# Solving Integral Equations (Math Problem Sample)

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

Math Test 00043

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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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