Find the vertex of the parabola y=2x^2-8x+7
We have that x=-b/2a=-(-8)/(2*2) = 2, and then y = 2(2^2)-8(2)+7=-1
Use implicit differentiation to find dy/dx for the given expression 3xy-2x^2*y = 1
First take the derivative of 3xy using the product rule in which (3xy)' = 3y+3x*dy/dx and then take the derivative of (2x^2y)' which equals 4xy + 2x^2*dy/dx so the following derived expression equals (3y+3x*dy/dx)-(4xy+2x^2*dy/dx) = 1. Next, I need to combine any like terms and isolate a few variables in which I need to isolate mainly dy/dx in which we are solving.
Determine the characteristic polynomial of the given ODE and find the roots, 3y"+2y'+2=0.
In order to determine and translate the given ODE into a characteristic polynomial, we must first consider the expression y=er^x. Next, take both the first and second derivative and then substitute both the expressions back into the ODE. Once that is done, the er^x is factored out and canceled out in which the characteristic polynomial is displayed as 3r^2+2r+2=0. From there, I have to solve for r and since this is the second order, I either go ahead and factor regularly or take the quadratic formula into consideration. Once that is all processed, I should receive my values.