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Y'' + y = f(x) - Variation of Parameters?

  • #1
FeDeX_LaTeX
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y'' + y = f(x) -- Variation of Parameters?

Homework Statement


Use variation of parameters to solve

##y'' + y = f(x), y(0) = y'(0) = 0.##


Homework Equations


A description of the method is here: http://en.wikipedia.org/wiki/Variation_of_parameters


The Attempt at a Solution


The complementary function is:

##y_{CF} = A\cos(x) + B\sin(x) = Ay_1 + By_2##, say.

If ##y = A(x)y_{1}(x) + B(x)y_{2}(x)## is the solution of the forced equation, then, if we choose ##A'y_1 + B'y_2 = 0##, ##y' = Ay'_{1} + By'_{2}## and ##y'' = A'y'_{1} + B'y'_{2}##.

Since ##y_{1,2}## satisfy the homogeneous equation, substitution will lead to ##A'y'_{1} + B'y'_{2} = f(x)##.

Solving these two equations for A' and B' yields

##A'(y_{2}y'_{1} - y_{1}y'_{2}) = f(x)y_{2}## and
##B'(y_{2}y'_{1} - y_{1}y'_{2}) = -f(x)y_{2}##

Now, if ##y_{1} = \cos(x), y_{2} = \sin(x)## then differentiating and substituting gives us

##A' = -f(x) \implies A = \int -f(x) dx## and
##B' = f(x) \implies B = \int f(x) dx##.

But from here, I'm stuck. Any advice?

EDIT: Am I going about this the wrong way? I'm following the example at the top of page 3 here.
 
Last edited:

Answers and Replies

  • #2
FeDeX_LaTeX
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Anyone? (Sorry, this homework is due pretty soon...)
 
  • #3
vanhees71
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Well, just solve your linear equations for [itex]A'[/itex] and [itex]B'[/itex] and then integrate. You are pretty close to the solution!
 
  • #4
FeDeX_LaTeX
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I've ended up with the particular integral:

##y = (\sin x - \cos x)\int f(x) dx##

But I think this is incorrect -- WolframAlpha is having sines and cosines in the integral itself. Where are those coming from?
 
  • #5
ehild
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Homework Statement


Use variation of parameters to solve

##y'' + y = f(x), y(0) = y'(0) = 0.##


The Attempt at a Solution


The complementary function is:

##y_{CF} = A\cos(x) + B\sin(x) = Ay_1 + By_2##, say.

If ##y = A(x)y_{1}(x) + B(x)y_{2}(x)## is the solution of the forced equation, then, if we choose ##A'y_1 + B'y_2 = 0##, ##y' = Ay'_{1} + By'_{2}## and ##y'' = A'y'_{1} + B'y'_{2}##.
Correctly: ##y'' = A'y'_{1} + B'y'_{2}+Ay''_{1}+By''_{2}##

Since ##y_{1,2}## satisfy the homogeneous equation, substitution will lead to ##A'y'_{1} + B'y'_{2} = f(x)##.

Solving these two equations for A' and B' yields

##A'(y_{2}y'_{1} - y_{1}y'_{2}) = f(x)y_{2}## and
##B'(y_{2}y'_{1} - y_{1}y'_{2}) = -f(x)y_{2}##

Now, if ##y_{1} = \cos(x), y_{2} = \sin(x)## then differentiating and substituting gives us

##A' = -f(x) \implies A = \int -f(x) dx## and
##B' = f(x) \implies B = \int f(x) dx##.


.
It is [tex]A' = \frac{f(x)y_2}{y_1'y_2-y_1y_2'}=-f(x)\sin(x)[/tex] and [tex]B' = -\frac{f(x)y_1}{y_1'y_2-y_1y_2'}=f(x)\cos(x)[/tex]


ehild
 
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