The document contains solutions to three math homework problems: 1) An integral problem involving trigonometric substitutions and integration by parts. The solution finds the integral I=1/2[x cos(ln(x))+xsin(ln(x))]+C. 2) A differential equation problem solved using an integrating factor. The solution finds the particular solution as Y=(2x+1)e^2x + ex + (x+1). 3) An eigenproblem with solutions for the eigenvectors and eigenvalues. The roots are found to be real and equal to 2.

1. Sample Math Homework | Sample Assignment | www.expertsmind.com
Solution:
(a) f(x)= Cos (ln(x))
ò
I= Cos(ln(x)) dx
Put ln(x) = t
X= et
Differentiating dx = et dt
Then
ò coste dt
t
I=
Integrating by parts we get
ò udv = uv - ò vdu dv= et dt u= cost
t
V= e du = -sint dt
ò e (-sint)dt
t t
I= e cost -
I= et cost + ò e (sint)dt
t
Further integrating by parts we get
dv= et dt u= sint
V= et du = Cost dt
ò e (cost)dt
t
I= et cost + et sint -
I= et cost + et sint – I
2I= et cost + et sint
1
I= [ et cost + et sint]
2
1
I= [ eln(x) cos(ln(x)) + eln(x) sin(ln(x))]
2

2. 1
I= [x cos(ln(x)) +xsin(ln(x))]
2
x
I= [cos(ln(x))+sin(ln(x))]+C
2
dy 1
(b) =
dx 3 + 5cos(x)
dx
y= ò3 + 5cos(x)
dx
y= ò é 2 x ù
ê 1 - tan 2 ú
3 + 5ê ú
ê1 + tan2 x ú
ê
ë 2ú
û
é 2 xù
ê1 + tan 2 údx
ë û
y= ò x
3 + 3tan2 + 5 - 5tan2
x
2 2
2 x
Sec dx
2
y= ò8 - 2tan2
x
2
x
Sec 2 dx
2 x
y= ò é xù
Put tan
2
= t; differentiating we get
2ê4 - tan2 ú
ë 2û
1 x
Sec 2 dx = dt
2 2
dt
y= ò
4 - t2
x
2 + tan
1 2-t +2+ t 1 é dt dt ù 1 2 + t 1 2 +c
y=
4 ò (2 - t)(2 + t) ò
dt = ê
4ë 2-t
+ ò = ln = ln
2 + tú 4 2 - t 4
û 2 - tan
x
2
dy
(c) = x 2 - 8x - 9
dx
dy
= x 2 - 8x + 16 - 16 - 9
dx
dy
= (x - 4)2 - 25
dx
y = ò (x - 4)2 - 25 dx Put x-4= 5 coshu
Differentiating dx= 5 sinhu du
y= ò ò
25cosh2u - 25 5sinhudu = 25sinh2udu

3. 2
y= ò
é eu - e -u ù
25ê
2
ú du =
25
4 ò
[e2u + e - 2u - 2]du =
8
[
25 2u
e - e - 2u - 4u ]
ê
ë ú
û
y=
25 u
8
[( )(
e - e -u eu + e -u - 4u ) ]
é x 2 - 8x - 9 x - 4ù
y=
25
[4sinhucosh u - 4u] = 25 ê x - 4 - cosh-1 ú+c
8 2 ê 5 5 5 ú
ë û
Solution:
(a)
I= ò ucoshudu dv= coshudu t= u
V= sinhu dt= du
ò tdv = tv - ò vdt
= usinhu - ò sinhudu
= usinhu - coshu + c
dy y
(b) + = (x2 + 4)Cosh(x 2 + 4)
dx x + 2
This is a linear differential equation.
1
P= and Q = (x2 + 4)Cosh(x 2 + 4)
x+2
Integrating factor is
dx
eò
pdx ò
= e x + 2 = eln(x + 2) = (x + 2)
Multiplying the given ODE by the integrating factor we get
dy
(x + 2) + y = (x + 2)(x2 + 4)Cosh(x 2 + 4)
dx
d(x + 2)y
= (x + 2)(x 2 + 4)Cosh(x 2 + 4)
dx
Integrating on both sides

