Answer to Question #279781 in Differential Equations for Khan

Question #279781

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).

1
Expert's answer
2021-12-15T16:21:47-0500

a)

ykyk1+2yk2=k²+5ky_k - y_{k-1} + 2y_{k-2} = k² + 5k


charasteristic equation:

r2r+2=0r^2-r+2=0


r=1±i72r=\frac{1\pm i \sqrt 7}{2}


yh=rk(acos(kθ)+bsin(kθ))y_h=r^k(acos(k\theta)+bsin(k\theta))


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

θ=arccos(1/(22))=1.2\theta=arccos(1/(2\sqrt 2))=1.2 rad


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


yt=Ak2+Bk+Cy_t=Ak^2+Bk+C


Ak2+Bk+C+A(k1)2+B(k1)+C+2A(k2)2+2B(k2)+2C=Ak^2+Bk+C+A(k-1)^2+B(k-1)+C+2A(k-2)^2+2B(k-2)+2C=

=k2+5k=k^2+5k


4A=1    A=1/44A=1\implies A=1/4

4B10A=5    B=1.8754B-10A=5\implies B=1.875

4C+9A5B=0    C=1.784C+9A-5B=0\implies C=1.78


yk=(2)k(acos(1.2k)+bsin(1.2k))+0.25k2+1.875k+1.78y_k=(\sqrt 2)^k(acos(1.2k)+bsin(1.2k))+0.25k^2+1.875k+1.78


b)

yk+24yk+1+yk=3k+2ky_{k+2} - 4y_{k+1} + y_k = 3k +2^k


charasteristic equation:

r24r+1=0r^2-4r+1=0


r=2±3r=2\pm \sqrt 3

yh=a(23)k+b(2+3)ky_h=a(2-\sqrt 3)^k+b(2+\sqrt 3)^k


yt1=Ak+By_{t1}=Ak+B

A(k+2)+B4(A(k+1)+B)+Ak+B=3kA(k+2)+B-4(A(k+1)+B)+Ak+B=3k

2A=3    A=1.5-2A=3\implies A=-1.5

2A2B=0    B=1.5-2A-2B=0\implies B=1.5


yt2=A2ky_{t2}=A2^k

A2k+24A2k+1+A2k=2kA2^{k+2}-4A2^{k+1}+A2^k=2^k

4A8A+A=14A-8A+A=1

A=1/3A=-1/3


yk=a(23)k+b(2+3)k1.5k+1.52k/3y_k=a(2-\sqrt 3)^k+b(2+\sqrt 3)^k-1.5k+1.5-2^k/3


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