Category: Engineering Mathematics Objective Questions

250+ TOP MCQs on Leibniz Rule and Answers

Engineering Mathematics Questions and Answers for Freshers focuses on “Leibniz Rule – 2”.

1. Let f(x) = ex sin(x2) ⁄ x Then the value of the fifth derivative at x = 0 is given by
a) 25
b) 21
c) 0
d) 5
Answer: b
Explanation: First expanding sin(x2) ⁄x into a Taylor series we have
sin(x2)=(frac{x^2}{1!}-frac{x^6}{3!}+frac{x^{10}}{5!}….infty)
(frac{sin(x^2)}{x}=frac{x}{1!}-frac{x^5}{3!}+frac{x^9}{5!}….infty)
Now applying the Leibniz rule up to the fifth derivative we have
(((e^x)(frac{sin(x^2)}{x})^{(5)} = c_0^5e^x(frac{x}{1!}-frac{x^5}{3!}+frac{x^9}{5!}…infty))
(+c_1^5e^x(frac{1}{1!}-frac{5x^4}{3!}+frac{9x^8}{5!}…infty)+…..+c_5^5e^x(frac{5!}{3!}+frac{(9.8.7.6.5)x^4}{5!}…infty))
Now substituting x=0 we get
(((e^x)(frac{sin(x^2)}{x}))^{(5)}=c_1^5(1)+frac{5!}{3!})
= 1 + 20 = 21.

2. Let f(x) = eex assuming all the nth derivatives at x =0 to be 1 the value of the (n + 1)th derivative can be written as
a) e – 1 + 2n
b) 0
c) 1
d) None
Answer: a
Explanation: Assume y = f(x)
Taking ln(x) on both sides The function has to be written in the form ln(y) = ex
Now computing the first derivative yields
y(1) = y * ex
Now applying the Leibniz rule up to nth derivative we have
y(n+1)=(c_0^ne^xy+c_1^ne^xy^{(1)}+….+c_n^ne^xy^{(n)})
We know that in the problem it is assumed that [y(1)=y(2)=…=y(n)=1]x=0
Now, substituting x=0 we get
y(n+1)=(c_0^ne+c_1^n+….+c_n^n)
From combinatorial results we know that 2n=(c_0^ne+c_1^n+….+c_n^n)
This gives us
y(n+1)=(e+(c_0^ne+c_1^n+….+c_n^n)-c_0^n)
y(n+1)=e-1+2n

3. Let f(x) = (sqrt{sin(x)}) and let yn denote the nth derivative of f(x) at x = 0 then the value of the expression 12y(5) y(1) + 30 y(4) y(2) + 20 (y(3))2 is given by
a) 0
b) 655
c) 999
d) 1729
Answer: a
Explanation: Assume y = f(x)
Rewriting the function as y2 = sin(x)
Now applying Leibniz rule up to the sixth derivative we have
(y2)(6) = c06 y(6) y + c16 y(5) y(1) + ………+ c66 y(6) y

(y2)(6) = 2 y(6) y + 12 y(6) y(1) + 30 y(4) y(2) + 20 (y(3))2

(sin(x))(6) = -sin(x)
Now substituting x = 0 and observing that y(0) = 0 we have
sin(0) = 0 = 12 y(6) y(1) + 30 y(4) y(2) + 20 (y(3))2.

4. The fourth derivative of f(x) = sin(x)sinh(x) ⁄ x at x = 0 is given by
a) 0
b) π⁄2
c) 45
d) 4
Answer: a
Explanation: First convert the function sinh(x)⁄x into its Taylor series expansion
(frac{sinh(x)}{x}=frac{frac{x}{1!}+frac{x^3}{3!}+frac{x^5}{5!}….infty}{x})
(frac{sinh(x)}{x}=frac{1}{1!}+frac{x^2}{3!}+frac{x^4}{5!}….infty)
Now pick up the whole function (((sin(x))(frac{sinh(x)}{x}))) and apply Leibniz rule up to the fourth derivative we have
(((sin(x))(frac{sinh(x)}{x}))^{(4)}=c_0^4sin(x)(frac{1}{1!}+frac{x^2}{3!}+frac{x^4}{5!}…..infty))
(-c_1^4cos(x)(frac{2x}{3!}+frac{4x^3}{5!}…..infty)+…..+c_4^4sin(x)(frac{4!}{5!}+frac{(7.6.5.4)x^3}{7!}…..infty))
Substituting x=0 we have
(((sin(x))(frac{sinh(x)}{x}))^{(4)}) = 0

5. The third derivative of f(x) = cos(x)sinh(x) ⁄ x at x = 0 is
a) 0
b) π⁄32
c) (π)2
d) cos(1)sinh(1)
Answer: a
Explanation: Assume y = f(x)
Rewriting the part sinh(x)⁄x as infinite series we have
(frac{sinh(x)}{x}=frac{1}{1!}+frac{x^2}{3!}+frac{x^4}{5!}….infty)
Now the function f(x) becomes
y=(cos(x)(frac{1}{1!}+frac{x^2}{3!}+frac{x^4}{5!}….infty))
Taking the third derivative of the above function using Leibniz rule we have
y(3)=(c_0^3sin(x)(frac{1}{1!}+frac{x^2}{3!}+frac{x^4}{5!}….infty)-c_1^3cos(x)(frac{2x}{3!}+frac{4x^3}{5!}….infty))
(-c_2^3sin(x)(frac{2}{3!}+frac{12x^2}{5!}….infty)+c_3^3cos(x)(frac{24x}{5!}…infty))
Now substituting x = 0 we have
y(3) = 0.

