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General Solution please help got exam tday

I'm stuck on this question, can someone tell me how do I find the particular integral of this:

Reply 1
Avatar for singh224
singh224
OP
8
I'm getting PI = 0 as f(x) = 0 however the answer is y(PI) = -2t.

(edited 11 years ago)
Reply 2
Avatar for Fat-Love
Fat-Love
15
In general the form of the particular integral will match that of the function at the end. In this case as the function 10 at the end is just a constant you can use a substitution of y= k where k is also a constant. How ever in this case your general solution from the auxiliary equation also has a constant in it. (A + Be^5x). A is a constant too. So you are required to shift your P.I form by one power of x by multiplying by x to make it y=kx. Hence y' = k and y'' =0. Substituting into the initial equation gives k =-2. so the P.I is y= -2x.

hence the overall solution becomes y = A + Be^5x -2x
(edited 11 years ago)
Reply 3
Avatar for davros
davros
16
Original post by singh224
I'm stuck on this question, can someone tell me how do I find the particular integral of this:



Note that since your equation has no term in y in it you can integrate w.r.t.t straight away to reduce it to a first-order DE!
Reply 4
Avatar for singh224
singh224
OP
8
Original post by Fat-Love
In general the form of the particular integral will match that of the function at the end. In this case as the function 10 at the end is just a constant you can use a substitution of y= k where k is also a constant. How ever in this case your general solution from the auxiliary equation also has a constant in it. (A + Be^5x). A is a constant too. So you are required to shift your P.I form by one power of x by multiplying by x to make it y=kx. Hence y' = k and y'' =0. Substituting into the initial equation gives k =-2. so the P.I is y= -2x.

hence the overall solution becomes y = A + Be^5x -2x


Thanks, I didn't notice that lol.

Thank you soo much :biggrin:

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