4. ò (x + 2)(x
2
Y(x+2)= + 4)Cosh(x2 + 4)dx =
ò x(x ò
2
+ 4)Cosh(x2 + 4)dx + 2 (x2 + 4)Cosh(x2 + 4)dx = I1+I2
In the first integral Put (x2+4) = t differentiating we get 2x dx= dt then it gets
transformed to
1 1 1
I1=
2 ò
tcoshtdt = {tsinht -cosht}+C= {( x2+4)Sinh (x2+4)-cosh(x2+4)}+C
2 2
I2 is not a standard integral. This has to be done by series method.
eu - e -u u2 u4
We know that Coshu= =1 + + ... , where * stands for factorial of a
2 2* 4*
number.
(x2 + 4)2 (x2 + 4)4
In our case u= x2+4, replacing we get =Cosh (x2+4) 1 + + ...
2* 4*
(x2 + 4)3 (x2 + 4)5
ò
I2= 1 +
2*
+
4*
... dx expanding and integrating we get
1 x 7 12x 5
I2=x+ [ + + 24x 2 + 64x ]…..
2 7 5
(c)
Area under region
p p
ò x coshx dx = 2ò x coshx dx
-p 0
due to its symmetry about y- axis.
= 2[xsinhx - coshx ]0 = 2[psinhp-coshp +1]
p
If the area density of the sheet is r, then the mass of the blade is 2r[psinhp-
coshp +1].
Center of mass
Consider a thin strip of thickness dx at a distance of x from the origin as shown in
the figure

5. Mass of the strip is rxcosh(x)dx
π
òx
2
coshx dx
Then the center of mass= -π
=
[x Sinhx - 2xcoshx - 2sinhx]
2 p
-p
(On
M M
integrating by Parts successively)
Center of mass = 0 as the numerator vanishes. The center of mass lies at the
origin which is true from its symmetry.
d2 y dy
Solution: (a) 2
-4 +4=0
dx dx
equation: m2 - 4m + 4 = 0 ; (m-2)2=0; m=2 the roots are real and equal.
Auxiliary
Hence y=(Ax+B)e 2x
Eigen vectors:
Eigen values:
d2 y dy
(b) -4 + 4 = e x + 4x
dx 2 dx
Auxiliary equation: m2 - 4m + 4 = 0 ; (m-2)2=0; m=2 the roots are real and
equal.
Hence
General solution: y=(Ax +B)e 2x
Particular Integral:
e x + 4x ex 4x ex
PI= 2
= + = +
D - 4D + 4 D 2 - 4D + 4 D2 - 4D + 4 (1)2 - 4(1) + 4
4x
4(1 - D + D 2 / 4)
=ex + (1-D+D2/4)-1(x)= ex + (x+1)
Hence Y= GS+PI= y=(Ax +B)e 2x + ex + (x+1)

6. Y(0) = 3; hence 3=B+2; B= 1
Y’= (Ax+B) 2 e2x +A e2x +ex +1
At x= 0 Y’=6
6= 2B +A +2; A= 2
2x
Substituting the values of A and B in Y= (Ax +B)e + ex + (x+1); we get
Y= (2x +1)e 2x + ex + (x+1)
Solution:
ò [cos x] dx
2
-1
(a) I= Put x= Cost; differentiating we get dx= sint dt
ò t (sint)dt
2
I= dv= -sint dt u= t2
V= cost du=2tdt
ò udv = uv - ò vdu
I = t cost - ò 2tcostdt
2
I = t 2cost - I1 where
I1= ò 2tcostdt dv= cost dt u=2t
V= sint du= 2dt
I1 = 2tsint - 2sintdtò
I1 = 2tsint + 2cost
I= t2cost-2tsint-2cost+c { x= Cost; 1 - x 2 = sint ; t= cos -1 x }
[ ]2
I= x cos -1x -2 cos -1 x 1 - x 2 -2x+C
(b) Trace of the given curve

7. Curve is symmetric about the Y axis
p p
2 2 p
Hence òp cosxdx = 2 cosxdx = 2[sinx ]0 =2[1-0]=2
ò 2
0
-
2
(b) Consider a small strip of thickness dx at a distance of x from the origin as
shown in the figure
To find the position on the Y axis,

8. Consider a thin horizontal strip of thickness dy at a distance of y from the origin;
the area of the strip is 2cos -1ydy. Hence the mass of the strip is r2cos -1ydy.
1 1
ò ρ2ycos ò ρ2ycos
-1 -1
ydy ydy 1
ò ycos
0 -1
Center of mass= 0
= = ydy =. Integrating by
M 2ρ
0
parts we get
After substituting y = cost we get
p p
1 2 2 p
1 1 p
ò
-1
ò
ycos ydy = cost sint tdt = ò tsin2tdt = [- 2tcos2t + sin2t]0 =
2
2 8 8
0 0 0
p
Hence the center of mass of the strip is (0, )
8
Microsoft Equation
3.0