6. Let f(x) = (x2 + x + 1)sinh(x) the (1097)th derivative at x = 0 is
a) 1097
b) 1096
c) 0
d) 1202313
Answer: d
Explanation: Expanding sinh(x) into a taylor series we have
sinh(x)=(x+frac{x^3}{3!}+frac{x^5}{5!}…infty)
f(x)=(x2+x+1)((x+frac{x^3}{3!}+frac{x^5}{5!}….infty))
On multiplication we get two series with odd exponents and one series with even exponent. The series with odd exponents are the only ones to contribute to the derivative at x=0
Hence it is enough to compute the derivative at for the following function
(x2+1)((x+frac{x^3}{3!}+frac{x^5}{5!}….infty)=frac{x}{1!}+x^3(frac{1}{3!}+1)+x^5(frac{1}{5!}+frac{1}{3!})….infty)
Taking the 1097thderivative of this function, we have
f(1097)(x)=((1097)!(frac{1}{(1097)!}+frac{1}{(1095)!})+(1099times 1098…4times 3)x^2(frac{1}{(1097)!}+frac{1}{(1097)!})+…infty)
Substituting x=0 we have
f(1097)(x)=((1097)!(frac{1}{(1097)!}+frac{1}{(1095)!}))
=(1+1097*1096)=(1+1202312)=1202313

7. The 7th derivative of f(x) = (x3 + x2 + x + 1) sinh(x) at x = 0 is given by
a) 43
b) 7
c) 0
d) 34
Answer: a
Explanation: Expanding sinh(x) into a Taylor series we have
sinh(x)=(frac{x}{1!}+frac{x^3}{3!}+frac{x^5}{5!}…infty)
Now rewriting the function we have
f(x)=(x3+x2+x+1)((frac{x}{1!}+frac{x^3}{3!}+frac{x^5}{5!}…infty))
For the 7th derivative observe that, only the odd termed powers contribute to the derivative at x=0
Hence it is enough for us to find seventh derivative for
(x2+1)((frac{x}{1!}+frac{x^3}{3!}+frac{x^5}{5!}…infty))
(frac{x}{1!}+x^3(frac{1}{3!}+frac{1}{1!})+x^5(frac{1}{5!}+frac{1}{3!})+…infty)
Taking the 7th derivative of this function we have
f(7)(x)=((7!)(frac{1}{7!}+frac{1}{5!}) + (9*8…4*3)x^2(frac{1}{7!}+frac{1}{9!}))
Now substituting x=0 yields
f(7)(0)=((7!)(frac{1}{5!}+frac{1}{7!}))=(1+7*6)=43.

8. The (1071729)th derivative of f(x) = (x6 + x4 + x2) cosh(x) at x = 0 is given by
a) 0
b) 1071
c) 1729
d) ∞
Answer: a
Explanation: Expanding cosh(x) into a Taylor series we have
cosh(x)=(frac{1}{1!}+frac{x^2}{2!}+frac{x^4}{4!}…infty)
Observe again, that the derivative in question is odd, and hence, only the odd powered terms contribute to the derivative at x = 0
Also note that, there are no odd powered terms and hence we can conclude that
The (1071729)th derivative must be 0.

9. The (17291728)th derivative of f(x) = (x2 + 1)tan-1 (x) at x = 0 is
a) 0
b) 1729
c) 1728
d) ∞
Answer: a
Explanation: Expanding the tan-1 (x) function into Taylor series we have
tan-1(x)=(frac{x}{1}-frac{x^3}{3}+frac{x^5}{5}-…infty)
Rewrite the function as
f(x)=(x2+1)((frac{x}{1}-frac{x^3}{3}+frac{x^5}{5}-…infty))
Now observe that the derivative in question is even. Hence, even terms are the only ones to contribute to the derivative at x = 0
Also note that there are no even powered terms in the function. One can conclude that the (17291728)th derivative at x = 0 is 0.

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250+ TOP MCQs on Polar Curves and Answers

Differential Calculus Multiple Choice Questions & Answers focuses on “Polar Curves”.

1. Polar equations of the circle for the given coordinate (x,y) which satisfies the equation given by (x-a)2+(y-b)2=r2 where (a,b) is the coordinates of the centre of the circle &r is the radius.
a) x = r cos⁡θ, y = r sin⁡θ
b) x = a+ r cos⁡θ, y = b + r sin⁡θ
c) y = a+r cos⁡θ, x = b + r sin⁡θ
d) x = r sin⁡θ, y = r cos⁡θ
View Answer

Answer: b
Explanation: option x = a+ r cos⁡θ, y = b + r sin⁡θ satisfies the equation (x-a)2+(y-b)2=r2
because LHS=(a+r cos⁡θ-a)2 + (b+ r sin⁡θ-b)2 = r2(cos2 θ + sin2 θ)= r2=RHS.

2. In an polar curve r=f (θ) what is the relation between θ & the coordinates (x,y)?
a) tan⁡θ = (frac{x}{y} )
b) (1+sin⁡θ) = (frac{y}{x} )
c) (1+sec2 θ) = (frac{y^2}{x^2} )
d) (1+cos⁡θ) = (frac{x}{y} )
View Answer

Answer: c
Explanation: w.k.t for the polar curve r=f (θ) x=rcos⁡θ, y=rsin⁡θ
dividing them we get (frac{y}{x} = frac{r sin⁡θ}{r cos⁡θ} = tan⁡θ)
squaring on both side (frac{y^2}{x^2} = tan^2 θ = (1+sec^2 θ)).

3. The angle between Radius vector r=a(1-cos⁡θ)and tangent to the curve is ∅ given by _______
a) ∅=(frac{π}{2})
b) ∅=π
c) ∅=(-frac{π}{2})
d) ∅=0
View Answer

Answer: a
Explanation: r= a(1-cos⁡θ)
taking logarithms on both sides we get,
log⁡r = log⁡a + log⁡(1-cos⁡θ)
differentiating w.r.t θ we get,
( frac{1}{r} frac{dr}{dθ} = 0 + frac{sin⁡θ}{1-cos⁡θ})
(frac{1}{r} frac{dr}{dθ} = frac{2 sin ⁡frac{θ}{2} cos⁡frac{θ}{2}}{2sin^2 frac{θ}{2}} = cot⁡frac{θ}{2}) ..(1),
but (cot⁡∅ = frac{1}{r} frac{dr}{dθ})….(2)
From (1)&(2)
∅=(frac{π}{2}).

4. Angle of intersection between two polar curves given by r=a(1+sin⁡θ) & r=a(1-sin⁡θ) is given by ________
a) (frac{π}{4})
b) (frac{π}{2})
c) Π
d) 0
View Answer

Answer: b
Explanation: r=a(1+sin⁡θ) : r=a(1-sin⁡θ)
taking logarithm on both the equations
log⁡r = log⁡a + log⁡(1+sin⁡θ) : log⁡r = log⁡a + log⁡(1-sin⁡θ)
differentiating on both side we get
( frac{1}{r} frac{dr}{dθ} = frac{cos⁡θ}{1+sin⁡θ} : frac{1}{r} frac{dr}{dθ} = frac{-cos⁡θ}{1-sin⁡θ})
(cot⁡∅1 = frac{cos⁡θ}{1+sin⁡θ} : cot⁡∅2 = frac{-cos⁡θ}{1-sin⁡θ})
where ∅1&∅2 are the angle between tangent & the vector respectively
(tan⁡∅1 = frac{1+sin⁡θ}{cos⁡θ} : tan⁡∅2 = frac{1-sin⁡θ}{cos⁡θ})
(tan⁡∅1 . tan⁡∅2 = frac{1+sin⁡θ}{cos⁡θ} . frac{1-sin⁡θ}{-cos⁡θ} = frac{1-sin^2 θ}{-cos^2 θ} = -frac{cos^2 θ}{cos^2 θ} = -1)
above is the condition of orthogonality of two polar curves thus
|∅1-∅2|=(frac{π}{2}).

