Recent content by middieman147

Homework Statement y''-4y'+4y=0 , y(1)=1 and y'(1)=1 The Attempt at a Solution Auxiliary equation: r2-4r+4=0 I tried factoring 2 different ways: (r-2)2=0 r=2,r=2 y1=e2t y2=y1 y(t)=c1e2t+c2e2t y(1)=c1e2+c2e2=1 ---eq(1)y'(t)=2c1e2t+2c2e2t ...c2=1/(2e2)-c1 ---eq(2) sub eq(2) into eq(1)...

isnt that irrelevant since there is an IC? wouldn't the x in that expression =1 and then just equal C multiplied by another constant?edit: Are you saying i should divide 10 by e^1 to find the constant?

wouldn't that just be another constant? so instead of writing Ce^sqrt(x), you could just consider it a new constant

yes I do. thanks.

Ok: y=-4sqrt(x)-4+C with C=10 is the answer given the IC y(1)=2?

Ok so now I have: y'-y/(2√(x))=2 p(x)=-1/(2√x) q(x)=2 u(x)=e∫(-1/(2√x))=e-√x y=e√x(2e-√x)+C y=2+C this is the solution for all values of x with C=0, correct? Or am I forgetting a step? Thanks

Or can I substitute in the y(1) for the 2 on the right side of the equation and make all the other x's equal to 1?

Given: Solve the initial value problem 2(√x)y'+y+4(√x) ; y(1)=2 I am having trouble separating the x's and y's in order to integrate. I keep coming up with: dy/dx +y/(2(√x))=2... What do I keep missing here? I am pretty sure you leave the y(1)=2 alone until you are finished with...