Third Order Differential Equation using the reduction of order method.Solve the following differential equation y‴−2y″+y′−2y=0 (1) using the reduction of order method and also by replacing y with z=z(x)=y′−2y (2). So z′=y″−2y′(3),z″=y‴−2y″(4) therefore (1) becomes z″+z=0 and we can easily find the general solution. So my question is... what about the function z′=y″−2y′.We can not plug that in (1) because the function z″=y‴−2y″ is the first two terms of the original differential equation and the function z=y′−2y is that last two terms.

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carlosegundoacyg · Accepted Answer

Answer:y‴−2y″+y′−2y=(y″−2y′⏟z′)′+y′−2y⏟z=0,so z″+z=0. So z=c1cos⁡(t)+c2sin⁡(t), and then you can use the assignment z=y′−2y to see you need only solve y′−2y=c1cos⁡(t)+c2sin⁡(t). The solutions to z=y′−2y provides the solutions to z′=y2y′, so we don&#039;t need to put any extra effort there.

Third Order Differential Equation using the reduction of

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