5. One among the following is the correct explanation of pedal equation of an polar curve, r=f (θ), p=r sin(∅) (where p is the length of the perpendicular from the pole to the tangent & ∅ is the angle made by tangent to the curve with vector drawn to curve from pole)is _______
a) It is expressed in terms of p & θ only
b) It is expressed in terms of p & ∅ only
c) It is expressed in terms of r & θ only
d) It is expressed in terms of p& r only
View Answer

Answer: d
Explanation: It is expressed in terms of p& r only
where p=r sin(∅) & (tan⁡∅ = frac{r}{frac{dr}{dθ}} = r (frac{dr}{dθ}))
& r=f (θ) or after solving we get direct relationship between p & r as
(frac{1}{p^2} = frac{1}{r^2} cosec^2∅.)

6. The pedal Equation of the polar curve rn=an cos⁡nθ is given by ______
a) rn=pan
b) rn-1=pan
c) rn+1=pan+1
d) rn+1=pan
View Answer

Answer: d
Explanation: Taking logarithm for the given curve we get
n log⁡r = n log⁡a + log⁡(cos⁡nθ)
differentiating w.r.t θ, we get
(frac{n}{r} frac{dr}{dθ} = frac{-n sin⁡nθ}{cos⁡nθ} rightarrow frac{1}{r} frac{dr}{dθ} = -tan⁡θ)
thus (cot⁡∅ = cot⁡(frac{π}{2} + nθ)rightarrow ∅ = frac{π}{2} + nθ……(1))
from the eqn w.k.t p=r sin ∅
substituting from (1)
p = r sin ((frac{π}{2}) + nθ) = r cos (nθ), but we have rn = an cos⁡nθ
hence dividing them we get (frac{p}{r^n} = frac{r cos (nθ)}{a^n cos⁡nθ})
rn+1=pan.

7. The length of the perpendicular from the pole to the tangent at the point θ=(frac{π}{2}) on the curve. r=a sec2((frac{π}{2})) is _____
a) (p = frac{2a}{sqrt{3}})
b) (p = frac{4a}{sqrt{3}})
c) (p = 2asqrt{2})
d) p = 4a
View Answer

Answer: c
Explanation: Taking Logarithm on both side of the polar curve
we get log⁡r = log⁡a + 2 log⁡sec⁡((frac{θ}{2}))
differentiating w.r.t θ we get
(frac{1}{r} frac{dr}{dθ} = frac{2 sec⁡(frac{θ}{2}).tan⁡(frac{θ}{2}) }{2 sec⁡(frac{θ}{2})} = tan⁡(frac{θ}{2}))
(cot⁡∅ = cot⁡(frac{π}{2} – frac{θ}{2}) rightarrow ∅ = frac{π}{2} -frac{θ}{2})
w.k.t length of the perpendicular is given by p=r sin ∅
thus substituting ∅ value we get (p = r ,sin(frac{π}{2} – frac{θ}{2}) = r ,cos(frac{θ}{2}))
at ( θ=frac{π}{4}, p = r ,cos frac{π}{4} = frac{r}{sqrt{2}}….(1))
but ( r=a ,sec^2 (frac{θ}{2}) ,at, θ = frac{π}{4}, r=a ,sec^2 (frac{π}{4}) = 4a…(2))
from (1) & (2) (p=frac{4a}{sqrt{2}} = 2asqrt{2}).

250+ TOP MCQs on Euler’s Theorem and Answers

Engineering Mathematics Question Bank focuses on “Euler’s Theorem – 2”.

1. In euler theorem x ∂z⁄∂x + y ∂z⁄∂y = nz, here ‘n’ indicates?
a) order of z
b) degree of z
c) neither order nor degree
d) constant of z
Answer: a
Explanation: Statement of euler theorem is “if z is an homogeneous function of x and y of order ‘n’ then x ∂z⁄∂x + y ∂z⁄∂y = nz”.

2. If z = xn f(y⁄x) then?
a) y ∂z⁄∂x + x ∂z⁄∂y = nz
b) 1/y ∂z⁄∂x + 1/x ∂z⁄∂y = nz
c) x ∂z⁄∂x + y ∂z⁄∂y = nz
d) 1/x ∂z⁄∂x + 1/y ∂z⁄∂y = nz
Answer: c
Explanation: Since the given function is homogeneous of order n, hence by euler’s theorem
x ∂z⁄∂x + y ∂z⁄∂y = nz.

3. Necessary condition of euler’s theorem is?
a) z should be homogeneous and of order n
b) z should not be homogeneous but of order n
c) z should be implicit
d) z should be the function of x and y only
Answer: a
Explanation:
Answer ‘z should be homogeneous and of order n’ is correct as statement of euler’s theorem is “if z is an homogeneous function of x and y of order ‘n’ then x ∂z⁄∂x + y ∂z⁄∂y = nz”
Answer ‘z should not be homogeneous but of order n’ is incorrect as z should be homogeneous.
Answer ‘z should be implicit’ is incorrect as z should not be implicit.
Answer ‘z should be the function of x and y only’ is incorrect as z should be the homogeneous function of x and y not non-homogeneous functions.

