# Ordinary Differential Equations Questions and Answers – Special Functions – 2 (Beta)

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This set of Ordinary Differential Equations Question Paper focuses on “Special Functions -2 (Beta)”.

1. β(m, n) = β(n, m). Is the statement true?
a) True
b) False

Explanation: L.H.S. = $$\beta(m, n) = \frac{\Gamma(m).\Gamma(n)}{\Gamma(m+n)}.$$
R.H.S. $$= \beta(n, m) = \frac{\Gamma(n).\Gamma(m)}{\Gamma(n+m)} = \frac{\Gamma(m).\Gamma(n)}{\Gamma(m+n)} =$$ L.H.S.
Therefore, $$\beta(m, n) = \beta(n, m).$$

2. Which of the following function is not called the Euler’s integral of the first kind?
a) $$\beta(m, n) = \int_0^1 x^{m-1} (1-x)^{n-1} dx (m>0, n>0)$$
b) $$\beta(m, n) = \int_0^{π/2} (sin⁡θ)^{2m-1} (cos⁡θ)^{2n-1} dθ$$
c) $$\beta(m, n) = \int_0^∞ \frac{y^{n+1}}{(1+y)^{m+n}} dy$$
d) $$\beta(m, n) = 2 \int_0^{π/2} (sinθ)^{2m-1} (cos⁡θ)^{2n-1} dθ$$

Explanation: Euler’s integral of the first kind is nothing but Beta function. So, here only $$\beta(m, n) = \int_0^{π/2}(sin⁡θ)^{2m-1} (cos⁡θ)^{2n-1} dθ$$ is not the definition of Beta function.

3. Which of the following is not the definition of Beta function?
a) $$\beta(m, n) = 2\int_0^1 x^{m-1} (1-x)^{n-1} dx (m>0, n>0)$$
b) $$\beta(m, n) = 2\int_0^{π/2}(sin⁡θ)^{2m-1} (cos⁡θ)^{2n-1} dθ$$
c) $$\beta(m, n) = \int_0^∞ \frac{y^{n+1}}{(1+y)^{m+n}} dy$$
d) $$\beta(m, n) = \int_0^1 \frac{x^{m-1}+x^{n-1}}{(1+x)^{m+n}} dx$$

Explanation: $$\beta(m, n)$$ can be written as either $$\int_0^1 x^{m-1} (1-x)^{n-1} dx \, (m>0, n>0)$$ (or)
$$= 2 \int_0^{π/2}(sin⁡θ)^{2m-1} (cos⁡θ)^{2n-1} dθ$$
(or)
$$= \int_0^∞ \frac{y^{n+1}}{\left(1+y\right)^{m+n}} dy \;or \int_0^1 \frac{x^{m-1}+x^{n-1}}{\left(1+x\right)^{m+n}} dx$$.
So the correct answer is $$2\int_0^1 x^{m-1} (1-x)^{n-1} dx (m>0, n>0)$$ which is actually not the formula for Beta function.

4. What is the value of $$\beta(m, \frac{1}{2})$$?
a) β(m, m)
b) 22m-1 β(m, m)
c) 22m+1 β(m, m)
d) 22m β(m, m)

Explanation: $$\beta(m, \frac{1}{2}) = 2\int_0^{π/2}(sin⁡θ)^{2m-1} dθ$$
$$\beta(m, m) = 2 \int_0^{π/2}(sin⁡θ)^{2m-1} (cos⁡θ)^{2m-1} dθ$$
$$= 2^{-2m+2} \int_0^{π/2}(2 sinθ cos⁡θ)^{2m-1} dθ$$
Substituting 2θ=φ,
$$= 2^{-2m+1} \int_0^π sin⁡φdφ$$
$$= 2^{-2m+1}.2.\int_0^{π/2}sin⁡φdφ$$
$$= \frac{1}{2^{2m-1}} \beta(m, \frac{1}{2}).$$

5. What is the value of β(3,2)?
a) $$\frac{1}{14}$$
b) $$\frac{1}{16}$$
c) $$\frac{1}{12}$$
d) $$\frac{1}{10}$$

Explanation: $$\beta(3, 2) = \frac{\Gamma(3).\Gamma(2)}{\Gamma(3+2)}$$
$$= \frac{2!1!}{4!} = \frac{1}{12}.$$

6. What is the value of $$\beta(\frac{1}{4},\frac{3}{4})$$?
a) $$\pi$$
b) $$\sqrt{2}\pi$$
c) $$\sqrt{2\pi}$$
d) $${2}\pi$$

