adam132
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How do I solve 4y'' + 12y' + 9 = 0?
It's not homogenous (I think?) but I'm stuck on finding the particular solution to f(x)=-9


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rayquaza17
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It's not homogeneous because not every term involves a y.
You can write it as 4y''+12y'=-9.

Can you solve 4y''+12y'=0?

Have you done something before where you would sub something like y=ax into the differential equation and try and work out what a is? [Method of undetermined coefficients is the proper name]
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adam132
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(Original post by rayquaza17)
It's not homogeneous because not every term involves a y.
You can write it as 4y''+2y'=-9.

Can you solve 4y''+2y'=0?

Have you done something before where you would sub something like y=ax into the differential equation and try and work out what a is? [Method of undetermined coefficients is the proper name]
Cheers for the reply

We did method of undetermined coefficients, thought not sure I recognise something like y=ax.

We did general solutions for if the right hand side is either a polynomial in x, an exponential, ksin(bx) + kcos(bx) or a product of these. Not sure what to do with a constant



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davros
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(Original post by adam132)
Cheers for the reply

We did method of undetermined coefficients, thought not sure I recognise something like y=ax.

We did general solutions for if the right hand side is either a polynomial in x, an exponential, ksin(bx) + kcos(bx) or a product of these. Not sure what to do with a constant



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A constant is just the simplest example of a polynomial you can have!

So the Particular Integral will be y = (some other constant)
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rayquaza17
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(Original post by adam132)
Cheers for the reply

We did method of undetermined coefficients, thought not sure I recognise something like y=ax.

We did general solutions for if the right hand side is either a polynomial in x, an exponential, ksin(bx) + kcos(bx) or a product of these. Not sure what to do with a constant



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The RHS is a polynomial in x: -9=0x-9

You might have y=ax+b in your notes, try doing that.
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adam132
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Oh right, if the solutions just a constant though (A) is there anyway to find it?

I got: y = c1 + c2e^(-3x) + A


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rayquaza17
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(Original post by adam132)
Oh right, if the solutions just a constant though (A) is there anyway to find it?

I got: y = c1 + c2e^(-3x) + A


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y=ax, y'=a, y''=0
Substitute this into the differential equation (4y''+12y'=-9) to get an equation that tells you what a should be.

So your final answer should be y=C_1+C_2e^{-3x}+ax, but replace a with what you work out above.
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adam132
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(Original post by rayquaza17)
y=ax, y'=a, y''=0
Substitute this into the differential equation (4y''+12y'=-9) to get an equation that tells you what a should be.

So your final answer should be y=C_1+C_2e^{-3x}+ax, but replace a with what you work out above.
Ah right. See I thought for the solution to a polynomial the orders are the same? So if it were 5x^2 the solution would be (A2)x^2 + (A1)x + A0

Or is it just for a constant that you include an x?


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rayquaza17
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(Original post by adam132)
Ah right. See I thought for the solution to a polynomial the orders are the same? So if it were 5x^2 the solution would be (A2)x^2 + (A1)x + A0

Or is it just for a constant that you include an x?


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Right tbh I'm not 100% sure, but I think it's because you have no y term in your differential equation.

If you let z=y', then you can write your differential equation as: 4z'+12z=-9. To determine the particular integral of this, we would set z=a. But then this is the same as y'=a, which you can integrate to get y=ax and I think this is why we use the y=ax in this case.

I think this is why, but like I said I'm not sure. Maybe davros knows?
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davros
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(Original post by adam132)
Ah right. See I thought for the solution to a polynomial the orders are the same? So if it were 5x^2 the solution would be (A2)x^2 + (A1)x + A0

Or is it just for a constant that you include an x?


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(Original post by rayquaza17)
Right tbh I'm not 100% sure, but I think it's because you have no y term in your differential equation.

If you let z=y', then you can write your differential equation as: 4z'+12z=-9. To determine the particular integral of this, we would set z=a. But then this is the same as y'=a, which you can integrate to get y=ax and I think this is why we use the y=ax in this case.

I think this is why, but like I said I'm not sure. Maybe davros knows?
Normally (when you have a y term) when there is a polynomial on the RHS, you set y to be a poly whose degree matches the degree on the RHS. That's because differentiation reduces the degree of a poly by one each time, so the only 'bit' of y that can match the highest power on the RHS is the highest power of y itself. In this case the y term is missing so setting y = constant gives you y' = 0 and y'' = 0 and you can't solve the equation! So the next best thing is to try y = ax so y' = a and y'' = 0 and you're just trying to match the constant from the y' term with the constant on the RHS.
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adam132
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(Original post by davros)
Normally (when you have a y term) when there is a polynomial on the RHS, you set y to be a poly whose degree matches the degree on the RHS. That's because differentiation reduces the degree of a poly by one each time, so the only 'bit' of y that can match the highest power on the RHS is the highest power of y itself. In this case the y term is missing so setting y = constant gives you y' = 0 and y'' = 0 and you can't solve the equation! So the next best thing is to try y = ax so y' = a and y'' = 0 and you're just trying to match the constant from the y' term with the constant on the RHS.
Cheers for this


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