4. If (z=e^{frac{x^2+y^2}{x+y}}) then, (x frac{∂z}{∂x} + y frac{∂z}{∂y}) is?
a) 0
b) zln(z)
c) z2 ln⁡(z)
d) z
Answer: b
Explanation:
Given (z=e^{frac{x^2+y^2}{x+y}}),let u=ln⁡(z)=(frac{x^2+y^2}{x+y}=frac{x(1+(frac{y}{x})^2)}{(1+frac{y}{x})}) = x f(y/x)
Hence u is homogeneous of order 1,
Hence,
(x frac{∂u}{∂x}+y frac{∂u}{∂y})=u
Putting, u = ln(z) we get,
(x frac{∂z}{∂x}+y frac{∂z}{∂y}) = zln(z)

5. If (z=sin^{-1}frac{x^3+y^3+z^3}{x+y+z}) then, (xfrac{∂z}{∂x}+yfrac{∂z}{∂y}).
a) 2 tan(z)
b) 2 cot(z)
c) tan(z)
d) cot(z)
Answer: a
Explanation:
Given (z=sin^{-1}⁡frac{x^3+y^3+z^3}{x+y+z}), put u=sin⁡(z)=(frac{x^3+y^3+z^3}{x+y+z}=x^2 f(frac{y}{x},frac{z}{x}))
Hence, (x frac{∂u}{∂x}+y frac{∂u}{∂y}=2u)
Putting u = sin(z), we get
(x frac{∂u}{∂x}+y frac{∂u}{∂y}=frac{2Sin(z)}{Cos(z)}=2Tan(z))

6. Value of (x frac{∂u}{∂x}+y frac{∂u}{∂y}) if (u=frac{Sin^{-1} (frac{y}{x})(sqrt{x}+sqrt{y})}{x^3+y^3}) is?
a) -2.5 u
b) -1.5 u
c) 0
d) -0.5 u
Answer: a
Explanation: Since the function can be written as,
u=(x^{frac{-5}{2}} frac{Sin^{-1} (frac{y}{x})(1+sqrt{frac{y}{x}})}{1+(frac{y}{x})^3}=x^n f(frac{y}{x})), by euler’s theorem,
(x frac{∂u}{∂x}+y frac{∂u}{∂y} = -frac{5}{2} u)

7. If f1(x,y) and f2(x,y) are homogeneous and of order ‘n’then the function f3(x,y) = f1(x,y) + f2(x,y) satisfies euler’s theorem.
a) True
b) False
Answer: a
Explanation: Since f1(x,y) and f2(x,y) are homogeneous and of order n hence,
(x frac{∂f_1}{∂x}+y frac{∂f_1}{∂y} = nf_1 (x,y))
(x frac{∂f_2}{∂x}+y frac{∂f_2}{∂y} = nf_2 (x,y))
Hence adding these two equations,
We get
(x frac{∂f_1+f_2}{∂x}+y frac{∂f_1+f_2}{∂y} = nf_2 (x,y)+nf_1 (x,y))
(x frac{∂f_3}{∂x}+y frac{∂f_3}{∂y} = nf_3 (x,y))
Hence f3 satisfies euler’s theorem.

8. If (z=ln⁡(frac{x^2+y^2}{x+y})-e^{frac{x^2+y^2}{x+y}}) then find (x frac{∂z}{∂x}+y frac{∂z}{∂y}).
a) (x frac{∂z}{∂x}+y frac{∂z}{∂y}=frac{x^2+y^2}{x+y} e^{frac{x^2+y^2}{x+y}})
b) (x frac{∂z}{∂x}+y frac{∂z}{∂y}=1-frac{x^2+y^2}{x+y} e^{frac{x^2+y^2}{x+y}})
c) (x frac{∂z}{∂x}+y frac{∂z}{∂y}=1+frac{x^2+y^2}{x+y} e^{frac{x^2+y^2}{x+y}})
d) (x frac{∂z}{∂x}+y frac{∂z}{∂y}=-frac{x^2+y^2}{x+y} e^{frac{x^2+y^2}{x+y}})
Answer: b
Explanation:
Given (z=ln⁡(frac{x^2+y^2}{x+y})-e^frac{x^2+y^2}{x+y})
Let, (u = ln⁡(frac{x^2+y^2}{x+y})) and (v=e^(frac{x^2+y^2}{x+y})) hence z=u-v
Now, let (u’ = e^u = frac{x^2+y^2}{x+y}=xf(frac{y}{x})) hence u’ satisfies euler’s theorem,
Hence,
(x frac{∂u’}{∂x}+y frac{∂u’}{∂y}=u’)
Hence, by putting u’=eu, we get
(x frac{∂u}{∂x}+y frac{∂u}{∂y}=frac{e^u}{e^u} = 1), ……(1)
Now, let v’ = ln(v)= (frac{x^2+y^2}{x+y}=xf(frac{y}{x})) hence v’ satisfies euler’s theorem,
Hence,
(x frac{∂v’}{∂x}+y frac{∂v’}{∂y}=v’)
Hence, by putting v’=ln(v),we get
(x frac{∂v}{∂x}+y frac{∂v}{∂y}=vln(v)), …… (2)
By subtracting eq(1) and eq(2), we get
(x frac{∂z}{∂x}-y frac{∂z}{∂y}=1-frac{x^2+y^2}{x+y} e^{frac{x^2+y^2}{x+y}})

9. If z = Sin-1 (x⁄y) + Tan-1 (y⁄x) then x ∂z⁄∂x + y ∂z⁄∂y is?
a) 0
b) y
c) 1 + x⁄y Sin-1 (x⁄y)
d) 1 + y⁄x Tan-1 (y⁄x)
Answer: a
Explanation: Given z = Sin-1 (x⁄y) + Tan-1 (y⁄x)
Let, u = Sin-1 (x⁄y) and v = Tan-1 (y⁄x) hence z = u + v
Now, let u’ = Sin(u) = x⁄y = f(x⁄y) hence u’ satisfies euler’s theorem,
Hence,
(x frac{∂u’}{∂x}+y frac{∂u’}{∂y}=0)
Hence, by putting u’=eu, we get
(x frac{∂u}{∂x}+y frac{∂u}{∂y}=0/e^u = 0) ,……(1)
Now, let v’= Tan(v)=y/x=f(y/x) hence v’ satisfies euler’s theorem,
Hence,
(x frac{∂v’}{∂x}+y frac{∂v’}{∂y}=0)
Hence, by putting v’=ln(v), we get
(x frac{∂v}{∂x}+y frac{∂v}{∂y}=0 ),……(2)
By adding eq(1) and eq(2), we get
(x frac{∂z}{∂x}+y frac{∂z}{∂y}=1+frac{x^2+y^2}{x+y} e^{frac{x^2+y^2}{x+y}})

