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I have some questions about writing a general solution, $y$, for $y''-y=0$ when $y_1 = e^x$ is a known solution.

I do not understand the logic of the method of reduction of order. How do we apply this to find a second, linearly independent solution?

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2 Answers 2

up vote 2 down vote accepted

I recommend you read this article which explains the general procedure.

However, in your particular case, since you have one solution, $y_1=e^x$ the method says to look for a second one of the form $y_2=v(x)y_1(x)$. Then \begin{align}y_2'&=v'e^x+ve^x\\ y_2''&=e^x(v''+2v'+v)\end{align} and since it is assumed that $y_2$ is a solution of $y''-y=0$, we conclude $$y_2''-y_2=e^x(v''+2v')=0\implies v''+2v'=0.$$

The next step is where the name reduction of order comes from. Let $u=v'$ so that $u'=v''$. Then the $v$ equation (a second order equation) becomes $u'+2u=0$ (a first order equation) which has general solution $u(x)=Ce^{-2x}$, i.e., $v'(x)=Ce^{-2x}$. Integrating we get $v(x)=-{1\over 2}Ce^{-2x}+D$.

Since we are only interested in a second linearly independent solution $y_2$, we can take $C=-2$ and $D=0$ for simplicity. Then $$y_2(x)=v(x)y_1(x)=e^{-2x}\cdot e^x=e^{-x},$$ and so $$y(x)=c_1y_1(x)+c_2y_2(x)=c_1e^x+c_2e^{-x}.$$

To verify that $y_1$ and $y_2$ are in fact linearly independent on an interval $I$, simply evaluate the Wronskian of $y_1$ and $y_2$: $$W(y_1,y_2)=\begin{vmatrix} y_1 & y_2\\ y_1' & y_2'\end{vmatrix}=\begin{vmatrix} e^x & e^{-x}\\ e^x & -e^{-x}\end{vmatrix}=e^x(-e^{-x})-e^{-x}(e^x)=-2\not=0,\quad \forall x\in I.$$

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that really worked. thank you. i got it. –  Yigit Can Jan 2 '13 at 20:35
    
oh sir @JohnD i could not understand how could you find c1y1(x) as e^x ?? –  Yigit Can Jan 2 '13 at 20:39
    
Well you know $y_1=e^x$ and we found $y_2=e^{-x}$. Since $y_1$ and $y_2$ are two linearly independent solutions of the equation, there is a theorem that says that any solution can be written as $c_1 y_1(x)+c_2 y_2(x)$ where $c_1,c_2$ are constants. (Think of $c_1$ and $c_2$ as parameters that aren't resolved until we have some auxiliary information such as initial conditions, boundary conditions, etc.) –  JohnD Jan 2 '13 at 20:43

It is a method used to find a second solution by knowing one solution. To find a second solution, assume that $y_2(x)=u(x)y_1(x)=u(x)e^{x}.$. Now substitute this in the differential equation and try to find $u(x)$ using the fact that $y''_1-y_1=0$.

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