Author Topic: TUT0801  (Read 313 times)

ZYR

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TUT0801
« on: October 11, 2019, 03:22:10 PM »
Verify that the function $y_1$ and $y_2$ are solutions of the given differential equation. Do they constitute a fundamental set of solutions?

    $x^2y''-x(x+2)y' + (x+2)y = 0$, $x >0$, $y_1 = x$, $y_2 = xe^x$
   
$W(y_1, y_2)(x) = \det{\begin{vmatrix}y_1 & y_2\\
   y_1' & y_2'\end{vmatrix}}
 = \det{\begin{vmatrix}x & xe^x \\
   1 & xe^x + e^x\end{vmatrix}}
 = x(xe^x + e^x) - xe^x
 = xe^x( x+1-1)
 = x^2e^x $

 since $x > 0$ and $e^x \neq 0$,  so $W = x^2e^x \neq 0$.
 Therefore, $y_1$ and $y_2$ constitute a fundamental set of solutions.

 
 Verify :
 1)Show $y_1 = x$ is one of the solutions
 We know $y_1 = x$, $y_1' = 1$, $y_1'' = 0$, and then use these substitute in the equation, we have
  LHS: $-x(x+2) + (x+2)x = 0$
  RHS : 0
  since LHS = RHS = 0, so $y_1 = x$ is one of the solutions

 2) Show $y_2 = xe^x$ is one of the solutions
 We know $y_2 = xe^x$, $y_2' = xe^x + e^x$, $y_2'' = e^x(x+2)$, also use thess substitute in the equation, we get
 LHS: $x^2e^x(x+2)-x(x+2)(xe^x + e^x) + (x+2)xe^x =  (x+2)e^x(x^2 -x(x+1) + x ) =  (x+2)e^x(x^2 -x^2 - x + x ) = (x+2)e^x 0 = 0$
 RHS : 0
 since LHS = RHS = 0, so $y_2 = xe^x$ is one of the solutions.