10. If f(x,y)is a function satisfying euler’ s theorem then?
a) (x^2 frac{∂^2 f}{∂x^2}+2xy frac{∂^2 f}{∂x∂y}+y^2 frac{∂^2 f}{∂y^2}=n(n-1)f)
b) (frac{1}{x}^2 frac{∂^2 f}{∂x^2}+2/xy frac{∂^2 f}{∂x∂y}+frac{1}{y}^2 frac{∂^2 f}{∂y^2}=n(n-1)f)
c) (x^2 frac{∂^2 f}{∂x^2}+2xy frac{∂^2 f}{∂x∂y}+y^2 frac{∂^2 f}{∂y^2}=nf)
d) (y^2 frac{∂^2 f}{∂x^2}+2xy frac{∂^2 f}{∂x∂y}+x^2 frac{∂^2 f}{∂y^2}=n(n-1)f)
Answer: a
Explanation: Since f satisfies euler’s theorem,
(x frac{∂z}{∂x}+y frac{∂z}{∂y}=nz)
Differentiating it w.r.t x and y respectively we get,
(x frac{∂^2 u}{∂x^2}+frac{∂u}{∂x}+y frac{∂^2 u}{∂x∂y}=n frac{∂u}{∂x}),
and
(x frac{∂^2 u}{∂y}∂x+frac{∂u}{∂y}+y frac{∂^2 u}{∂y^2}=n frac{∂u}{∂y})
Multiplying with x and y respectively,
(x^2 frac{∂^2 u}{∂x^2}+x frac{∂u}{∂x}+xy frac{∂^2 u}{∂x∂y}=nx frac{∂u}{∂x}),
and
(xy frac{∂^2 u}{∂y}∂x+y frac{∂u}{∂y}+y^2 frac{∂^2 u}{∂y^2}=ny frac{∂u}{∂y})
Adding above equations we get
(x^2 frac{∂^2 u}{∂x^2}+y^2 frac{∂^2 u}{∂y}+2xy frac{∂^2 u}{∂x∂y}=n(n-1)u)

11. If (u = Tan^{-1} (frac{x^3+y^3}{x+y})) then, (x^2 frac{∂^2 u}{∂x^2}+y^2 frac{∂^2 u}{∂y}+2xy frac{∂^2 u}{∂x∂y}) is?
a) Sin(4u) – Cos(2u)
b) Sin(4u) – Sin(2u)
c) Cos(4u) – Sin(2u)
d) Cos(4u) – Cos(2u)
Answer: b
Explanation:
Let, v = Tan(u) = x2 f(y/x)
By euler’s theorem,
g(u) = (x frac{∂u}{∂x}+y frac{∂u}{∂y} = 2frac{Tan(u)}{Sec^2 (u)} = Sin(2u))
Hence,
(x^2 frac{∂^2 u}{∂x^2}+y^2 frac{∂^2 u}{∂y}+2xy frac{∂^2 u}{∂x∂y} = g(u)[g’(u)-1])
(x^2 frac{∂^2 u}{∂x^2}+y^2 frac{∂^2 u}{∂y}+2xy frac{∂^2 u}{∂x∂y}) = Sin(2u)[2Cos(2u)-1] = Sin(4u)-Sin(2u)

12. If (u = e^{frac{(x^2+y^2)}{x+y}}) Then, (x^2 frac{∂^2 u}{∂x^2}+y^2 frac{∂^2 u}{∂y}+2xy frac{∂^2 u}{∂x∂y})=?
a) u ln⁡(u)
b) u ln⁡(u)2
c) u [1+ln⁡(u)]
d) 0
Answer: b
Explanation: Let, v = ln(u) = (frac{x^2+y^2}{x+y} = x f(frac{y}{x}))
Hence by applying euler theorem,
(x frac{∂v}{∂x}+y frac{∂v}{∂y}=v)
Hence,
g(u) = (x frac{∂u}{∂x}+y frac{∂u}{∂y}=u ln⁡(u))
Hence,
(x^2 frac{∂^2 u}{∂x^2}+y^2 frac{∂^2 u}{∂y}+2xy frac{∂^2 u}{∂x∂y}) = g(u)[g’(u)-1]
(x^2 frac{∂^2 u}{∂x^2}+y^2 frac{∂^2 u}{∂y}+2xy frac{∂^2 u}{∂x∂y}) = u ln(u)[1+ln⁡(u)-1] = u ln⁡(u)2

To practice Engineering Mathematics Question Bank,

250+ TOP MCQs on Improper Integrals and Answers

Engineering Mathematics Multiple Choice Questions on “Improper Integrals – 2”.

1. Find the value of ∫tan-1⁡(x)dx.
a) sec-1 (x) – 1⁄2 ln⁡(1 + x2)
b) xtan-1 (x) – 1⁄2 ln⁡(1 + x2)
c) xsec-1 (x) – 1⁄2 ln⁡(1 + x2)
d) tan-1 (x) – 1⁄2 ln⁡(1 + x2)
Answer: b
Explanation: Add constant automatically
Given, ∫tan-1⁡(x)dx
Putting, x = tan(y),
We get, dy = sec2(y)dy,
∫ysec2(y)dy
By integration by parts,
ytan(y) – log⁡(sec⁡(y)) = xtan-1 (x) – 1⁄2 ln⁡(1 + x2).

2. Integration of (Sin(x) + Cos(x))ex is?
a) ex Cos(x)
b) ex Sin(x)
c) ex Tan(x)
d) ex (Sin(x) + Cos(x))
Answer: b
Explanation: Add constant automatically
Let f(x) = ex Sin(x)
∫ex Sin(x)dx = ex Sin(x) – ∫ex Cos(x)dx
∫ex Sin(x)dx + ∫ex Cos(x)dx = ∫ex [Cos(x) + Sin(x)]dx = ex Sin(x).

3. Find the value of ∫x3 Sin(x)dx.
a) x3 Cos(x) + 3x2 Sin(x) + 6xCos(x) – 6Sin(x)
b) – x3 Cos(x) + 3x2 Sin(x) – 6Sin(x)
c) – x3 Cos(x) – 3x2 Sin(x) + 6xCos(x) – 6Sin(x)
d) – x3 Cos(x) + 3x2 Sin(x) + 6xCos(x) – 6Sin(x)
Answer: d
Explanation: Add constant automatically
Let f(x) = x3 Sin(x)
∫x3 Sin(x)dx = – x3 Cos(x) + 3∫x2 Cos(x)dx
∫x2 Cos(x)dx = x2 Sin(x) – 2∫xSin(x)dx
∫xSin(x)dx = – xCos(x) + ∫Cos(x)dx = – xCos(x) + Sin(x)
=> ∫x3 Sin(x)dx = – x3 Cos(x) + 3[x2 Sin(x) – 2[ – xCos(x) + Sin(x)]]
=> ∫x3 Sin(x)dx = – x3 Cos(x) + 3x2 Sin(x) + 6xCos(x) – 6Sin(x).

