Undetermined Coefficients to solve

kuahji
Messages
390
Reaction score
2
Undetermined Coefficients to solve.

y'''-6y''=3-cosx

So I set everything up, get my system & I get an answer of

y=c1+c2x+c3e^(6x)-6/37cosx+1/37sinx

the book has y=c1+c2x+c3e^(6x)-6/37cosx+1/37sinx-1/4x^2.

So I'm not sure where the book gets the last term, any ideas?
 
Physics news on Phys.org
The book's answer is correct, but I have no idea where you are going wrong since you haven't shown your work...
 
Ok, I have yp= A + B cosx + c sinx
y'p=-Bsinx+C cosx
y''p=-B cosx - C sinx
y'''P=Bsinx - C cosx

After I have plugged these back into the LHS, I get a system of
B+6C=0
6B-C=-1

These give me my coefficients -6/37cosx & 1/37sinx.
 
Oh & yh=r^2(r-6), which is where I get c1+c2x+c3e^6c.
 
kuahji said:
Ok, I have yp= A + B cosx + c sinx

Well that's your problem right there, the constant term 'A' is not linearly independent to your homogeneous solution, which also contains a constant term... to account for the constant term (3) on the RHS of your DE, you need to add a term of the form Ax^r where r is the smallest non-negative integer such that no terms of the same order are present in your homogeneous solution...

kuahji said:
where I get c1+c2x+c3e^6x
 
Thanks. It works out nicely now.
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
Back
Top