# Tutor profile: Sumeet C.

## Questions

### Subject: Arduino Programming

What is a way of implementing a delay without using the delay() function?

Solution: using millis() function //define time period int T = 10; //define previous timestamp unsigned long previousTime = 0; void setup() { ... } void loop() { //store current timestamp unsigned long currentTime = millis(); //check if time period has been passed and update previousTime if(currentTime - previousTime >=T) { previousTime = currentTime; //execute some function functionX(); } }

### Subject: Algebra

Partial Fraction Expansion (one of the most useful things to know, in my opinion, that comes up a lot in engineering problems) Decompose $$\frac{17x-53}{x^2-2x-15} $$

Straight off the bat, we can factor the denominator: $$ \frac{17x-53}{(x+3)(x-5)} $$ If your factoring is a bit rusty, remember to check that the sum is equal to the second term, and the product to the third term ( $$3-5=-2$$ and $$ 3\times-5 = -15)$$ Next, we write out the decomposition with unknown constants: $$\frac{17x-53}{(x+3)(x-5)} = \frac{A}{x+3} + \frac{B}{x-5}$$ Add the two fractions using LCD: $$\frac{17x-53}{(x+3)(x-5)} = \frac{A(x-5) + B(x+3)}{(x+3)(x-5)}$$ Set the numerators equal to each other: $$17x-53 = A(x-5) + B(x+3)$$ Pick values of $$x$$ we can use to determine A and B: Using $$x=5$$, we can get rid of A and solve for B: $$17(5)-53 = A(0)+B(5+3)$$ $$32 = 8B $$, $$B = 4$$ Using $$x=-3$$, we can get rid of B and solve for A: $$17(-3)-53 = A(-3-5) + B(0) $$ $$ -104 = -8A , A = 13$$ Putting this all together: $$\frac{17x-53}{(x+3)(x-5)} = \frac{13}{x+3} + \frac{4}{x-5}$$

### Subject: Differential Equations

Solve the ODE $$\frac{dx}{dt} = 2x-3 $$ with initial condition x(1) = 1.

First, we separate $$x$$ and $$t$$ to two sides. $$\frac{dx}{2x-3} = dt$$ Then, integrate both sides. $$\frac{1}{2} \log |2x-3| = t + C_1 $$ Solve for $$x$$. First, we take the exponential of both sides: $$ 2x-3 = \pm \exp (2t + 2C_1)$$ $$x = \pm \frac{1}{2} \exp (2t + 2C_1) + \frac{3}{2}$$ We can further simplify the constant by letting $$C = \frac{1}{2}exp(2C_1)$$ $$x(t) = Ce^{2t} + \frac{3}{2} $$ Using the initial condition, we can solve for C. $$x(1) = 1$$ $$ 1 = Ce^{2} + \frac{3}{2} $$ $$ 1 - \frac{3}{2} = Ce^{2}$$ $$ -\frac{1}{2} e^{-2} = C$$ Therefore, our final solution is: $$ x(t) = -\frac{1}{2} e^{2(t-1)} + \frac{3}{2} $$

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