Variation of parameter

**mitch_nufc** · April 6th 2009, 02:28 AM

Find the general solution of this differential equation, given that a solution of the corresponding homogeneous equation is y = x,

(x^2)y'' + x(x-2)y' - (x-2)y = x^3

I can sorta deal with these problems but i havnt came across them with functions of x attached to the y's, I can also deal with them in the case of y'' + y' + y = f(x) :s... do i still find an auxillary equation or what? Can someone show me how these type of questions are done please? Thanks

**chisigma** · April 6th 2009, 06:22 AM

Let's start with the homogeneous equation...

$x^{2}\cdot y^{''} + x\cdot (x-2)\cdot y^{'} - (x-2)\cdot y = 0$ (1)

... one solution of which we know to be $y=x$ . If u(x) and v(x) are solutions of (1) then...

$u^{''} + (1-\frac{2}{x})\cdot u^{'} - \frac{x-2}{x^{2}}\cdot u=0$

$v^{''} + (1-\frac{2}{x})\cdot v^{'} - \frac{x-2}{x^{2}}\cdot v=0$ (2)

Multipling the first equation by v, the second by u an taking the difference we obtain...

$(u^{''}\cdot v - u\cdot v^{''}) + (1-\frac{2}{x})\cdot (u^{'}\cdot v - u\cdot v^{'}) =0$ (3)

... that with the substitution $\gamma= u^{'}\cdot v - u\cdot v^{'}$ becomes...

$\gamma^{'} + (1-\frac{2}{x})\cdot \gamma =0$ (4)

The (4) is easily solved and gives...

$\gamma= c\cdot x^{2}\cdot e^{-x}$ (5)

where c is a constant that can be set =1 without losing anything. Deviding both terms of (5) by $v^{2}$ we obtain the new equation...

$\frac {u^{'}\cdot v - u\cdot v^{'}}{v^{2}}= \frac{d}{dx}(\frac{u}{v})= \frac{x^{2}}{v^{2}}\cdot e^{-x}$ (6)

Since $y=x$ is solution we can set $v(x)=x$ in (6) obtaining...

$\frac {d}{dx}(\frac{u}{x})= e^{-x}$ (7)

... the solution of which is...

$u(x)= x\cdot e^{-x}$ (8)

The general solution of (1) is then...

$y(x)= c_{1}\cdot x + c_{2}\cdot x\cdot e^{-x}$ (9)

Let's consider now the 'complete' equation...

$x^{2}\cdot y^{''} + x\cdot (x-2)\cdot y^{'} - (x-2)\cdot y = x^{3}$ (10)

With a little of patience you find that $y=x^{2}$ is a particular solution so that the general solution of (10) is...

$y(x)= c_{1}\cdot x + c_{2}\cdot x\cdot e^{-x} + x^{2}$ (11)

Kind regards

$\chi$ $\sigma$

**mitch_nufc** · April 6th 2009, 06:54 AM

Thanks for the work, but I dont really get what's going on though... I havn't been taught this at university yet however.

What I'm familiar with is the following:

ay'' + by' + cy = 0, where i obtain a general solution with either complex, repeated or real distinct

ay'' + by' + cy = f(x)

And I assume that the next installment of this would be the variation of parameters... i.e. what i've posted. Thanks again for the working, but i don't understand how ya got there.

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