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Suppose $$y^{'}+p(x)y=q(x),\,\,\,y(x_0)=y_0$$ where $p$ and $q$ are continuous functions in some interval $I$ containing $x_0$. Show that the particular solution is $$y(x)=e^{\int_{x_o}^{x}{p(t)}dt}[\int_{x_0}^{x}{e^{\int_{x_0}^{t}{p(\xi)}d\xi}q(t)dt+y_0}]$$ I have no idea where the $\xi$ comes from. I can only get the general solution $$y(x)=\frac{1}{I(x)}{\int{}I(x)q(x)dx+C} ,$$ where $I(x)$ is an integrating factor

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isn't this first order linear differential equation? –  Santosh Linkha Jan 9 '13 at 15:11
    
@experimentX: yes, but somehow I cannot get the particular solution. –  Idonknow Jan 9 '13 at 15:13
    
@NgHongWai ξ doesn't matter, it's just a dummy variable ... what matters is $ e^{\int_{x_0}^t p( ξ ) d ξ }$ which is your integrating factor –  Santosh Linkha Jan 9 '13 at 15:19

2 Answers 2

up vote 1 down vote accepted

Searching for the method called Variation of Parameters, we will find out, for the linear 1-order differential equation $y'+p(x)y=q(x)$ where the functions $p(x), q(x)$ have the conditions as you gave them above; there is a solution like $y_1(x)=\text{e}^{\int-p(x)dx}$.(You know all of these)

The method goes further and tells us that the one-parameter family of solutions of our equation is as the form you noted above as well. In fact we set $y(x)=v(x)y_1(x)$ into the equation to find another part of solution which is free of any constant. This is the particular solution $$y_p(x)=\text{e}^{\int-p(x)dx}\int\text{e}^{\int-p(x)dx}f(x)dx$$. In 2-order linear equation we can easily understand why this approach was made. Now I make an example to see why that formula arisen. Let we have $y'=f(x,y),\; y(x_0)=y_0$. You surely accept that if $f(x,y)$ be continuous in a region containing the point $(x_0,y_0)$ then by integrating from both sides of our latter OE, we have $$y(x)=c+\int_{x_0}^{x}f(t,y(t))dt$$ and certainly $$y(x_0)=c+\int_{x_0}^{x_0}f(t,y(t))dt=c$$ and so $$y(x)=y_0 +\int_{x_0}^{x}f(t,y(t))dt$$. I hope you got the point. Moreover @experimentX gave you additional points.

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+1 Well done, as usual :-) –  amWhy Feb 20 '13 at 0:30

if you multiply $e^{\int p(x)\text{d}x}$ to both sides of the equation,get: $$y'e^{\int p(x)\text{d}x}+p(x)ye^{\int p(x)\text{d}x}=q(x)e^{\int p(x)\text{d}x}$$ the above equation can be written as $$\text{d}\left(ye^{\int p(x)\text{d}x}\right)=\text{d}\left(\int q(x)e^{\int p(x)\text{d}x}\right)$$ then you can get the answer.

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