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Second order equations

Posted February 10th, 2008 by vic777
in

What is the solution for:
$ \frac{d^2u}{dx ^2 }-4u=-e^{-x} $ with u(0)=1 and $ u(\infty) $=0

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Comments

Bounded solution

On February 10th, 2008 Structure says:

Homogeneous equation has exponential solution of form

$$u(x)=e^{rx}$$

So you have

$$u'(x)=re^{rx}$$
$$u"(x)=r^2e^{rx}$$

Let

$$P(D)u(x)=u"(x)-4u(x)$$
$$P(D)(e^{rx})=(r^2-4)e^{rx}$$
$$P(D)(e^{rx})=(r^2-4)e^{rx}=0\: if\: and \:only \:if\:r^2-4=(r-2)(r+2)=0$$

Solutions of the homogeneous equation are in a two dimensional space spanned by $ u_1(x)=e^{2x}\:u_2(x)=e^{-2x} $ so $ u_{0}(x)=c_1e^{2x}+c_2e^{-2x} $
Non homogeneous equation has a solution of form

$$u_p(x)=Ae^{-x}$$

From

$$P(D)(e^{rx})=(r^2-4)e^{rx}$$

we have

$$P(D)(Ae^{-x})=A(1-4)e^{-x}=-3Ae^{-x}=-e^{-x}$$

so we have

$$A=\frac{1}{3}$$

Then we have

$$u(x)=u_{0}(x)+u_p(x)=c_1e^{2x}+c_2e^{-2x}+\frac{1}{3}e^{-x}$$

As we want u(x) bounded at infinite $ c_1 $ has to be zero.
Then

$$u(x)=c_2e^{-2x}+\frac{1}{3}e^{-x}$$

.
As $ u(0)=1 $ we have

$$c_2=\frac{2}{3}$$

Then we have the solution

$$u(x)=\frac{2}{3}e^{-2x}+\frac{1}{3}e^{-x}$$

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