# find y as a function of x if y'''+36y'=0 y(0)=-6 , y'(0)=-18, y''(0)=-36?

### 1 Answer

- BrianLv 77 years agoFavorite Answer
The characteristic equation is m^3 + 36*m = 0 ----> m*(m^2 + 36) = 0,

so m = 0, m = 6i and m = -6i are the roots.

Thus the general solution is y(x) = c*e^(6i*x) + k*e^(-6i*x) + C.

Now y(0) = c + k + C = -6.

Next, y'(x) = 6i*c*e^(6i*x) - 6i*k*e^(-6i*x), so

y'(0) = 6i*c - 6i*k= -18 ----> c - k = -3/i = 3i.

Finally, y''(x) = -36*c*e^(6i*x) - 36*k*e^(-6i*x), so

y''(0) = -36*(c + k) = -36 ----> c + k = 1.

Now since c - k = 3i and c + k = 1, we have 2c = 1 + 3i ----> c = (1/2) + (3/2)*i.

Thus k = (1/2) - (3/2)*i, and C = -6 - c - k = -7.

So y(x) = ((1/2) + (3/2)*i)*e^(6i*x) + ((1/2) - (3/2)*i)*e^(-6i*x) - 7 ---->

y(x) = (1/2)*(e^(6i*x) + e^(-6i*x)) + (3/2)*(e^(6i*x) - e^(-6i*x))*i - 7.

Now since e^(iz) = cos(z) + i*sin(z) the solution simplifies to

y(x) = (1/2)*(2*cos(6x)) - (3/2)*(2*sin(6x)) - 7 = cos(6x) - 3*sin(6x) - 7.

Edit: It would have probably been easier to go with

y(x) = c*e^(6i*x) + k*e^(-6i*x) + C =

c*(cos(6x) + i*sin(6x)) + k*(cos(-6x) + i*sin(-6x)) + C =

(c + k)*cos(6x) + (c - k)*sin(6x)*i + C = a*cos(6x) + b*sin(6x)*i + C.

Then y(0) = a + C = -6.

Next, y'(x) = -6a*sin(6x) + 6b*cos(6x)*i, so

y'(0) = 6b*i = -18 ----> b = -3/i = 3i.

Finally, y''(x) = -36a*cos(6x) - 36b*sin(6x)*i, so

y''(0) = -36a = -36 -----> a = 1, and so C = -7.

Thus y(x) = cos(6x) + 3i*sin(6x)*i - 7 = cos(6x) - 3*sin(6x) - 7.

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