Recent content by shiri

well I divide both sides by x, so...

Find the particular solution of the differential equation xy' + 4y = -20xcos(x^5) satisfying the initial condition y(pi) = 0.Solution y' + 4(y/x) = -20cos(x^5) p(x) = 4/x q(x) = -20cos(x^5) u(x) = int(4/x) = 4lnx e^{u(x)} = x^4 1/e^{u(x)} int(e^{u(x)}*q(x)dx 1/(x^4)...

thanks vela! I got another question I want to ask. It is same as the first post but this question had a different equation y'' - 16y = 4cos(4x) + 4 +e^{4x} The particular equation for this question is C1cos(4x) + C2sin(4x) + C3 + C4xe^{4x} right?

oops...the first term in C) was suppose to be cosine C) C1xcos(2x) + C2xsin(2x) + C3 + C4e^{2x} so according to my work (post #8), the answer is C) right? yh = Axcos(2x) + Bxsin(2x) + C + De^{2x}

anyone?

derive yp = Asin(2t) + Bcos(2t) until you solve the second derivative, y''p

yh = Axcos(2x) + Bxsin(2x) + C + De^{2x} y'h = Acos(2x) - 2Axsin(2x) + Bsin(2x) + 2Bxcos(2x) + 2De^{2x} y''h = -2Asin(2x) - 2Asin(2x) - 4Axcos(2x) + 2Bcos(2x) + 2Bcos(2x) -4Bxsin(2x) + 4De^{2x} y''h = -4Asin(2x) - 4Axcos(2x) + 4Bcos(2x) - 4Bxsin(2x) + 4De^{2x} y'' + 4y = 2sin(2x) + 2...

Yup, my mistake didn't you do a second derivative for yp? furthermore, both yp and y'p are wrong.

It appears you didn't do a characteristic equation. eg. y'' + 0y' -1y = -578sin(4t) r^2 - 1 = 0

so it should be like this yh = c1e^{2x}+c2e^{-2x} ? Update: I made an error on the equation from a 1st post. It was suppose to be y''+4y=2sin(2x)+e^{2x}+2 I forgot to put an addition sign between 2 and e^{2x}

Bold: Changes Solution: y^2 + 4 = 0 y = -2i, 2i yh = c1cos(x) + c2sin(x) yp = Acos(2x) + Bsin(2x) + C + De^{2x} y'p = -2Asin(2x) + 2Bcos(2x) + 2De^{2x} y''p = -4Acos(2x) - 4Bsin(2x) + 4De^{2x} y'' + 4y = 2sin(2x) + 2 e^{2x} [-4Acos(2x) - 4Bsin(2x) + 4De^{2x}] + 4[Acos(2x) + Bsin(2x) + C +...

Consider the equation y''+4y=2sin(2x)+e^{2x}+2. According to the method of Undetermined Coefficients, the particular solution has the form: C1 = 1st constant C2 = 2nd constant C3 = 3rd constant . . . A) C1xsin(2x) + C3x + C4xe^{2x} + C5xe^{-2x} B) C1sin(2x) + (C2)2 + C3xe^{-2x} C) C1xcos(2x) +...

Just to be sure, the answer is y(x) = -3x/35 + 178/(35x^{6}) right?

Find the particular solution of the differential equation dy/dx = (3x+42y)/7x satisfying the initial condition y(1) = 5. Attempt: dy/dx = 3/7 + 6y/x dy/dx - 6y/x = 3/7 p(x) = -6/x q(x) = 3/7 u(x) = -6 ∫(1/x)dx = -6ln|x| e^(u(x)) = -x^6 1/e^(u(x)) ∫e^(u(x))q(x)dx =...

Let f(x) = sqrt(49-x^2) + 7arccos(x/7). Then f'(x) can be written in the simplified form -((x+c)/(mx+n))^p What are the values of c, m, n and p? So far what I got in simplified form is (-x-7)/sqrt(49-x^2) How can I make my simplified form into that simplified form?