- #1
cragar
- 2,552
- 3
If I have [itex] y''+y'+2y=sin(x)+cos(x) [/itex]
can I just say [itex] y=Ae^{ix} [/itex]
and then find y' and y'' and then plug them in and solve for A.
so I get that [itex] A= \frac{1}{1+i} [/itex]
then i multiply and divided by the complex conjugate.
then I back substitute in Eulers formula.
now since I have my original equation has both a real and an imaginary part.
when I multiply out A times Eulers formula, I will take both real and imaginary parts.
and I get that y=sin(x) and tested this and it works.
But is what i did with Eulers formula ok.
can I just say [itex] y=Ae^{ix} [/itex]
and then find y' and y'' and then plug them in and solve for A.
so I get that [itex] A= \frac{1}{1+i} [/itex]
then i multiply and divided by the complex conjugate.
then I back substitute in Eulers formula.
now since I have my original equation has both a real and an imaginary part.
when I multiply out A times Eulers formula, I will take both real and imaginary parts.
and I get that y=sin(x) and tested this and it works.
But is what i did with Eulers formula ok.