TutorMe homepage
Subjects
PRICING
COURSES
Start Free Trial
Manoj P.
10 years of experience
Tutor Satisfaction Guarantee
Calculus
TutorMe
Question:

Find two solutions to Y’-9y=0

Manoj P.

This is the same differential equation that we looked at in the first example. This time however, let’s not just guess. Let’s go through the process as outlined above to see the functions that we guess above are the same as the functions the process gives us. First write down the characteristic equation for this differential equation and solve it. r^2 -9y=0 r=+-3 The two roots are 3 and -3. Therefore, two solutions are Y1(t)=e3t and y2 (t)=e-3t These match up with the first guesses that we made in the first example.

Calculus
TutorMe
Question:

Find a and b such that both g(x) given below and its first derivative are continuous? G(x)={a2 + bx if x greater equal to 2 -2x-2 if x is less than 2

Manoj P.

For x > 2, g(x) = a x 2 + b is a polynomial function and therefore continuous. For x < 2, g(x) = -2 x + 2 is a polynomial function and therefore continuous. let L1 = lim x→ 2 - g(x) = a (2) 2 + b = 4 a + b L2 = lim x→ 2 + g(x) = -2(2) + 2 = -2 For continuity of g at x = 2, we need to have L1 = L2 = g(2) Which gives 4 a + b = -2 Continuity of the derivative g' For x > 2, g '(x) = 2 a x is a polynomial function and therefore continuous. For x < 2, g '(x) = -2 is a constant function and therefore continuous. Let l1 = lim x→ 2 - g'(x) = 2 a (2) = 4 a l2 = lim x→ 2 + g'(x) = - 2 For continuity of g' at x = 2, we need to have l1 = l2 or 4 a = - 2 The last equation gives: a = - 1 / 2. And substitute a by - 1 / 2 in the equation 4 a + b = -2 obtained above, we obtain b = 0.

Algebra
TutorMe
Question:

Solve the equation 5(-3x - 2) - (x - 3) = -4(4x + 5) + 13

Manoj P.

Given the equation 5(-3x - 2) - (x - 3) = -4(4x + 5) + 13 Multiply factors. -15x - 10 - x + 3 = -16x - 20 +13 Group like terms. -16x - 7 = -16x - 7 Add 16x + 7 to both sides and write the equation as follows 0 = 0 The above statement is true for all values of x and therefore all real numbers are solutions to the given equation.

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