anonymous
  • anonymous
Find a solution to f′′− 2f′− 3f = 2 cos(x) which also satisfies f(0) = 2 and f′(0) = 3. -4/11sinx-2/11cox solves the equation without the conditions..
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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amistre64
  • amistre64
\[\begin{vmatrix}W_1&W&W_2\\y_1&0&y_2\\ y_1^{'}&2cos(x)&y_{2}^{'}\end{vmatrix}\]
amistre64
  • amistre64
if you take the cross product/ determinant of that you will get your Wronskian values; and y1 y2 are the solutions to your homogenous part
amistre64
  • amistre64
only difference is you dont need to negate the W part :)

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anonymous
  • anonymous
ok thank you!! how would i go about finding the solutions to the homogeonous part?!
amistre64
  • amistre64
by applying the characteristic equation associated with a second order homogenous linear diffyQ
amistre64
  • amistre64
assume e^rx is an answer and apply it to the problem in place of y y' and y''
amistre64
  • amistre64
what you end up with is something looking like a quadratic equation to find the roots to: \[e^{r}(r^2-2r-3)=0\] solve for "r"
anonymous
  • anonymous
ohh thank you!!!
amistre64
  • amistre64
yep
anonymous
  • anonymous
sorry can i ask how to go abotu solving e^r = 0?
amistre64
  • amistre64
since e^r is never zero; its not really important. find the zeros of the quadratic part and you are good to go
anonymous
  • anonymous
so r = -3, 1? and those are the solutions to put into the matrix?!
amistre64
  • amistre64
close, those are solutions of e^rx that are put into the matrix (r-3)(r+1) = 0 ; r = -1,3 \[y=y_h+y_p\]\[y_h=c_1e^{3x}+c_2e^{-x}\]\[y_p=e^{3x}\int \frac{W_1}{W}+e^{-x} \int \frac{W_2}{W}\]

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