4. Value of ∫uv dx,where u and v are function of x.
a) (sum_{i=1}^n(-1)^i u_i v^{i+1})
b) (sum_{i=0}^nu_i v^{i+1})
c) (sum_{i=0}^n(-1)^i u_i v^{i+1})
d) (sum_{i=0}^n(-1)^i u_i v^{n-i})
Answer: c
Explanation: Add constant automatically
Given, f(x)=(int uvdx=sum_{i=0}^n (-1)^i u_i v^{i+1})

5. Find the value of ∫x7 Cos(x) dx.
a) x7 Sin(x) + 7x6 Cos(x) + 42x5 Sin(x) + 210x4 Cos(x) + 840x3 Sin(x) + 2520x2 Cos(x) + 5040xSin(x) + 5040Cos(x)
b) x7 Sin(x) – 7x6 Cos(x) + 42x5 Sin(x) – 210x4 Cos(x) + 840x3 Sin(x) – 2520x2 Cos(x) + 5040xSin(x) – 5040Cos(x)
c) x7 Sin(x) + 7x6 Cos(x) + 42x5 Sin(x) + 210x4 Cos(x) + 840x3 Sin(x) + 2520x2 Cos(x) + 5040xSin(x) + 5040Cos(x)
d) x7 Sin(x) + 7x6 Cos(x) + 42x5 Sin(x) + 210x4 Cos(x) + 840x3 Sin(x) + 2520x2 Cos(x) + 5040xSin(x) + 10080Cos(x)
Answer: a
Explanation: Add constant automatically
By, f(x)=(int uvdx=sum_{i=0}^n (-1)^i u_i v^{i+1})
Let, u = x7 and v = Cos(x),
∫x7 Cos(x) dx = x7 Sin(x) + 7x6 Cos(x) + 42x5 Sin(x) + 210x4 Cos(x) + 840x3 Sin(x) + 2520x2 Cos(x) + 5040xSin(x) + 5040Cos(x)

6. Find the value of ∫x3 ex e2x e3x….enx dx.
a) (frac{2}{n(n+1)} e^{frac{n(n+1)}{2}x} left [x^3+3x^2 [frac{2}{n(n+1)}]^1+6x[frac{2}{n(n+1)}]^2 +6[frac{2}{n(n+1)}]^3right ])
b) (frac{2}{n(n+1)} e^{frac{n(n+1)}{2}x} left [x^3+3x^2 [frac{2}{n(n+1)}]^1+6x[frac{2}{n(n+1)}]^2 +6[frac{2}{n(n+1)}]^3right ])
c)(frac{2}{n(n+1)} e^{frac{n(n+1)}{2}x} left [x^3+3x^2 [frac{2}{n(n+1)}]^1+6x[frac{2}{n(n+1)}]^2 +6[frac{2}{n(n+1)}]^3right ])
d)(frac{2}{n(n+1)} e^{frac{n(n+1)}{2}x} left [x^3+3x^2 [frac{2}{n(n+1)}]^1+6x[frac{2}{n(n+1)}]^2 +6[frac{2}{n(n+1)}]^3right ])
Answer: a
Explanation: Add constant automatically
By, f(x)=(int uvdx=sum_{i=0}^n (-1)^i u_i v^{i+1})
Let, u = x3 and v=ex e2x e3x…..enx=ex(1+2+3+…n)=(e^{frac{n(n+1)x}{2}}),
(int x^3 e^x e^2x e^3x……..e^nx dx)
(=x^3 frac{2}{n(n+1)} e^{frac{n(n+1)}{2}x}+3x^2 [frac{2}{n(n+1)}]^2 e^{frac{n(n+1)}{2}x})
(+6x[frac{2}{n(n+1)}]^3 e^{frac{n(n+1)}{2}x}+6[frac{2}{n(n+1)}]^4 e^{frac{n(n+1)}{2}x})
=(frac{2}{n(n+1)} e^{frac{n(n+1)}{2}x} left [x^3+3x^2 [frac{2}{n(n+1)}]^1+6x[frac{2}{n(n+1)}]^2+6[frac{2}{n(n+1)}]^3right])

7. Find the area of a function f(x) = x2 + xCos(x) from x = 0 to a, where, a>0.
a) a2⁄2 + aSin(a) + Cos(a) – 1
b) a3⁄3 + aSin(a) + Cos(a)
c) a3⁄3 + aSin(a) + Cos(a) – 1
d) a3⁄3 + Cos(a) + Sin(a) – 1
Answer: c
Explanation: Given, f(x) = x2 + xCos(x)
Hence, F(x) = ∫x2 + xCos(x) dx = x3⁄3 + xSin(x) + Cos(x)
Hence, area inside f(x) is,
F(a) – F(0) = a3⁄3 + aSin(a) + Cos(a) – 1.

8. Find the area ln(x)⁄x from x = x = aeb to a.
a) b2⁄2
b) b⁄2
c) b
d) 1
Answer: a
Explanation:
Let, F(x)=(int frac{ln⁡(x)}{x} dx)
Let, z=ln⁡(x)=>dz=dx/x
=F(x)=∫ zdz=(frac{z^2}{2}=frac{ln^2⁡(x)}{2})
Area inside curve from 4a to a is,
(F(ae^b)-F(a)=frac{ln^2⁡(ae^b )}{2}-frac{ln^2⁡(a)}{2}=frac{ln^2⁡(frac{ae^b}{a})}{2}=frac{ln^2⁡(e^b)}{2}=frac{b}{2})

9. Find the area inside a function f(t) = ( frac{t}{(t+3)(t+2)} dt) from t = -1 to 0.
a) 4 ln⁡(3) – 5ln⁡(2)
b) 3 ln⁡(3)
c)3 ln⁡(3) – 4ln⁡(2)
d) 3 ln⁡(3) – 5 ln⁡(2)
Answer: d
Explanation:
Now, F(t)=(int frac{t}{(t+3)(t+2)} dt)
F(t)=(int frac{t}{(t+3)(t+2)} dt)
=(int [frac{3}{t+3}-frac{2}{t+2}]dx)
=(int [frac{3}{t+3}]dx-int [frac{2}{t+2}]dx)
=3 ln⁡(t+3)-2ln⁡(t+2)
Now area inside a function is, F(0) – F(-1),
hence, F(0)-F(-1)=3 ln⁡(3)-2 ln⁡(2)-3 ln⁡(2)+2 ln⁡(1)=3 ln⁡(3)-5ln⁡(2)

