# FP2 2nd ODE's

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#1
I understand first ode's where you use separation of variables, use substitutions and also use the integrating factor y=e^integral(P)dx ... Which you derive ( correct me if I'm wrong) because we wanted a way of expressing it as a derivative of two variables.

When it comes to 2nd ODE's...
I do not understand where the general solutions from when b^2 is greater, smaller or equal to 4ac from the auxiliary equation... Is one of them y=e^mx(a+bx) because the auxiliary eqn could be expressed as (a+bx)^2 because of the equal roots or something...

So yah... I don't get the other 2 general solutions as well.
Neither do I understand why y=PI+CF, and why the in particular PI's ( they change order depending on order of f(x). ) ...are relevant to the particular f(x).

I would be greatful for any help.
Thank you
0
5 years ago
#2
(Original post by demigawdz)
I understand first ode's where you use separation of variables, use substitutions and also use the integrating factor y=e^integral(P)dx ... Which you derive ( correct me if I'm wrong) because we wanted a way of expressing it as a derivative of two variables.

When it comes to 2nd ODE's...
I do not understand where the general solutions from when b^2 is greater, smaller or equal to 4ac from the auxiliary equation... Is one of them y=e^mx(a+bx) because the auxiliary eqn could be expressed as (a+bx)^2 because of the equal roots or something...

So yah... I don't get the other 2 general solutions as well.
Neither do I understand why y=PI+CF, and why the in particular PI's ( they change order depending on order of f(x). ) ...are relevant to the particular f(x).

I would be greatful for any help.
Thank you
For second order ODE's at A-level, you 'guess' a solution of the form y=e^mx and sub this into your ODE (I'm sure there are more rigorous university methods). You should then get a quadratic which you can solve. Take the example y''+5y+6=0. Guessing y=e^mx and subbing in, we get e^mx(m^2+5m+6)=0. As e^mx =/= 0, m=3 or m=2. This means y=e^3x and y=e^2x are both solutions. In fact, y=Ae^3x and y=Be^2x where A and B are constants are both solutions. It follows that y=Ae^3x + Be^2x is the general solution (don't believe me, sub it into the ODE).

When the auxiliary equation has equal roots, eg y''-2y+1=0, y=Ae^x is a solution. However, it turns out (by guessing the next 'most' complicated function) that y=Bxe^x is also a solution, hence the general solution is y=e^x(A+Bx).

In the case that the auxiliary equation has non-real roots, using the identity e^ix=cos(x)+isin(x) will result in you finding the general solution

On the topic of y=PI+CF, note that the CF will be 0 when subbed into the ODE. The PI's are just guesses (again!). Although, there is a method for determining the PI you learn at uni, however I can't remember the name of it (I'm sure someone much smarter than me will say it)
0
5 years ago
#3
All you really have to remember is that when b^2-4ac for the auxiliary equation > 0, the general solution of the second order ODE is of the form y=Ae^(mx) + Be^(mx)
If b^2-4ac = 0 then the general solution is y=(e^(mx))(Ax+b)
And if b^2-4ac <0 then the general solution is of the form (e^(qx))(cos(px) + sin(px)) where the roots of the auxiliary equation are of the form
q +/- pi Where i is the square root of -1. This is all assuming that the second order ODE is = 0
0
5 years ago
#4
(Original post by demigawdz)
I understand first ode's where you use separation of variables, use substitutions and also use the integrating factor y=e^integral(P)dx ... Which you derive ( correct me if I'm wrong) because we wanted a way of expressing it as a derivative of two variables
The reason why you find an integrating factor is because when you multiply both sides of the differential equation by the integrating factor it transforms the equation into an 'exact' differential equation. Where the left hand side as the exact derivative of a product of yf(x). This allows you to integrate both sides but since you now know that the left hand side of the differential equation is if the form d/dx[yf(x)] you will find that when you integrate it you end up with just yf(x) - as the integral of a derivative of a function leaves just the function. This means all you have to do then is integrate the right hand side of the differential equation and then you have the general solution.
1
#5
(Original post by B_9710)
The reason why you find an integrating factor is because when you multiply both sides of the differential equation by the integrating factor it transforms the equation into an 'exact' differential equation. Where the left hand side as the exact derivative of a product of yf(x). This allows you to integrate both sides but since you now know that the left hand side of the differential equation is if the form d/dx[yf(x)] you will find that when you integrate it you end up with just yf(x) - as the integral of a derivative of a function leaves just the function. This means all you have to do then is integrate the right hand side of the differential equation and then you have the general solution.
Thanks
0
#6
(Original post by Gome44)
For second order ODE's at A-level, you 'guess' a solution of the form y=e^mx and sub this into your ODE (I'm sure there are more rigorous university methods). You should then get a quadratic which you can solve. Take the example y''+5y+6=0. Guessing y=e^mx and subbing in, we get e^mx(m^2+5m+6)=0. As e^mx =/= 0, m=3 or m=2. This means y=e^3x and y=e^2x are both solutions. In fact, y=Ae^3x and y=Be^2x where A and B are constants are both solutions. It follows that y=Ae^3x + Be^2x is the general solution (don't believe me, sub it into the ODE).

