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7 pages/≈1925 words
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Mathematics & Economics
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Math Problem
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# Solving Integral Equations (Math Problem Sample)

Instructions:

1: Integrate by parts ∫xsin3x dx
2: SOLVE [ ■(cosα&sinα@ sinα&-cosα) ] =[■(1&0@0&1)]
3: find the value of m using equation y=mx+1 which is the tangent to curve y=4x
4:FIND the general solution of equation (x^2-y^2 )dx+2xydy=0
5:We will solve system of linear equations, using matrix method.
x − y +2z =7
3x +4y −5z =−5
2x − y + 3z = 12

source..
Content:

Math Test 00043
Students Name
Institution Affiliated
Instructor’s Name
Question 1 Solution
We will evaluate by integration by parts.
If the equationcan be written in form of udvdxdx=uv-vdudxdx, then
let u=x and dudx=sin3x
The above equation can be re written in form of uv-vdudxdx
Therefore,xsin3xdx=x-13cos3x-x-13cos3x1dx
=-x3cos3x+13cos3xdx
=x3cos3x+19sin3x+c
Hence, the solution to equation is x3cos3x+19sin3x+c
Question 2 Solution
cosαsinα sinα-cosα =1001
Since, matrix A is equal to identity matrix then,
coxα=1 , sinα=0
sinα=0 ,-cosα=1
α=cos-11
α=0
α=sin-1(0)
α=0,π
When α=0 , A is identity matrix
α=2nπ, nϵZ
Question 3 Solution
We will find the value of m using equation y=mx+1 which is the tangent to curve y=4x
Because, the equation of the tangent to the given curve is y=mx+c,
we can substitute equation y=mx+1 into the equation y2=4x and we will get:
mx+12=4x
=m2x2+1+2mx-4x=0
=m2x2+x2m-4+1=0………………….(1)
The roots of equation (1) must be equal to the tangent as the tangent touches the curve at one point.
Thus, we have:
Discriminants=0
2m-4-4m21=0
=4m2+16-16m-4m2=0
=16-16m=0
m=1
Hence, the needed value of m is 1
Question 4 Solution
We will work out dydx by placing y=vx
Place y=vx and differentiate it with respect to x
dydx=xdvdx+vdxdx
=dydx=xdvdx+v……………………(1)
Afterwards, we place the value of dydx together with y=vx within equation (1)
dydx=y2-x22xy
xdvdx+v=xv2-x22xxv
xdvdx=x2v2-x22x2v-v
xdvdx=x2v2-x2-2x2v22x2v
xdvdx=-x2v2-x22x2v
dvdx=1x(v2+12v)
2vdvv2+1=-dxx
We will integrate both sides of the equation
2vv2+1dv=-dxx
2vv2+1dv=-logx+c……………………….(2)
Next, we place t=v2+1 and differentiate with respect to v in equation (2).
ddvv2+1=dtdv
2v=dtdv
dv=dt2v
Using equation (2), it follows that:
2vtdt2v=-logx+c
dtt=-logx+c
logt=-logx+c
Place t=v2+1
logv2+1=-logx+c
logv2+1+logx=c
log⁡|xv2+1|=c
Placing v=yx, it follows that:
logyx2+1x=c
Placing c=logc
logy2+x2x2=logc
y2+x2=cx
Hence, the general solution of equation x2-y2dx+2xydy=0 is y2+x2=cx
Question 5 Solution
We will solve system of linear equations, using matrix method.
x − y +2z =73x +4y −5z =−52x − y + 3z = 12
solution
Let B= 1-1234-5

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