Solution of second order differential eqauation $ D''y + (k)^2 y=0$

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What is the need to write two different solutions of the following equation ( although they are same solutions as I think) -

$$ D''y + (k)^2 y=0$$

( where D" means ordinary derivative of y with respect to x, and symbol ^ denotes power raised to )

Solution one $$ y = A sin(kx) + B cos(kx)$$

Solution two $$ y = A exp(ikx) + B exp( -ikx)$$

When I told my teacher from where these solutions came from,she said the equation you are showing is a standard equation and you have to remember the solutions. But I want to know how did we reach at these solutions. I mean when we have first order differential equation, we obtain the solutions by Integration or by some any method using integration. I want somewhat that kind of solution which have been obtained by integration or I mean I don't want to cram these solutions but want to know from where they came from and what is the actual way to obtain them?

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You are looking for a solution to $$y''+k^2 y =0$$

That means you are looking for a function whose second derivative is a constant multiple of the original function.

What types of functions have the property that when you take second derivative you get a constant multiple of your function?

Well $$y= e^{\lambda x}$$ is a good candidate.

Plug in your equation and solve for $\lambda$ and you will get the correct answers.

In the process you need to know $$ e^{ix} = \cos x + i \sin x $$

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Solution two is equal to one $$ y = A exp(ikx) + B exp( -ikx)=A(cos kx+i sin kx) + B (cos kx-i sin kx)=(A+B) cos kx+(A-iB) sin kx)= A_1 sin(kx) + B_1 cos(kx)$$