How Do I Find the Second Linearly Independent Solution of This ODE?

[Benny]

Hi can someone please help me out with the following DE question. I'm reading ahead so as to enable myself to keep up to date so I don't how the part in italics(below) can be used so could someone please explain it to me. I'm also having trouble with the ODE itself.

Q. Given that y = x^7 is a solution of x^2 y'' - 12xy' + 42y = 0...(1), find the general solution of x^2 y'' - 12xy' + 42y = 280x^2 + 150x - 168...(2).

Hence find a second linearly independent solution of (1) and a particular solution of (2).

Here is what I've tried. I think this is one of those DEs with a 'partially known complimentary solution.'

Let y = u(x)v(x) = (x^7)v. Then:

<br /> y&#039; = 7x^6 v + x^7 \frac{{dv}}{{dx}}<br />

<br /> y&#039;&#039; = 42x^5 v + 7x^6 \frac{{dv}}{{dx}} + 7x^6 \frac{{dv}}{{dx}} + x^7 \frac{{d^2 v}}{{dx^2 }} = 42x^5 v + 14x^6 \frac{{dv}}{{dx}} + x^7 \frac{{d^2 v}}{{dx^2 }}<br />

Substituting into equation (2) and simplifying I get:

<br /> x^9 \frac{{d^2 v}}{{dx^2 }} + 2x^8 \frac{{dv}}{{dx}} = 280x^2 + 150x - 168<br />

<br /> \frac{{d^2 v}}{{dx^2 }} + \frac{2}{x}\frac{{dv}}{{dx}} = 280x^{ - 7} + 150x^{ - 8} - 168x^{ - 9} <br />

This is a first order linear ODE in dv/dx.

<br /> IF = \mu \left( x \right) = \exp \left( {\int {\frac{2}{x}dx} } \right) = x^2 <br />

<br /> \frac{d}{{dx}}\left( {\mu \left( x \right)\frac{{dv}}{{dx}}} \right) = \mu \left( x \right)\left( {280x^{ - 7} + 150x^{ - 8} - 168x^{ - 9} } \right)<br />

<br /> \Rightarrow x^2 \frac{{dv}}{{dx}} = \int {\left( {280x^{ - 5} + 150x^{ - 6} - 168x^{ - 7} } \right)} dx<br />

<br /> \Rightarrow \frac{{dv}}{{dx}} = - 70x^{ - 6} - 30x^{ - 7} + 28x^{ - 6} + c_1 x^{ - 2} <br />

Hmm nevermind about this part...I realized that when I did this on paper I wrote (d/dx)(IFv) rather than (d/dx)(IF(dv/dx)). Anyway the general solution I get, and according to answer it is correct, is:

<br /> y = 14x^2 + 5x - 4 + c_2 x^6 + c_3 x^7 <br />

Can someone tell me how to find a second linearly independent solution of (1) and a particular solution of (2) from the general solution that I have found? Any help would be good thanks.

Edit: A while ago when I did questions where you substitute y = Ae^(rx) as a solution, ie. the ones where you have an auxillary/characteristic equation and you solve for the roots to find the solution to the DE, the particular solution was the 'non-complimentary' of the general solution.

Comparing those sorts of questions with the general solution I found - The arbitrary constants suggest to me that (c_2)(x^6) + (c_3)(x^7) is a solution to the homogeneous equation while 14x^2 + 5x - 4 is a particular solution. Is that how the I am supposed to deduce a particular solution to (2)? Also, how would I deduce a second linearly independent solution of (1)? When I did the easier questions involving characteristic equations I simply multiplied the particular solution, I think it was, by x whenever there was a repeated root, not sure if that is related to this though.

 

[HallsofIvy]

Can someone tell me how to find a second linearly independent solution of (1) and a particular solution of (2) from the general solution that I have found? Any help would be good thanks.

You write, correctly, that y = 14x^2 + 5x - 4 + c_2 x^6 + c_3 x^7.

I don't know what you mean by a "second linearly independent solution". You have two constants, c_2 and c_3. The TWO linearly independent solutions are their coefficients: x^6 and x^7 (the solution set only forms a vector space, and so the term "linearly independent" only applies, for homogenous equations). To find a specific solution, take whatever values you want for c_2 and c_3- c_2= c_3= 0 would be simple.

 

[lurflurf]

y = 14x^2 + 5x - 4 + c_2 x^6 + c_3 x^7
This is simple
for a particular solution chose any c2,c3
the typical choice would be c2=c3=0 giving
14x^2 + 5x - 4
for a linearly independent solution take the difference of two particular solutions and make sure to use different c2 for each
the typical choice would be c2first-1=c2second c3first=c3second giving
x^6
this is also clear by inspection
for a particular solution take the stuff without constants
14x^2 + 5x - 4
for an independent homogeneous solution take the suff with a constant that you did not get the first time
x^6
Now do this simple one for practice
y''+y=exp(x)
y=exp(x)/2+c1 cos(x)+c2 sin(x)
find two linearly independent solutions to the homogeneous problem
and a particular solution

 

[saltydog]

For the record, this is a particular case of the Euler-Cauchy Equation:

x^2y^{&#039;&#039;}+axy^{&#039;}+by=0

In general, powers y=x^m decrease by 1 when we differentiate, y^{&#039;}=mx^{m-1},y^{&#039;&#039;}=m(m-1)x^{m-2}. Hence they should solve linear differential equations in x, xy', and x^2y&#039;&#039;.

So, letting:

y=x^m

and substituting into the ODE:

m(m-1)x^m-12mx^m+42x^m=0

Avoiding the case x=0 and dividing through by x^m[/tex] yields:<br /> <br /> m^2-13m+42=0<br /> <br /> For which the roots are 7 and 6. That is, the general solution for the homogeneous equation is:<br /> <br /> y(x)=c_1x^7+c_2x^6

 

[Benny]

Thanks for the help guys. I'll need to go over some definitions. At the moment it's probably best if I just stick with solving some DEs until the theory is covered.