Find a general solution to :y'' + 2y' + 4y = 0.

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hala718 | High School Teacher | (Level 1) Educator Emeritus

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First we will rewrite into auxiliary equation form.

==> r^2 + 2r + 4 = 0

Now we will calculatet the roots.

==> r1= [ -2+ sqrt(4--16) / 2 = -1+sqrt3*i

==> r2= -1-sqrt3*i

Since the roots are not real, then we know that:

==> a = -1 ==> B= sqrt3

Then the solution is given by :

y(x)= c1e^-x* cos(sqrt3 x) + c2*e^-x * sin(sqrt3*x)

Then the general solution is given by :

==> y(x) = e^-x [ c1*cos(sqrt3*x) + c2*sin(sqrt3* x)].

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