# Tutor profile: Jacob S.

## Questions

### Subject: Calculus

Find the derivative of y = ((9x^2)+1)arctan(3x)

Now here we an equation that consists of two equations multiplied. We will need the multiplication rule. (g(x)h(x))' = g'(x) h(x) + g(x) h'(x) Thus: y' = (9x^2 + 1)'arctan(3x) + ((9x^2) +1)(arctan(3x))' so lets solve each derivative separately g'(x) = 9*2x + 0 = 18x solving h'(x) is more challenging because has a function within it. Solving it requires chain rule: h'(x)= f'(x)*F'(f(x)) Thus: h'(x) = (f(x))' (arctan(f(x)) which gives h'(x) = f'(x)(1/[(f(x)^2) +1]) in which f(x) = 3x and f'(x) = 3, thus: h'(x) = 3 (1/[((3x)^2) +1]) which can be simplified to: h'(x) = 3/ ((9x^2) + 1) Now that we know the derivatives they can be substituted back into the original derivatives: y' = 18x(arctan(3x)) + ((9x^2) +1)*(3/ ((9x^2) + 1)) Parts can be canceled out and we get: y' = 18x(arctan(3x)) + 3

### Subject: Differential Equations

Solve the following higher-order differential equation: y'' + 5y' +6y = 3x

So the given equation is a linear differential equation to which we can apply the standard linear solving method. First, we must substitute the auxiliary equation y=e^mx for y. In the substitution, we leave the 3x out and will solve it later. y''+5y'+6y = 0 -> (e^mx)'' + 5(e^mx)' + 6(e^mx) = 0 After taking the derivatives of the auxliliary equation we get: (m^2)(e^mx) + (5m)(e^mx) + (6)(e^mx) = 0 We can pull out the common multiple and divide it over to the other side ((m^2) + 5m + 6)(e^mx) = 0 -> (m^2) + 5m + 6 = 0 We can now solve for the values of m, which factors into: (m+3)(m+2) = 0 in which m = -2,-3 so we can have y1 = e^(-2x) and y2 = e^-3x Which can be plugged into yc = C1 y1 + C2 y2 the general solution is thus: yc= C1(e^-2x) + C2 (e^-3x) Now we need to solve for the particular solution. This can be done with the method of undetermined coefficients. We can pick yp = Ax + B and find its derivatives: yp' = A yp'' = 0 These can then be plugged into the differential equation: 0 + 5(A) + 6 (Ax + B) = 3x The constant terms and x terms can be grouped into separate equations: 6Ax = 3x -> A = 1/2 5A + 6B = 0 -> 5*(1/2) +6B = 0 -> 6B = -5/2 -> B = -5/12 Knowing A and B the particular solution is: yp = (1/2)x - 5/12 We know the general and particular solutions which can be combined using the following equation: y = yc + yp So the final solution is: y = C1(e^-2x) + C2 (e^-3x) + (1/2)x - 5/12

### Subject: Algebra

Expand the following polynomial and find the values of x at which y = 0 (the horizontal intercepts): y = x^5 - x

While the large exponent is intimidating at first, I always recommend looking for and isolating common multiples. In the equation, x is a common multiple and can be factored out: y=(x^4 - 1)*x The portion of the equation within the parenthesis looks very similar to the common factorable quadratic a^2 - b^2 = (a-b)(a+b). In this case, a = x^2 and b = 1, after factoring the equation becomes: y = (x^2 - 1)*(x^2 +1)*x Now (x^2-1) can be factored like before, (x^2+1) cannot be factored into real terms. The equation is now completely factored in: y=(x+1)(x-1)(x)(x^2+1) Now we need to determine where y = 0. If any of the factored portions of the equation becomes 0 then the whole expression becomes = 0. Thus 0 = x+1 -> x=-1 0= x-1 -> x=1 0 = x -> x =0 0 = x^2 + 1 -> x = sqrt(-1) this is an unreal value so we will be ignoring this portion. Finally the values of x that cause y = 0 are x =0,-1,1

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