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Second Order ODE Method Question

x2d2ydx22y=r(x) x^2 \frac{d^2y}{dx^2}-2y=r(x)

Part a) was to show y=x2y=x^2 is a solution for r(x)=0.

Part b) asks to use the method of variation of parameters to find the general solution when r(x)=3x2r(x)=3x^2.

In my notes, it says we use the method of variation of parameters when we know both solutions to the homogeneous equation (so both solutions when r(x)=0).

However, I don't know how you would find the other solution? I was going to use the method of reduction of order to find the second solution, but wouldn't this construct the general solution that I need to find?

TLDR: For the ODE above, given I know one solution for r(x)=0, should I use the method of reduction of order to find a solution for r(x)=3x2r(x)=3x^2?
Reply 1
Original post by rayquaza17
x2d2ydx22y=r(x) x^2 \frac{d^2y}{dx^2}-2y=r(x)

Part a) was to show y=x2y=x^2 is a solution for r(x)=0.

Part b) asks to use the method of variation of parameters to find the general solution when r(x)=3x2r(x)=3x^2.

In my notes, it says we use the method of variation of parameters when we know both solutions to the homogeneous equation (so both solutions when r(x)=0).

However, I don't know how you would find the other solution? I was going to use the method of reduction of order to find the second solution, but wouldn't this construct the general solution that I need to find?

TLDR: For the ODE above, given I know one solution for r(x)=0, should I use the method of reduction of order to find a solution for r(x)=3x2r(x)=3x^2?

Yes, I imagine that would work fine. (The other complementary function is also very simple.)
Reply 2
Original post by Smaug123
Yes, I imagine that would work fine. (The other complementary function is also very simple.)


How would you work out the other cf?
Reply 3
Original post by rayquaza17
How would you work out the other cf?

y2=uy1=ux2y_2 = u y_1 = u x^2, and work from there, substituting it into the DE. (It came out for me in about five lines.)
Reply 4
Original post by Smaug123
y2=uy1=ux2y_2 = u y_1 = u x^2, and work from there, substituting it into the DE. (It came out for me in about five lines.)


But is this not the method of variation of parameters?

Thanks for the help anyway, I'm going to email my lecturer. I can't rep you, PRSOM. Thumbs up instead.:gthumb:
Reply 5
Original post by rayquaza17
But is this not the method of variation of parameters?

Thanks for the help anyway, I'm going to email my lecturer. I can't rep you, PRSOM. Thumbs up instead.:gthumb:

Thanks :smile: no, variation of parameters involves knowing all the complementary functions and constructing a solution for a particular integral. I described a way of finding one complementary function if we already know only one.
Reply 6
Original post by rayquaza17
x2d2ydx22y=r(x) x^2 \frac{d^2y}{dx^2}-2y=r(x)

Part a) was to show y=x2y=x^2 is a solution for r(x)=0.

Part b) asks to use the method of variation of parameters to find the general solution when r(x)=3x2r(x)=3x^2.

In my notes, it says we use the method of variation of parameters when we know both solutions to the homogeneous equation (so both solutions when r(x)=0).

However, I don't know how you would find the other solution? I was going to use the method of reduction of order to find the second solution, but wouldn't this construct the general solution that I need to find?

TLDR: For the ODE above, given I know one solution for r(x)=0, should I use the method of reduction of order to find a solution for r(x)=3x2r(x)=3x^2?


Observe that in the homogeneus DE you are differentiating twice, multiplying by x^2, and getting out 2y. This means that differentiating y is a bit like dividing by x^2. This should hint that the solution should be some power of x. Plug in x^n and you will get a quadratic for n, with one solution being n=2. You just need the other one.

Alternatively you can do reduction of order on the homogeneus equation.
Reply 7
Original post by Smaug123
Thanks :smile: no, variation of parameters involves knowing all the complementary functions and constructing a solution for a particular integral. I described a way of finding one complementary function if we already know only one.


I meant to say the method of reduction of order! (Such similar names...)
Reply 8
Original post by rayquaza17
I meant to say the method of reduction of order! (Such similar names...)

Yep, it is :smile:
Alternatively, as james22 pointed out, the equation is equidimensional so has complementary functions a power of x.

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