7D (Further Complex Methods)

Substituting

y (z) = \int_{γ} e^{z t} f (t) d t, y^{(3)} (z) = \int_{γ} e^{z t} t^{3} f (t) d t,

and using

z e^{z t} = \partial_{t} e^{z t}

gives

z y^{(3)} + 2 y = [e^{z t} t^{3} f (t)]_{\partial γ} + \int_{γ} e^{z t} {2 f (t) - (t^{3} f (t))^{'}} d t .

It is therefore enough to require

(t^{3} f)^{'} = 2 f

and choose

γ

so that the integral converges and

e^{z t} t^{3} f (t)

vanishes at its endpoints. Solving the amplitude equation,

\frac{f ^{'}}{f} = \frac{2}{t ^{3}} - \frac{3}{t}, f (t) = C t^{- 3} e^{- 1/ t^{2}} .

b

For

x > 0

, take the real contour

γ = (- \infty, 0)

. At

t \to 0^{-}

the factor

e^{- 1/ t^{2}}

kills every power, while at

t \to - \infty

the factor

e^{x t}

makes the integrand and endpoint term vanish. The laplace integral solution of a singular third-order equation is therefore

y (x) = C \int_{- \infty}^{0} e^{x t - 1/ t^{2}} t^{- 3} d t .

With

u = - 1/ t

, and absorbing a minus sign into the arbitrary constant, this becomes

y (x) = C_{1} \int_{0}^{\infty} u exp (- u^{2} - \frac{x}{u}) d u .

There are no discussions about this article yet.

There are currently no matching articles.