Unit 36 Applied Mathematics: Calculus Solutions for Engineering HND

This assignment provides detailed solutions to a range of calculus problems within the context of applied mathematics. Task 1 focuses on forming and identifying the order of differential equations from given functions, followed by finding general solutions for first-order differential equations. The task concludes with finding auxiliary equations and solutions for second-order differential equations. Task 2 involves an optimization problem, determining the dimensions of a window frame that maximizes light passage, subject to a constraint on the length of the metal strip. Task 3 covers integration techniques, including integration by parts to solve ∫ e-3x cos2x dx and integration using partial fractions to solve ∫ (2x2 + x + 1)/ {(x – 1) ( x 2+ 1 )} dx. Finally, Task 4 focuses on calculating the area under a curve and the area enclosed by a curve and a straight line, utilizing definite integrals. The document is available on Desklib, a platform providing students with access to a wide range of educational resources and study tools.

1
Calculus Questions
Applied Mathematics

2
Calculus Questions
Task-1
a.i. Given : y = 4Asinx + 2Bcosx
Differentiating wrt x-
y’= 4Acosx – 2Bsinx
Again Differentiating wrt x-
y’’= - 4Asinx - 2Bcosx
y’’= -y or
y’’+ y = 0
Since we differentiate it twice, it is 2nd order differential equation
a.ii. Given : y = x + A
x
Differentiating wrt x-
y’= 1 - A
x2
y’= 1 - 1
x (y – x)
y’+ y
x - 1 = 1
y’+ y
x = 2
It is 1st Order Differential Equation
a.iii. Given : y = Ax2 – Bx (1)
Differentiating wrt x-
y’= 2Ax – B (2)
Again Differentiating wrt x-
y’’= 2A or A = y’’/2 (3)
Put (3) in (2)-

3
Calculus Questions
y’= y’’x – B or B = y’’x – y’ (4)
Put (3) and (4) in (1)-
y = y ' '
2 x2 – y’’x2 – y’x
y ' '
2 x2 + y’x + y = 0
It is a 2nd Order Differential Equation.
b.i. Given : dy
dx = y
x
dy
y = dx
x
Integrating both sides-
∫ dy
y =¿∫ dx
x ¿
ln(y) = ln(x) + ln(c) where ln(c) is the constant of integration
ln(y) = ln(xc) or
y = xc is the general solution of the given differential equation.
b.ii. Given : cosx
3+ y
dy
dx = sinx
Separating the variables-
dy
3+ y = tanx dx
Integrating both sides-
∫ dy
3+ y =∫tanx dx
ln(3 + y) = ln(secx) + ln(c) where ln(c) is the constant of integration
ln(3 + y) = ln(c*secx)
3 + y = c*secx

4
Calculus Questions
y = c*secx – 3 is the general solution of the given differential
equation.
b.iii. Given : dy
dx = (2y + 1)(x – 3)
Separating the variables-
dy
2 y +1 = (x – 3)dx
Integrating both sides-
∫ dy
2 y+1=∫( x – 3)dx
1
2 ln(2y + 1) = x2
2 – 3x + C
ln(2y + 1) = x2 – 6x + 2C
2y + 1 = ex2−6 x+2 C
y = 1
2 (e x2−6 x+2 C )− 1
2
c.i. Given : -3 d2 y
d x2 + 36 dy
dx - 108y = 0
Putting d
dx ≡ D∧d2
d x2 ≡ D2
Then, (-3D2 + 36D – 108)y = 0
Solving the above auxiliary equation which is quadratic.
D = −b ± √b2−4 ac
2a = −36 ± √1296−1296
−6 = 6 or 6
Solution of the differential equation is : y = (A + Bx)e6x
c.ii. Given : d2 y
d x2 + 6 dy
dx + 9y = 0

