Undetermined Coefficients, just a piece i with

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Undetermined Coefficients, just a piece i need help with please

In one of the last steps in trying to find the Coefficients of Yp (particular solution), I am supposed to set coefficients of similar terms equal to each other. Here's where i need help:
i have
A(e^x) + (-2B-2C+2F+E)sinx + (2B+F-2C-2E)cosx + (B+2C) xsinx + (C-2B)xcosx = (e^x)+xcosx

I set: A=1, -2B-2C+2F+E = 0, 2B+F-2C-2E=0, B+2C=0, and C-2B=1

is this the correct method?

OR

am i supposed to join the sinx term and xsinx terms (second and fourth terms from my first equation) together like this:

(-2B-2C+2F+E+Bx+2Cx)sinx
?
 
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rygza said:
In one of the last steps in trying to find the Coefficients of Yp (particular solution), I am supposed to set coefficients of similar terms equal to each other. Here's where i need help:
i have
A(e^x) + (-2B-2C+2F+E)sinx + (2B+F-2C-2E)cosx + (B+2C) xsinx + (C-2B)xcosx = (e^x)+xcosx

I set: A=1, -2B-2C+2F+E = 0, 2B+F-2C-2E=0, B+2C=0, and C-2B=1

is this the correct method?
Yes, this is correct. The equation above has to be identically true for all x. The only way this can happen is for the coefficients of the e^x terms on each side to be equal, the coefficients of the sinx terms have to be equal, and so on.

The method below is not correct.
rygza said:
OR

am i supposed to join the sinx term and xsinx terms (second and fourth terms from my first equation) together like this:

(-2B-2C+2F+E+Bx+2Cx)sinx
?
 


Mark44 said:
Yes, this is correct. The equation above has to be identically true for all x. The only way this can happen is for the coefficients of the e^x terms on each side to be equal, the coefficients of the sinx terms have to be equal, and so on.

The method below is not correct.

thanks!
 
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...

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