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1. Ok, just wanted to quickly see if somebody could give a couple of my answers a quick check and point me in the right direction for one question.

The questions are:

Find the general solution of each of the following differential equations.

a) y'' - y = exp(2x) (incase it isn't showing up properly the first term, y'' is d^2y/dx^2)

b) y'' - y = x^2

c) y'' - y = 2exp(x)

And here is what I've done.

I know that for all three to get the general solution you need to find the complentary function and a particular integral, and then the general solution is simply the sum of the two.

a) For the complementary function we let y'' - y = 0, and then we need to figure out the roots of the equation so we take m^2 - 1 = 0 and this gives us m = 1 or m = -1
Then from the roots we know that the complementary function is Aexp(x) + Bexp(-x)

Then we move onto the particular integral. So we need to take a trial function which we define as aexp(2x) and so we now need to find A, so we differentiate it to get y'' = 4aexp(2x) and then substitue this into the first equation to give 4aexp(2x) - aexp(2x) = exp(2x) and from that we find that a = 1/3 and so the particular integral is 1/3exp(2x)

As a result of this the general solution is Aexp(x) + Bexp(-x) + 1/3exp(2x)

b) As the left hand side is the same as in part a we already know that the complementary function is Aexp(x) + Bexp(-x)

So, all we need to find is the particular integral. We take the trial function to be ax^2 + b and differentiating it gives us y'' = 2a, which we then substitue back in to the first equation to give us 2a - (ax^2 + b) = x^2 which we can work through to give a = -1 and b = -2 so when we put it all together we get:

Aexp(x) + Bexp(-x) - x^2 - 2

c) This is the one that is completely stumping me. We have the same complementary function as before. Then the trial function is 2aexp(x) which we differentiate to get y'' = 2aexp(x) but this then gives me the problem that when we substitute it back in we get 2aexp(x) - 2aexp(x) = 2exp(x) and so the left side becomes 0 and it all goes wrong.

Can somebody please give part a & b a quick check for me please, and let me know where I'm going wrong with the third part.

Thanks very much.
2. (Original post by mackemforever)
Ok, just wanted to quickly see if somebody could give a couple of my answers a quick check and point me in the right direction for one question.

The questions are:

Find the general solution of each of the following differential equations.

a) y'' - y = exp(2x) (incase it isn't showing up properly the first term, y'' is d^2y/dx^2)

b) y'' - y = x^2

c) y'' - y = 2exp(x)

And here is what I've done.

I know that for all three to get the general solution you need to find the complentary function and a particular integral, and then the general solution is simply the sum of the two.

a) For the complementary function we let y'' - y = 0, and then we need to figure out the roots of the equation so we take m^2 - 1 = 0 and this gives us m = 1 or m = -1
Then from the roots we know that the complementary function is Aexp(x) + Bexp(-x)

Then we move onto the particular integral. So we need to take a trial function which we define as aexp(2x) and so we now need to find A, so we differentiate it to get y'' = 4aexp(2x) and then substitue this into the first equation to give 4aexp(2x) - aexp(2x) = exp(2x) and from that we find that a = 1/3 and so the particular integral is 1/3exp(2x)

As a result of this the general solution is Aexp(x) + Bexp(-x) + 1/3exp(2x)

b) As the left hand side is the same as in part a we already know that the complementary function is Aexp(x) + Bexp(-x)

So, all we need to find is the particular integral. We take the trial function to be ax^2 + b and differentiating it gives us y'' = 2a, which we then substitue back in to the first equation to give us 2a - (ax^2 + b) = x^2 which we can work through to give a = -1 and b = -2 so when we put it all together we get:

Aexp(x) + Bexp(-x) - x^2 - 2

c) This is the one that is completely stumping me. We have the same complementary function as before. Then the trial function is 2aexp(x) which we differentiate to get y'' = 2aexp(x) but this then gives me the problem that when we substitute it back in we get 2aexp(x) - 2aexp(x) = 2exp(x) and so the left side becomes 0 and it all goes wrong.

Can somebody please give part a & b a quick check for me please, and let me know where I'm going wrong with the third part.

Thanks very much.

You could try something like y=Ax*e^x for c). Looks like that would work as you'd remove any x*e^x terms when differentiating.

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