# Variation of parameters method (1 Viewer)

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#### djeitnstine

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

$$y''+y=tan(x)+e^{3x}-1$$

2. Relevant equations

homogeneous solution:
$$y_{hom..}=C_{1}cos(x)+C_{2}sin(x)$$

particular solution:
$$y_{parti..}=v_{1}' cos(x)+v_{2}' sin(x)$$

3. The attempt at a solution

$$v_{1}' cos(x)+v_{2}' sin(x)=0$$ (1)

$$-v_{1}' sin(x)+v_{2}' cos(x) = tan(x)+e^{3x}-1$$ (2)

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

and solve for v2

$$sin(x)+e^{3x}cos(x)-cos(x)=v_{2}'$$

$$-cos(x)+ \frac{3}{10}e^{3x}cos(x)-\frac{1}{10}sin(x)-sin(x)=v_{2}$$

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 $$y_{p}=\frac{1}{2}cos(x)ln(sin(x)+1)-\frac{1}{2}cos(x)ln(sin(x)-1)-\frac{1}{10}e^{3x}+1$$

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

#### gabbagabbahey

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

$$v_{1}' cos(x)+v_{2}' sin(x)=0$$ (1)

$$-v_{1}' sin(x)+v_{2}' cos(x) = tan(x)+e^{3x}-1$$ (2)
Where are these equations coming from?

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

#### HallsofIvy

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

Gold Member
Ok thanks I finally got the answer.

#### djeitnstine

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

Where are these equations coming from?

Your particular solution should satisfy $y_p''+y_p=\tan(x)+e^{3x}-1$ not $y_p=0$ 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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