begin{matrix}i1(t)\\i2(t)\\i3(t)\\\end{matrix}\right)$  and $A=\left(\begin{matrix}2&2&1\\1&3&1\\1&2&2\\\end{matrix}\right)$

Then the system of DE is x’ = Ax

Let’s find the eigenvalues and eigenvectors of matrix A.

det(A-λI) = 0, where I is a identity matrix.

$\left|\begin{matrix}2-\lambda&2&1\\1&3-\lambda&1\\1&2&2-\lambda\\\end{matrix}\right|=-\lambda^3+7\lambda^2-11\lambda+5$

-λ³+7λ²-11λ+5 = - (λ-5)(λ-1)² = 0  

So roots or eigenvalues are λ₁ = 5, λ_2,3 = 1

For λ=1 the associated eigenvectors satisfy

$\left(\begin{matrix}2&2&1\\1&3&1\\1&2&2\\\end{matrix}\right)\left(\begin{matrix}x\\y\\z\\\end{matrix}\right)=\left(\begin{matrix}x\\y\\z\\\end{matrix}\right)$

from which x+2y+z = 0, x+2y+z = 0, x+2y+z = 0

and we’ll get two eigenvectors $v_2=\left(\begin{matrix}3\\-1\\-1\\\end{matrix}\right),\ v_3=\left(\begin{matrix}1\\-1\\1\\\end{matrix}\right)$

For λ=1 the associated eigenvectors satisfy

$\left(\begin{matrix}2&2&1\\1&3&1\\1&2&2\\\end{matrix}\right)\left(\begin{matrix}x\\y\\z\\\end{matrix}\right)=5\left(\begin{matrix}x\\y\\z\\\end{matrix}\right)$

from which -3x+2y+z = 0, x-2y+z = 0, x+2y-3z = 0

and eigenvector $v_1=\left(\begin{matrix}1\\1\\1\\\end{matrix}\right)$

From these eigenvectors as columns matrix $S=\left(\begin{matrix}1&3&1\\1&-1&-1\\1&-1&1\\\end{matrix}\right)$  transform matrix A to diagonal

D = S^-1AS, $D=\left(\begin{matrix}5&0&0\\0&1&0\\0&0&1\\\end{matrix}\right)$

Thus A = SDS^-1 system of DE may be rewritten:

x’(t) = SDS^-1   =>  (S^-1x(t))’ = DS^-1x(t) or for y(t) = S^-1x(t) y’(t) = Dy(t) with diagonal matrix D.

Therefore $y(t)=\left(\begin{matrix}ae^{5t}\\be^t\\ce^t\\\end{matrix}\right)$  with arbitrary constants a,b,c.

$x(t)=Sy(t)=\left(\begin{matrix}ae^{5t}+3be^t+ce^t\\ae^{5t}-be^t-ce^t\\ae^{5t}-be^t+ce^t\\\end{matrix}\right)$

Finally i1(t) = ae^5t+3be^t+ce^t, i2(t) = ae^5t-be^t-ce^t, i3(t) = ae^5t-be^t+ce^t