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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 [tex]x^2 y'' - 12xy' + 42y = 0[/tex]...(1), find the general solution of [tex]x^2 y'' - 12xy' + 42y = 280x^2 + 150x - 168[/tex]...(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:

[tex]

y' = 7x^6 v + x^7 \frac{{dv}}{{dx}}

[/tex]

[tex]

y'' = 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 }}

[/tex]

Substituting into equation (2) and simplifying I get:

[tex]

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

[/tex]

[tex]

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

[/tex]

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

[tex]

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

[/tex]

[tex]

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

[/tex]

[tex]

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

[/tex]

[tex]

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

[/tex]

Hmm nevermind about this part...I realised 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:

[tex]

y = 14x^2 + 5x - 4 + c_2 x^6 + c_3 x^7

[/tex]

Can someone tell me how to find a second linearly independent solution of (1) and a particular solution of (2) from the genral 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.

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

[tex]

y' = 7x^6 v + x^7 \frac{{dv}}{{dx}}

[/tex]

[tex]

y'' = 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 }}

[/tex]

Substituting into equation (2) and simplifying I get:

[tex]

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

[/tex]

[tex]

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

[/tex]

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

[tex]

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

[/tex]

[tex]

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

[/tex]

[tex]

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

[/tex]

[tex]

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

[/tex]

Hmm nevermind about this part...I realised 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:

[tex]

y = 14x^2 + 5x - 4 + c_2 x^6 + c_3 x^7

[/tex]

Can someone tell me how to find a second linearly independent solution of (1) and a particular solution of (2) from the genral 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.

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