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quasar987

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## Main Question or Discussion Point

If we have a constant coefficient second order homogeneous ODE, the way to solve this is to suppose a solution of the exponential type. This yields a second order polynomial equation (the "characteristic equation") that the exponent must satisfy. In case the solutions of the characteristic equation are complex, we get a solution of the form [itex]c1e^{wx}+c2e^{\overline{w}x}[/itex] (call this

To my eyes, forms 1 and 2 are equivalent. So then, how come trying to solve an initial value problem with form 1 does not work?!?

Consider for instance y''-2y+2=0 with y(0)=1, y'(0)=0. The roots of the char. equ. are 1+i and 1-i. If we try to find c1 and c2 using the initial values given we are lead to the impossible system c1 + c2 = 1, c1 + c2=0, c1 - c2=0. With form 2 however, we find A=1, B=-1.

Thanks, I'm confused!

**form 1**). The general solution of the ODE is usually presented in the form e^x(Acos(bx)+Bsin(bx)) (call this**form 2**) by using Euler's formula for e^{a+ib}.To my eyes, forms 1 and 2 are equivalent. So then, how come trying to solve an initial value problem with form 1 does not work?!?

Consider for instance y''-2y+2=0 with y(0)=1, y'(0)=0. The roots of the char. equ. are 1+i and 1-i. If we try to find c1 and c2 using the initial values given we are lead to the impossible system c1 + c2 = 1, c1 + c2=0, c1 - c2=0. With form 2 however, we find A=1, B=-1.

Thanks, I'm confused!