2.2.3 Example C The inhomogeneous equation Yk+1 – kyk 1 (2.21) has Pk k and qk = 1. Since %3D k I[i = k!, (2.22) i=1 then k-1 qi Pr (2.23) i=1 r=1 and equation (2.21) has the solution k-1 Yk = A(k –...

Explain this 2.2.3 Example C The inhomogeneous equation Yk+1 – kyk 1 (2.21) has Pk k and qk = 1. Since %3D k I[i = k!, (2.22) i=1 then k-1 qi Pr (2.23) i=1 r=1 and equation (2.21) has the solution k-1 Yk = A(k – 1)! + (k – 1)! ) (2.24) where A is an arbitrary constant. ーに

2.2.3<br>Example C<br>The inhomogeneous equation<br>Yk+1 – kyk<br>1<br>(2.21)<br>has Pk<br>k and qk = 1. Since<br>%3D<br>k<br>I[i = k!,<br>(2.22)<br>i=1<br>then<br>k-1<br>qi<br>Pr<br>(2.23)<br>i=1<br>r=1<br>and equation (2.21) has the solution<br>k-1<br>Yk =<br>A(k – 1)! + (k – 1)! )<br>(2.24)<br>where A is an arbitrary constant.<br>ーに<br>

Extracted text: 2.2.3 Example C The inhomogeneous equation Yk+1 – kyk 1 (2.21) has Pk k and qk = 1. Since %3D k I[i = k!, (2.22) i=1 then k-1 qi Pr (2.23) i=1 r=1 and equation (2.21) has the solution k-1 Yk = A(k – 1)! + (k – 1)! ) (2.24) where A is an arbitrary constant. ーに

Jun 04, 2022

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2.2.3 Example C The inhomogeneous equation Yk+1 – kyk 1 (2.21) has Pk k and qk = 1. Since %3D k I[i = k!, (2.22) i=1 then k-1 qi Pr (2.23) i=1 r=1 and equation (2.21) has the solution k-1 Yk = A(k –...

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