5
Calculus Questions
Putting d
dx ≡ D∧d2
d x2 ≡ D2
Then, (D2 + 6D + 9)y = 0
Solving the above auxiliary equation which is quadratic1
D = −b ± √b2−4 ac
2a = −6 ± √36−36
2 = -3 or -3
Solution of the differential equation is : y = (A + Bx)e-3x
Task-2
Let the length of the rectangular window be a, breadth be b and radius of the
semicircle surmounted on rectangle be r, then, the shape would be like-
b
a
Here, r = a/2
According to question,
2b + a + πa
2 = 10 (a and b are in metre)
b = 5 – a
2 - πa
4 (1)
Now, area of the window will be the sum of area of rectangle and semicircle i.e.,
A = ab + π a2
4

6
Calculus Questions
Putting equation (1)-
A = a(5 – a
2 - πa
4 ) + π a2
4
A = 5a - a2
2
For Area to be maximum, we have to differentiate wrt a,
dA
dx = 5 – a = 0 (For Maxima or Minima)
a = 5
Also, d2 A
d x2 = - 1
So, a = 5 is a point of Maxima for Area.
Maximum Amount of light pass through the window when length of the rectangle is
5m, breadth of the rectangle is ( 5
2 - 5 π
4 )m and radius of the semicircle is 2.5m
Task-3
i. y = ∫ e−3 x cos 2 xdx (1)
Using Integration by parts, considering e-3x as the 1st function and cos2x as the 2nd
function.
y = e-3x ∫cos 2 xdx + 3
2 ∫e−3 x sin 2 x dx
y = e−3 x sin2 x
2 + 3
2( e-3x∫sin 2 xdx - 3
2 ∫e−3 x cos 2 xdx )
y = e−3 x sin2 x
2 + 3 e−3 x cos 2 x
4 - 9
4 y (From (1))
13
4 y = e−3 x sin2 x
2 + 3 e−3 x cos 2 x
4
y = 4
13 ( e−3 x sin2 x
2 + 3 e−3 x cos 2 x
4 )
ii. Given : y = ∫ 2 x2+ x +1
( x−1 ) ( x2+1 ) dx

7
Calculus Questions
Separating Partial Fractions2-
Let 2 x2 + x +1
( x−1 ) ( x2 +1 ) = A
( x−1 ) + Bx +C
( x2+1 )
2x2 + x + 1 = (A + B)x2 + (-B + C)x + (A-C)
Comparing the coefficients-
A = 2, B = 0, C = 1
So, 2 x2 + x +1
( x−1 ) ( x2 +1 ) = 2
( x−1 ) + 1
( x2+1 )
Now,
y = ∫ ( 2
( x−1 ) + 1
( x2+ 1 ) )dx
= 2ln(x – 1) + tan-1x + C
Task-4
i. Function of the curve : y = 5ex
Area under the curve = ∫ ydx
Area is to be calculated from x = 0 to x = 3
Area = ∫
0
3
5 e x
= 5[e3 – e0]
= 5[20.085 – 1]
= 95.43 sq.unit
ii. Given : Equation of straight line is y = -2x + 28
And Equation of curve is y = 36 – x2
For finding the points of intersection of the line and the curve, we have to
equate both equations-
36 – x2 = -2x + 28
x2 – 2x – 8 = 0
(x + 2)(x – 4) = 0
Therefore, x = -2 or 4 are the points at which both the line and the curve
intersect.

8
Calculus Questions
Area enclosed by the curve = ∫
−2
4
ycurve dx−∫
−2
4
yline dx
= ∫
−2
4
( 36−x2 ) dx−∫
−2
4
( −2 x+28 ) dx
= 36 (4 + 2) – 64+8
3 + (16 – 4) + 28 (4 + 2)
= 216 – 24 + 12 + 168
= 12 sq. unit
References-
1Sharma R.D. (2018). Mathematics (2019 edition). Dhanpat Rai Publication, New Delhi
2Aggarwal, R.S. (2013). Senior Secondary School Mathematics (2018 edition). Bharti
Bhawan Publishers, New Delhi