Answer to Question #279781 in Differential Equations for Khan

Question

Solve the differential equation:

a) yk - y(k-1) + 2y(k-2) = k² + 5k

b) y(k+2) - 4y(k+1) + yk = 3k +2^k

In LHS yk , y(k-1) and so on is not in multiply but the k part is written in down to y( like yk is not y×k but the k is written in right down of y).

Answer 1

a)

$y_k - y_{k-1} + 2y_{k-2} = k² + 5k$

charasteristic equation:

$r^2-r+2=0$

$r=\frac{1\pm i \sqrt 7}{2}$

$y_h=r^k(acos(k\theta)+bsin(k\theta))$

$r=\sqrt{(1/2)^2+(\sqrt 7/2)^2}=\sqrt 2$

$\theta=arccos(1/(2\sqrt 2))=1.2$ rad

$y_h=(\sqrt 2)^k(acos(1.2k)+bsin(1.2k))$

$y_t=Ak^2+Bk+C$

$Ak^2+Bk+C+A(k-1)^2+B(k-1)+C+2A(k-2)^2+2B(k-2)+2C=$

$=k^2+5k$

$4A=1\implies A=1/4$

$4B-10A=5\implies B=1.875$

$4C+9A-5B=0\implies C=1.78$

$y_k=(\sqrt 2)^k(acos(1.2k)+bsin(1.2k))+0.25k^2+1.875k+1.78$

b)

$y_{k+2} - 4y_{k+1} + y_k = 3k +2^k$

charasteristic equation:

$r^2-4r+1=0$

$r=2\pm \sqrt 3$

$y_h=a(2-\sqrt 3)^k+b(2+\sqrt 3)^k$

$y_{t1}=Ak+B$

$A(k+2)+B-4(A(k+1)+B)+Ak+B=3k$

$-2A=3\implies A=-1.5$

$-2A-2B=0\implies B=1.5$

$y_{t2}=A2^k$

$A2^{k+2}-4A2^{k+1}+A2^k=2^k$

$4A-8A+A=1$

$A=-1/3$

$y_k=a(2-\sqrt 3)^k+b(2+\sqrt 3)^k-1.5k+1.5-2^k/3$