ASSIGNMENT 6

PROBLEMS

1. Find the approximation value of √ 5 according to Heron’s method!

2. Determine the Mac Laurin series of function; f(x) = ln(x+1)

3. Find the integration factor of y’=e^2x + y – 1 and solve those differential equation!

4. Find the solution of ∫ (x-7)/(x²-x-12) dx

5. Find the solution of ∫ e^2x sinx dx

ANSWERS

1. Given √ n with n=5, to find the approximation value; at first we have to change n be multiplication form that is n = 2a → a = n/2 so, we find a = 2.5

Then, we must subtitute it to Heron’s formula;

a₁ = (a+2)/2 , so we have a₁ = 2.25

a₂ = (a₁ + (n:a₁))/2 , so we have a₂ = 2.23611

a₃ = (a₂ + (n:a₂))/2 , so we have a₂ = 2.23606

From the calculation above, we can conclude that the approximation value of √ 5 according to Heron’s method is equal to 2.23606.

2. The basic form of Mac Laurin series function is

p(x) = f(0) + f’(0)x + f’’(0)x²/2! + f’’’(0)x³/3! + …+ f ^k (0)x^k/k!

Given a function that is;

f(x) = ln(x+1) then f(0) = 0

f’(x) = 1/ (x+1) then f’(0) = 0

f’’(x) = -1/(x+1)² then f’’(0) = 1

f’’’(x) = 2/(x+1)³ then f’’’(0) = 2

f’^v (x) = -6/(x+1)4 then f’^v (0) = -6

f^k(x)=(-1)^k+1 (k-1)! / (x+1)^k then f ^k(0)=(-1)^k+1 (k-1)!

Substitute those value above to the Mac Laurin basic form series.

So, the Mac Laurin form series of function f(x) = ln(x+1) is equal to

p(x) = x – x²/3 + x³/3 – x⁴/4 + … + (-1)^k+1 x^k/k!

3. Given a differential equation y’= e^2x + y - 1

Step 1 : find the integration factor

Remember the formula; if given y’+p(t)=q(t) the integration factor is µ(t) = exp ∫ p(t) dt

So, y’= e^2x + y – 1 ↔ y’ - y= e^2x – 1

µ(x) = exp ∫ -1 dx = e^-x

We have found the integration factor.

Step 2: find the solution of these differential equation

y’ - y= e^2x – 1

Multiplying each parts of the equation with the integration factor

y’e^-x – ye^-x = e^x – e^-x

[ye^-x ]’ = ∫ (e^x – e^-x) dx

ye^-x = e^x + e^-x+c

y=e^2x + 1 + ce^x

So the solution is y=e^2x + 1 + ce^x

4. Given ∫ (x-7)/(x²-x-12) dx

The form above is one of the partial integration of a function which have different linear factor.

To find the solutions, we must change the function to be partial fraction;

∫ (x - 7)/(x²-x-12) dx = ∫ (x-7)/(x – 4)(x + 3) dx

Then

(x -7)/(x – 4)(x + 3) = (A/(x – 4)) + (B/(x + 3))

↔ (x -7)/(x – 4)(x + 3) = (A(x + 3) + (B(x - 4))/(x – 4)(x + 3)

↔ x - 7 = A(x + 3) + B(x – 4)

↔ x - 7 = Ax + 3A + Bx – 4B

↔ x - 7 = (A + B)x + (3A – 4B)

From the last equation above we find that (A+B) = 1 and (3A - 4B) = -7

With substitution or elimination method we will find that A = -3/7and B = 10/7

So, the solution is:

∫ (x-7)/(x – 4)(x + 3) dx

= ∫ (-3/7)/(x – 4) dx + ∫ (10/7)/(x + 3) dx

= (-3/7) ∫ (x – 4) dx + (10/7) ∫ (x + 3) dx

= (-3/7) ln │x - 4│+ (10/7) ln│x + 3│+ c

5. Remember that d(uv) = u dv + v du ↔ uv = ∫ u dv + ∫ v du

So, we have a formula that ∫ u dv =uv - ∫ v du

Given ∫ e^2x sinx dx = I

Choose;

u = e^2x

du = 2e^2x dx

dv = sin x dx

v = - cos x

Remember the formula above;

∫ u dv = uv - ∫ v du

∫ e^2x sinx dx = -e^2x cos x + 2 ∫ e^2x cos x dx

I = -e^2x cos x + 2 ∫ e^2x cos x dx

Seen that ∫ e^2x cos x dx is have a same form with ∫ e^2x sinx dx.

So, we will do the same procedure;

Choose;

u = e^2x

du = 2e^2x dx

dv = cos x dx

v = sin x

Subtitute to the formula, then we have;

∫ e^2x cos x dx = e^2x sin x - 2 ∫ e^2x sin x dx

Subtitute to the previous equation, we have;

I = -e^2x cos x + 2 ∫ e^2x cos x dx

I = -e^2x cos x + 2 (e^2x sin x - 2 ∫ e^2x sin x dx)

I = -e^2x cos x + 2e^2x sin x - 4 ∫ e^2x sin x dx

I = -e^2x cos x + 2e^2x sin x - 4I + c

5I = -e^2x cos x + 2e^2x sin x + c

5I = (2 sin x – cos x) e^2x + c

I = (1/5) (2 sin x – cos x) e^2x + c

So, the solution of ∫ e^2x sinx dx =(1/5) (2 sin x – cos x) e^2x + c