Expressing the Solution to a DE in the Form ce\alphax*sin(\betax+\gamma)

[hover]

Homework Statement

Solve the DE:
y''-2y'+2y=0

and express your answer in the form

ce^$$\alpha$$x*sin($$\beta$$x+$$\gamma$$)

where alpha = ;
and beta =

Homework Equations

y = e^$$\lambda$$x

The Attempt at a Solution

When you plug in the values of y into the DE and solve for lambda you get
$$\lambda$$ = 1+ i or 1-i

A particular solution to the DE is
y(x)= e^(1+i)x

when this is expanded using Euler's formula to solve the general solution to the DE, I get

y(x) = k₁e^xcos(x)+k₂e^xsin(x)

When looking at the form of the solution they want me to express my answer in I feel like that are asking for

k₂e^xsin(x) = ce^$$\alpha$$x*sin($$\beta$$x+$$\gamma$$)

alpha = 1 but how do I get beta? I don't think I can express my answer with gamma

Thanks!

 

[fzero]

If you rewrite

$$y(x) = \sqrt{k_1^2 + k_2^2}~ e^x \left( \frac{k_1}{\sqrt{k_1^2 + k_2^2}} \cos(x)+\frac{k_2}{\sqrt{k_1^2 + k_2^2}} \sin(x) \right),$$

can you guess at how the angle $$\gamma$$ is defined?

 

[Delta2]

To me it seems that beta=1 and gamma is such that cos(gamma)=k2, sin(gamma)=k1. All these having the trigonemetric identity sin(bx+g)=sin(g)cos(bx)+cos(g)sin(bx).

 

[tiny-tim]

isn't γ an entirely arbitrary constant, exactly like c ?