Mathematics Assignment | EssayIvy.com

Running head: MATHEMATICS

ASSIGNMENT 1

Mathematics Assignment

Student Name

Institutional Affiliation

MATHEMATICS ASSIGNMENT 2

Provide solution to the following questions:

Question 1. Evaluate the following:

Solution

Integrating by parts, we apply the formula ∫ udv = uv - ∫ vdu such that u = x and dv = sin (3x).

Thus x(





cos (3x)) - ∫ -





cos (3x) dx –





- -∫





. However, since -1 is a constant

with respect to x, we integrate 





with respect to x to get - ∫





























Multiplying -1 by -1 we get 1, thus













 







Multiply







 by 1, to get









Thus



 









∫











is a constant with respect to x in both equations, thus, the integral of





with respect to x is

















 

























.

Let u = 3x, and du = 3dx, thus





  . The equation can then be rewritten using u and du

such that,

 



















 





























Since

























 







 























 







 

















 







 

















MATHEMATICS ASSIGNMENT 3

This becomes 

 















  

Simplifying the expression, the answer becomes;





























 

2. If , then for what value of α is A an identity matrix?

Solution

identity matrix of A is given by AI = 1 0

0 1

Thus Cosα -sinα = 1 0 Thus, cos α = 1, –sin = 0, sin α = 0, and cos α = 1

Sinα cosα 0 1

Therefore the value of α in the identity matrix A is 0 or 1

3. The line y = mx + 1 is a tangent to the curve y

= 4x .Find the value of m.

Solution

If y = mx + 1 is the tangent to curve y

= 4x, then the derivative of the curve equals the slope of

the line y = mx + 1.

= 4x 2yy’ = 4 y’ = 2/y m = 2/y

Solving the equation simultaneously,

Y= mx + 1 x = (y-1)/m

= 4x x = y

x=x

MATHEMATICS ASSIGNMENT 4

(y-1)/m = y²/4

4(y-1) = y²m

4(y-1) = y²(2/y) substituting the value of m

4y - 4 = 2y

2y = 4

y = 2

Therefore

m = 2/y = 2/2 = 1

The value of m is 1

4. Solve the following differential equation: (x

− y

) dx + 2 xy dy = 0.

Solution

From the equation = (x

− y

) dx + 2 xy dy = 0.

y' + p(x)y=q(x)yα.

Substitute 2xy dy on both sides to get 2xy dy=(x

)dx, then divide by dx to

get 2xydydx=x

Dividing by 2xy to get dydx=x

y−1+12xy.

Putting this in the form y'+p(x)y=q(x)yα as follows: y'−12xy=x

y−1.

Substituting v=y1−α

ddx[v=y1−α]dvdx=(1−α)y−αdydx.

So dydx=11−αyαdvdx.

Making y the subject,

Since v=y1−α, v=y1y−α.

So y=vyα

Sub-backing into the form y'+p(x)y=q(x)yα we get 11−αyαdvdx+p(x)vyα=q(x)yα.

To make it a linear equation, we divide through to get; dvdx+(1−α)p(x)v=(1−α)q(x).

MATHEMATICS ASSIGNMENT 5

Having formed a first order linear differential equation, we then solve using an integrating factor,

Looking back at the actual problem,

We have y'−12xy=x2y−1.

α=−1, substituting; v=y1−(−1)v=y2.

Since v=y2, then dvdx=2ydydx, so dydx=12y−1dvdx.

We know that y=vy−1.

Considering that y'−12xy=x2y−1 we plug in the equation such that

12y−1dvdx−12xy−1v=x2y−1.

Thus we get,12dvdx−12xv=x2.

divide by 12 to standardize the linear equation.

v'−1xv=x.

solving;

I(x)=e∫p(x)dx, where p(x)=−1x.

∫p(x)dx=∫−1xdx=−lnx.

e−lnx=1x.

Multiply by 1x to get 1xv'−1xv=1.

ddx[1xv]=1, so 1xv=∫1dx=x+C.

Solve for v to get v=x

+Cx.

But v=y

Thus,

+Cx.

The answer = y

+Cx.

MATHEMATICS ASSIGNMENT 6

5. Solve system of linear equations, using matrix method.

x − y +2z =7

3x +4y −5z =−5

2x − y + 3z = 12

Solution

1 -1 2 x 7

A= 3 4 -5 X = y B = -5

2 -1 3 z 12

1 -1 2 a11 a12 a13

2 4 -5 = a21 a22 a23

2 -1 3 a31 a32 a33

A11 = (-1)

1+1

a22 a23 = 4 -5

a32 a33 -1 3 = {4x3-(-1x5)} =7

A12 = (-1)

1+2

a21 a23 3 -5

a31 a33 = 2 3 = -{3x3-(2x-5)} = 19

A13 = (-1)

1+3

a21 a22 3 4

a31 a32 = 2 -1 = {3x-1-(2x4)} = -11

A21 = (-1)

2+1

a12 a13 -1 2

a32 a33 = -1 3 = {-1x3-(-1x2)} =1

MATHEMATICS ASSIGNMENT 7

A22 = (-1)

2+2

a11 a13 1 2

a31 a33 = 2 3 ={1x3- (2x2)} =-1

A23 = (-1)

2+3

a11 a12 = 1 -1

a31 a32 2 -1 = {1x-1-(2x2)} =-1

A31 = (-1)

3+1

a12 a13 = -1 2

a22 a23 4 -5 = {-1x-5-(4x2)} = -3

A32 = (-1)

3+2

a11 a13 1 2

a21 a23 = 3 -5 = -{1x-5-(3x2) = 11

A33 = (-1)

3+3

a11 a12 1 -1

a21 a22 = 3 4 = {1x4 –(3x-1)} = 7

7 1 -3

Adjacency matrix -19 -1 11

-11 -1 7

A : a11 x A11 + a12 x A12 + a13 x A13

A : 1x7 + -1 x -19 + 2 x -11

A = 4

Using the formula

X =









 7 1 -3 7

we plug the value of







to get X =





-19 -1 11 -5

-11 -1 7 12

49 -5 -36

X =





-133+ 5+132

-77+5+84

MATHEMATICS ASSIGNMENT 8

8 8/4

X =





4 = 4/4

12 12/4

x 2

thus, y = 1

z 3

Thus X= 2, Y= 1, and Z = 3

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Deadline

Page count

550 words

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