Enable contrast version

# Tutor profile: Joseph W.

Inactive
Joseph W.
15-year Professional Tutor in Electrical Engineering and Math
Tutor Satisfaction Guarantee

## Questions

### Subject:Calculus

TutorMe
Question:

Find the general solution to the linear homogeneous equation y'' + 5y' + 6y = 0.

Inactive
Joseph W.

First, we assume that any solution of the above differential equation will be of the form y = e^rx where r is any real number. We then know that y' = r*e^rx and y'' = (r^2)*e^rx Thus, our differential equation will take the form (r^2)*e^rx + 5*(r*e^rx) + 6*e^rx = 0. Dividing both sides by e^rx, which we can do because e^rx will never equal 0, we get what is called the characteristic polynomial of the differential equation r^2 + 5r +6 = 0 Realizing that this is an easily factorization problem, we factor the left hand side to become (r + 3)*(r + 2) = 0 -> r = -3, -2 Thus, we have two solutions to the above given differential equation, and they take the form of y1 = e^(-3x) y2 = e^(-2x) and we know that the total solution will just be some linear combination of the two solutions, so we know the homogeneous solution of the given differential equation will be of the form yh = A*e^(-3x) + B*e^(-2x) Note that we cannot solve for A or B unless we have two initial conditions, which we do not have for this problem.

### Subject:Calculus

TutorMe
Question:

Find the derivative of y = tan(ln(ax+b))

Inactive
Joseph W.

Breaking the above into three functions, we have y(u) = tan (u) u(v) = ln(v) v(x) = ax+b The derivative of the above three equations according to the chain rule will be dy(u)/dx = (sec^2 u)*du/dx du(v)/dx = (1/v)*dv/dx dv(x)/dx = a Thus, working backwards, we know that du(v)/dx = (1/(ax+b))*a y' = sec^2 (ln(ax+b))*(a/(ax+b))

### Subject:Electrical Engineering

TutorMe
Question:

How are power, voltage, and current related in a basic circuit involving current sources, voltage sources, and resistors?

Inactive
Joseph W.

So, mathematically, we can calculate power in several different ways. If we're talking about a basic circuit with just a set of current sources, voltage sources, and resistors. the power of each element can be found by multiplying the voltage across an element by the current across the element. If, after finding the voltage and current across each element and labeling each element with the conventional notation you note that the current is leaving the positive terminal of the voltage, the element is supplying power. If the current is leaving the negative terminal of the voltage across an element, the element is absorbing power. A resistor will always absorb power, but current sources and voltage sources can either supply or absorb power.

## Contact tutor

Send a message explaining your
needs and Joseph will reply soon.
Contact Joseph