10. Find the area inside integral f(x)=(frac{sec^4⁡(x)}{sqrt{tan⁡(x)}}) from x = 0 to π.
a) π
b) 0
c) 1
d) 2
Answer: b
Explanation:
Given,F(x)=(int frac{sec^4⁡ (x)}{sqrt{tan⁡(x)}} dx)
F(x)=(int frac{sec^2⁡ (x) sec^2⁡ (x)}{sqrt{tan⁡(x)}} dx)
=(int frac{1+t^2}{sqrt{t}} dt)
=(int [frac{1}{sqrt{t}}+t^{3/2}]dt)
=(2sqrt{t}+frac{2}{5} t^{5/2})
F(x)=(frac{2}{5} sqrt{tan⁡(x)} [5+tan^2⁡(x)])
Now area inside a function f(x) from x=0 to π, is
F(π)-F(0)=0-0=0

11. Find the area inside function (frac{(2x^3+5x^2-4)}{x^2}) from x = 1 to a.
a) a2⁄2 + 5a – 4ln(a)
b) a2⁄2 + 5a – 4ln(a) – 11⁄2
c) a2⁄2 + 4ln(a) – 11⁄2
d) a2⁄2 + 5a – 11⁄2
Answer: b
Explanation: Add constant automatically
Given,
f(x) = (frac{(2x^3+5x^2-4)}{x^2}),
Integrating it we get, F(x) = x2⁄2 + 5x – 4ln⁡(x)
Hence, area under, x = 1 to a, is
F(a) – F(1)=a2⁄2 + 5a – 4ln(a) – 1/2 – 5=a2⁄2 + 5a – 4ln(a) – 11⁄2

12. Find the value of ∫(x4 – 5x2 – 6x)4 4x3 – 10x – 6 dx.
a) (frac{(x^4-5x^2-6x)^4}{4})
b) (frac{(x^4-5x^2-6x)^5}{5})
c) (frac{(4x^3-10x-6)^5}{5})
d) (frac{(4x^3-10x-6)^4}{4})
Answer: b
Explanation: Add constant automatically
Given, (int (x^4-5x^2-6x)^4 4x^3-10x-6 dx)
putting, (x^4-5x^2-6x=z), we get, (dz=4x^3-10x-6 dx)
(int z^4 dz=frac{z^5}{5}=frac{(x^4-5x^2-6x)^5}{5})

13. Temperature of a rod is increased by moving x distance from origin and is given by equation T(x) = x2 + 2x, where x is the distance and T(x) is change of temperature w.r.t distance. If, at x = 0, temperature is 40 C, find temperature at x=10.
a) 473 C
b) 472 C
c) 474 C
d) 475 C
Answer: a
Explanation: Temperature at distance x is,
T = ∫T(x) dx = ∫x2 + 2x dx = x3⁄3 + x2 + C
At x=0 given T = 40 C
C = T(x = 0) = 40 C
At x= 10,
T(x = 10) = 1000⁄3 + 100 + 43 = 473 C.

14. Find the value of (int frac{1}{16x^2+16x+10}dx).
a) 1⁄8 sin-1(x + 1⁄2)
b) 1⁄8 tan-1(x + 1⁄2)
c) 1⁄8 sec-1(x + 1⁄2)
d) 1⁄4 cos-1(x + 1⁄2)
Answer: b
Explanation: Add constant automatically
Given, (int frac{1}{16x^2+16x+10}dx=frac{1}{2}int frac{1}{4x^2+4x+5}dx)
=(int frac{1}{8(x^2+x+frac{5}{4}+frac{1}{4}+frac{1}{4})}dx=int frac{1}{8[(x+frac{1}{2})^2+1^2]}dx=frac{1}{8}tan^{-1}(x+frac{1}{2}))

250+ TOP MCQs on Formation of Ordinary Differential Equations by Elimination of Arbitrary Constants and Answers

Ordinary Differential Equations Questions and Answers for Entrance exams focuses on “Formation of Ordinary Differential Equations by Elimination of Arbitrary Constants”.

1. What is the slope of the equation, y= x2+8?
a) 2x
b) 0
c) 8
d) x
Answer: a
Explanation: The slope of the given equation, y= x2+8, is given by,
Slope= (frac{dy}{dx}=2x )

2. Which of the following is true with respect to formation of differential equation by elimination of arbitrary constants?
a) The given equation should be differentiated with respect to independent variable
b) Elimination of the arbitrary constant by replacing it using derivative
c) If ‘n’ arbitrary constant is present, the given equation should be differentiated ‘n’ number of times
d) To eliminate the arbitrary constants, the given equation must be integrated with respect to the dependent variable
Answer: d
Explanation: Consider a general equation, f(x,y,c)=0 ……………………………………… (1)
To form a differential equation by elimination of arbitrary constant, the following steps need to be followed:

  • Differentiate (1) with respect to x
  • In case of ‘n’ arbitrary constants, the equation should be differentiated ‘n’ number of times
  • Eliminate the arbitrary constant using (1) and the derivatives

3. In the formation of differential equation by elimination of arbitrary constants, after differentiating the equation with respect to independent variable, the arbitrary constant gets eliminated.
a) False
b) True
Answer: a
Explanation: In the formation of differential equation by elimination of arbitrary constants, the first step is to differentiate the equation with respect to the dependent variable. Sometimes, the arbitrary constant gets eliminated after differentiation.

4. What is the differential equation of a family of parabolas with the foci at the origin and axis along the X-axis?
a) 2xy’+ 4y(y’)2-y=0
b) xy’+ y(y’)2-y=0
c) 2xy’+ y(y’)2-y=0
d) 2xy’+2y(y’)2-y=0
Answer: c
Explanation: The equation is, y2=4ax+4a2……………………………………. (1)
Differentiating (1) with respect to x, we get,
2yy’=4a ………………………………………………………………………………………….. (2)
Therefore, substituting the value of 4a in (1), we get,
y2=2yy’x+(yy’)2
So, the required differential equation is given by,
2xy’+y(y’)2-y=0

5. What is the nature of the equation, (xy^3 (frac{dy}{dx})^2+yx^2+frac{dy}{dx}=0)?
a) Second order, third degree, linear differential equation
b) First order, third degree, non-linear differential equation
c) First order, third degree, linear differential equation
d) Second order, third degree, non-linear differential equation
Answer: b
Explanation: Since the equation has only first derivative, i.e. ((frac{dy}{dx}),) it is a first order equation.
Degree is defined as the highest power of the highest order derivative involved. Hence it is 2.
The equation has one/more terms having a variable of degree two/higher; hence it is non-linear.