When the auxiliary equation has equal roots, eg y''-2y+1=0, y=Ae^x is a solution. However, it turns out (by guessing the next 'most' complicated function) that y=Bxe^x is also a solution, hence the general solution is y=e^x(A+Bx).

In the case that the auxiliary equation has non-real roots, using the identity e^ix=cos(x)+isin(x) will result in you finding the general solution

On the topic of y=PI+CF, note that the CF will be 0 when subbed into the ODE. The PI's are just guesses (again!). Although, there is a method for determining the PI you learn at uni, however I can't remember the name of it (I'm sure someone much smarter than me will say it)
I do not understand this 0
5 years ago
#7
If the right hand side of the second order ODE is not 0 then you have to find a particular integral. To do this you have to try a function that is similar to the function on the right hand side. For example if the right hand side was
x^2 + 4x
You would try a particular integral say y=px^2 + qx + r. Then you would differentiate it to find dy/dx and then differentiate it again to find d^2y/dx^2
Then you would sub these values of y, dy/dx and d^2y/dx^2 into the left hand side of the equation. Then you would try to find values of p, q and r such that the left hand side = right hand side. This often involves solving simple simultaneous equations. Then the general solution is y=CF + PI where PI is the particular integral.
If the right hand side of the equation is e^2x then try a particular integral like y=pe^2x
If the right hand side is cos(2x) then try a PI y=pcos(2x) + qsin(2x)

This can be quite difficult to explain and more difficult to understand when just reading. Go on to videos on YouTube such as examsolutions and look at this it should be easier to follow and understand
0
5 years ago
#8
(Original post by Gome44)
On the topic of y=PI+CF, note that the CF will be 0 when subbed into the ODE. The PI's are just guesses (again!). Although, there is a method for determining the PI you learn at uni, however I can't remember the name of it (I'm sure someone much smarter than me will say it)
Note that if you have a linear equation ay'' + by' + cy = f(x). then if you have 2 solutions to this, let's say y = u(x) and y = v(x), then consider the difference between them w(x) = u(x) - v(x).

Because the DE is linear, aw'' + bw' + cw = (au'' + bu' + cu) - (av'' + bv' + cv) = f(x) - f(x) = 0. In other words, the difference between any 2 general solutions of the original equation will just be a solution of the homogeneous version of the DE. i.e. it will be a CF.

This means that IF you can solve the equation ay'' + by' + cy = 0 to get the CF, and IF you can guess just one PI y = u(x), then all the solutions are of the form CF + PI.
1
5 years ago
#9
(Original post by demigawdz)
I do not understand this Which part? All of it?
0
5 years ago
#10
(Original post by davros)
Note that if you have a linear equation ay'' + by' + cy = f(x). then if you have 2 solutions to this, let's say y = u(x) and y = v(x), then consider the difference between them w(x) = u(x) - v(x).

Because the DE is linear, aw'' + bw' + cw = (au'' + bu' + cu) - (av'' + bv' + cv) = f(x) - f(x) = 0. In other words, the difference between any 2 general solutions of the original equation will just be a solution of the homogeneous version of the DE. i.e. it will be a CF.

This means that IF you can solve the equation ay'' + by' + cy = 0 to get the CF, and IF you can guess just one PI y = u(x), then all the solutions are of the form CF + PI.
Ah very neat explanation. I would rep but I can't .

I think the method I was couldn't remember is called variation of parameters (although I have no idea how to actually do it, or how it works)
0
#11
(Original post by davros)
Note that if you have a linear equation ay'' + by' + cy = f(x). then if you have 2 solutions to this, let's say y = u(x) and y = v(x), then consider the difference between them w(x) = u(x) - v(x).

Because the DE is linear, aw'' + bw' + cw = (au'' + bu' + cu) - (av'' + bv' + cv) = f(x) - f(x)
Wow thanks davros!
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