Reduce Order | Find Second Solution for $$y''+4y=0$$

[QuantumCurt]

Homework Statement

Given - $$y_1(x)=sin(2x)$$, find a second linearly independent solution to $$y''+4y=0$$

Homework Equations

I'm using reduction of order to write a second solution as a multiple of the first solution. I think I've gotten to the right answer, my main question is in how I should write the general solution of the equation.

The Attempt at a Solution

Using the fact that $y_1(x)=sin(2x)$, and that $y_2(x)=vy_1(x)$ I'm writing a second solution as $$y_2(x)=vsin(2x)$$
Now taking the first and second derivatives I get $$y_2'(x)=v'sin(2x)+2vcos(2x)$$
and $$y_2''(x)=v''sin(2x)+4v'cos(2x)-4vsin(2x)$$

Now I substitute back into the original equation and get, $$v''sin(2x)+4v'cos(2x)=0$$

Now to reduce order; $z=v' and z'=v''$

Then substituting - $$z'sin(2x)+4zcos(2x)=0$$

Which leads to $$\frac{dz}{z}=-4cot(2x)dx$$

Integrating this to logarithms and then exponentiating etc. leads me to - $$z=csc^2(2x)$$

Since $z=v'$ I can substitute again
$$v=\int csc^2(2x)dx$$
$$v=-\frac{1}{2}cot(2x)$$

Now since $y_2(x)=vy_1(x)$, I can say that,
$$y_2(x)=[-\frac{1}{2}cot(2x)][sin(2x)]$$
which simplifies to
$$y_2(x)=-\frac{1}{2}cos(2x)$$

The formula for the general solution is $y(x)=c_1y_1(x)+c_2y_2(x)$. Given this fact, my general solution is written as - $$y(x)=c_1sin(2x)-\frac{1}{2}c_2cos(2x)$$

Now...my question -

Since $-\frac{1}{2}c_1$ is really just an arbitrary constant, can I write my general solution in terms of just a constant rather than a fraction times a constant? This would give me the solution of -
$$y(x)=c_1sin(2x)+c_2cos(2x)$$

Is this essentially the same solution, or would it be better to write it in terms of the negative fractional constant?

Any help would be much appreciated. :)

 

[vela]

Just absorb the -1/2 into c₂.

 

[QuantumCurt]

Thank you. I figured that was the case, but I wanted to be sure.