6. Which of the following is a type of Iterative method of solving non-linear equations?
a) Graphical method
b) Interpolation method
c) Trial and Error methods
d) Direct Analytical methods
Answer: b
Explanation: There are 2 types of Iterative methods, (i) Interpolation methods (or Bracketing methods) and (ii) Extrapolation methods (or Open-end methods).

7. The half-interval method in numerical analysis is also known as __________
a) Newton-Raphson method
b) Regula Falsi method
c) Taylor’s method
d) Bisection method
Answer: d
Explanation: The Bisection method, also known as binary chopping or half-interval method, is a starting method which is used, where applicable, for few iterations, to obtain a good initial value.

8. What is the solution of the given equation?
x6y6 dy + (x7y5 +1) dx = 0
a) (frac{(xy)^6}{6} + lnx = c)
b) (frac{(xy)^5}{6} + lny = c)
c) (frac{(xy)^5}{5} + lnx = c)
d) (frac{(xy)^6}{6} + lny = c)
Answer: a
Explanation: Given: (x6y6 + 1) dy + x7y5dx = 0, is an example of non-exact differential equation.
Dividing the equation by x we get,
x5y6 dy + x6y5dx + (frac{dx}{x} = 0)
x5y5 (ydy + xdx) + (frac{dx}{x} = 0 )
(xy)5(d(xy)) + (frac{dx}{x} = 0)
(frac{(xy)^6}{6} + lnx = c)

9. A rectangular frame is to be made of 240 cm long. Determine the value of the length of the rectangle required to maximize the area.
a) 24 cm
b) 60 cm
c) 240 cm
d) 120 cm
Answer: b
Explanation: Let us consider ‘x’ as length and ‘y’ as the breadth of the rectangle.
Given: Perimeter 2(x + y) = 240 cm
x + y = 120
y = 120 – x
Area of the rectangle, a = x*y = x(120-x) = 120x – x2
Finding the derivative, we get, (d(a))/dx = (d(120x – x2))/dx=120-2x
To find the value of x that maximizes the area, we substitute (d(a))/dx = 0.
Therefore, we get, 120 – 2x =0
2x = 120
x = 60 cm
To check if x = 60 cm is the value that maximizes the area, we find the second derivative of the area,
(d2 (a))/(dx2)= -2 < 0 …………………. (i)
We know that the condition for maxima is (d2 (f(x)))/(dx2)<0, which is satisfied by (i), therefore, x = 60 cm maximizes the area of the rectangle.

10. In the equation, y = x2+c,c is known as the parameter and x and y are known as the main variables.
a) True
b) False
Answer: a
Explanation: Given: y = x2+c, where c is known as an arbitrary constant. It is also referred to as the parameter to differentiate it from the main variables x and y.

Global Education & Learning Series – Ordinary Differential Equations.

To practice all areas of Ordinary Differential Equations for Entrance exams,

250+ TOP MCQs on Laplace Transform of Periodic Function and Answers

Ordinary Differential Equations Multiple Choice Questions on “Laplace Transform of Periodic Function”.

1. Find the laplace transform of f(t), where
f(t) = 1 for 0 < t < a
-1 for a < t < 2a
a) (frac{1}{s} coth⁡(frac{as}{2}))
b) (frac{1}{s} sinh⁡(frac{as}{2}))
c) (frac{1}{s} e^{-as})
d) (frac{1}{s} tanh⁡(frac{as}{2}))
Answer: d
Explanation: In the given question
f(t) is a periodic function having a period 2a
The formula for Laplace Transform is given by:
(L(f(t))=frac{1}{1-e^{-2as}} int_{0}^{2a}e^{-st} f(t)dt)
(L(f(t))=frac{1}{1-e^{-2as}} int_{0}^{a}e^{-st} (1)dt + frac{1}{1-e^{-2as}} int_{a}^{2a}e^{-st}(-1)dt)
=(begin{bmatrix}frac{1}{1-e^(-2as)}×frac{e^{-as}}{-s} – frac{1}{1-e^{-2as}} × frac{-1}{s}end{bmatrix} – begin{bmatrix}frac{1}{1-e^{-2as}} × frac{e^{-2as}}{-s} – frac{1}{1-e^{-2as}} × frac{e^{-as}}{-s}end{bmatrix})
= (frac{1}{1-e^{-2as}}×frac{1}{s}×(1-e^{-as})^2)
= (frac{1}{s}(frac{1+e^{-as}}{1-e^{-as}}))
Dividing both numerator and denominator by (e^{frac{-as}{2}})
= (frac{1}{s} tanh⁡(frac{as}{2}))
Thus, the correct answer is (frac{1}{s} tanh⁡(frac{as}{2})).

2. Find the laplace transform of f(t), where f(t) = |sin(pt)| and t>0.
a) (frac{p}{s^2+p^2}×cosh⁡(frac{spi}{2p}))
b) (frac{p}{s^2+p^2}×sinh⁡(frac{spi}{2p}))
c) (frac{p}{s^2+p^2}×coth⁡⁡(frac{spi}{2p}))
d) (frac{p}{s^2+p^2}×tanh⁡⁡(frac{spi}{2p}))
Answer: c
Explanation: From this question, we know –
Period of sin(t)=2π
Period of sin⁡(pt)=(frac{2pi}{p})
Period of |sin⁡(pt)|=(frac{pi}{p})
(L(f(t))=frac{1}{1-e^{frac{-pi}{ps}}} int_{0}^{frac{pi}{p}}e^{-st} f(t)dt)
Since |sin⁡(pt)| is positive in all quadrants
(L(f(t))=frac{1}{1-e^{frac{-pi}{ps}}} int_{0}^{frac{pi}{p}}e^{-st} sin⁡(pt)dt)
=(frac{1}{1-e^{frac{-pi}{ps}}}begin{bmatrix}frac{e^{frac{-sπ}{p}}}{s^2+p^2}×pend{bmatrix}-begin{bmatrix}frac{1}{s^2+p^2}×(-p)end{bmatrix})
=(frac{1}{1-e^{frac{-pi}{ps}}}×frac{p}{s^2+p^2}×(1+e^{frac{-π}{ps}}))
=(frac{p}{s^2+p^2}×coth⁡(frac{spi}{2p})), (Multiplying and dividing by (e^{frac{-sπ}{2p}}))
Thus, the answer is (frac{p}{s^2+p^2}×coth⁡⁡(frac{spi}{2p})).

Global Education & Learning Series – Ordinary Differential Equations.

To practice all areas of Ordinary Differential Equations,