Variation of parameters method (1 Viewer)

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djeitnstine

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1. The problem statement, all variables and given/known data

[tex]y''+y=tan(x)+e^{3x}-1[/tex]

2. Relevant equations

homogeneous solution:
[tex]y_{hom..}=C_{1}cos(x)+C_{2}sin(x)[/tex]

particular solution:
[tex]y_{parti..}=v_{1}' cos(x)+v_{2}' sin(x)[/tex]

3. The attempt at a solution

[tex]v_{1}' cos(x)+v_{2}' sin(x)=0[/tex] (1)

[tex]-v_{1}' sin(x)+v_{2}' cos(x) = tan(x)+e^{3x}-1 [/tex] (2)

(1)* by sin (2)* by cos

and solve for v2

[tex]sin(x)+e^{3x}cos(x)-cos(x)=v_{2}'[/tex]

[tex]-cos(x)+ \frac{3}{10}e^{3x}cos(x)-\frac{1}{10}sin(x)-sin(x)=v_{2}[/tex]

Now finding v1 is the problem here. if I reverse the multiplication and do (1)* by cos (2)* by sin it seems even worse =(. I saw the complete answer and if I remember correctly its [tex]y_{p}=\frac{1}{2}cos(x)ln(sin(x)+1)-\frac{1}{2}cos(x)ln(sin(x)-1)-\frac{1}{10}e^{3x}+1[/tex]

I really just need to know what to multiply by to get going
 

gabbagabbahey

Homework Helper
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particular solution:
[tex]y_{parti..}=v_{1}' cos(x)+v_{2}' sin(x)[/tex]
Why the primes? Why not just use [tex]y_p=v_{1}\cos(x)+v_{2}\sin(x)[/tex] as your ansatz (assumed form)?

[tex]v_{1}' cos(x)+v_{2}' sin(x)=0[/tex] (1)

[tex]-v_{1}' sin(x)+v_{2}' cos(x) = tan(x)+e^{3x}-1 [/tex] (2)
Where are these equations coming from?:confused:

Your particular solution should satisfy [itex]y_p''+y_p=\tan(x)+e^{3x}-1[/itex] not [itex]y_p=0[/itex] or (2)
 

HallsofIvy

Science Advisor
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As gabbagabbahey said, drop the primes from you solution- they are confusing.
y(x)= v1(x)cos(x)+ v2(x)sin(x)
Differenitate:
y'= v1' cos(x)- v1 sin(x)+ v2' sin(x)+ v2 cos(x)
Now, we can simplify this by "restricting" our search. Out of the infinitely many v1 and v2 that would satisfy the equation, we only look for those such that
v1' cos(x)+ v2' sin(x)= 0

So now
y'= -v1 sin(x)+ v2 cos(x)
Differentiate again: y"= -v1' sin(x)- v1 cos(x)+ v2' cos(x)- v2 sin(x).

Put that and the equation for y' into the original differential equation and it gives one equation for v1' and v2' (all of the v1 and v2 terms without ' will cancel). That, together with v1' cos(x)+ v2' sin(x)= 0 gives to equations to solve (algebraically) for v1' and v2' and then you integrate to find v1 and v2.
 

djeitnstine

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Ok thanks I finally got the answer.
 

djeitnstine

Gold Member
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Why the primes? Why not just use [tex]y_p=v_{1}\cos(x)+v_{2}\sin(x)[/tex] as your ansatz (assumed form)?



Where are these equations coming from?:confused:

Your particular solution should satisfy [itex]y_p''+y_p=\tan(x)+e^{3x}-1[/itex] not [itex]y_p=0[/itex] or (2)
These equations come from a proof shown by various texts including my professor. One example is at http://tutorial.math.lamar.edu/ under the Diff Eq. pdf if you'd like to see.
 

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