Explanation: $$\beta(\frac{1}{4}, \frac{3}{4}) = \frac{\Gamma(\frac{1}{4}).\Gamma(\frac{3}{4})}{\Gamma(\frac{1}{4}+\frac{3}{4})}$$
$$= \frac{\pi}{sin⁡ \frac{π}{4}} = \sqrt{2}\pi.$$

7. What is the value of $$\int_0^{π/2}\sqrt{sin⁡θdθ} + \int_0^{π/2}\sqrt{cos⁡θ⁡dθ}$$?
a) $$8\sqrt{\pi} \frac{\Gamma(\frac{3}{4})}{\Gamma(\frac{1}{4})}$$
b) $$4\sqrt{\pi} \frac{\Gamma(\frac{3}{4})}{\Gamma(\frac{1}{4})}$$
c) $$8\sqrt{\pi} \frac{\Gamma(\frac{1}{4})}{\Gamma(\frac{3}{4})}$$
d) $$4\sqrt{\pi} \frac{\Gamma(\frac{1}{4})}{\Gamma(\frac{3}{4})}$$

Explanation: $$\int_0^{π/2}\sqrt{sin⁡θdθ} + \int_0^{π/2}\sqrt{cos⁡θ⁡dθ}$$
$$= \beta(\frac{3}{4}, \frac{1}{2}) + \beta(\frac{3}{4}, \frac{1}{2})$$
$$= 2 \beta(\frac{3}{4}, \frac{1}{2})$$
$$= 2\frac{\Gamma(\frac{1}{2}).\Gamma(\frac{3}{4})}{\Gamma(\frac{1}{2}+\frac{3}{4})}$$
$$= 2\sqrt{\pi} \frac{\Gamma(\frac{3}{4})}{\Gamma(\frac{1}{4})} . \frac{1}{(\frac{1}{4})}$$
$$= 8\sqrt{\pi} \frac{\Gamma(\frac{3}{4})}{\Gamma(\frac{1}{4})}.$$

8. What is the value of $$\int_0^∞ \frac{1}{(1+y)^5} dy$$?
a) 12
b) 13
c) 14
d) 15

Explanation: $$\int_0^∞ \frac{y^{1-1}}{(1+y)^5} dy$$
n=1 and m + n = 5 which implies m=4.
$$\beta(4, 1) = \frac{\Gamma(4).\Gamma(1)}{\Gamma(4+1)}$$
$$\frac{3!0!}{4!} = \frac{1}{4}$$.

9. What is the value of $$\int_0^1 \frac{dx}{\sqrt{1+x^4}}$$?
a) $$\beta(\frac{1}{4}, \frac{1}{2})$$
b) $$\frac{1}{4\sqrt{2}}\beta(\frac{1}{4}, \frac{1}{2})$$
c) $$\frac{1}{3\sqrt{2}}\beta(\frac{1}{3}, \frac{1}{2})$$
d) $$\frac{1}{4\sqrt{3}}\beta(\frac{1}{4}, \frac{1}{3})$$

Explanation: Substitute x2 = tan⁡θ
Therefore θ varies from 0 to $$\frac{\pi}{4}.$$
$$\int_0^{π/4} \frac{(secθ)^2}{2 sec⁡θ \sqrt{tan⁡θ}} dθ$$
$$= \int_0^{π/4}\frac{dθ}{2\sqrt{sin⁡θ cos⁡θ}}$$
$$= \frac{1}{4\sqrt{2}} \beta(\frac{1}{4}, \frac{1}{2}).$$

10. What is the value of $$\int_0^1\frac{(y^5+y^2)}{(1+y)^9} dy$$?
a) $$\frac{1}{158}$$
b) $$\frac{2}{167}$$
c) $$\frac{1}{146}$$
d) $$\frac{1}{168}$$

Explanation: m-1 = 5 => m=6 and n-1 = 2 => n=3.
$$\beta(6, 3) = \frac{\Gamma(6).\Gamma(3)}{\Gamma(6+3)}$$
$$= \frac{5!2!}{8!} = \frac{1}{168}.$$

11. Solve using the Beta function. $$\int_0^1 x^{-2} (1-x)^{-3} dx.$$
a) Can be solved using a Beta function with m = -1 and n = -2
b) Can be solved using a Beta function with m = 1 and n = -2
c) Can be solved using a Beta function with m = -1 and n = 2
d) Can’t be solved using the Beta function

Explanation: This function can’t be solved using the Beta function as m and n have negative values. Beta function can’t be solved if m and n are negative numbers.

12. What is the value of $$\int_0^∞ (sech x)^5 dx$$?
a) $$\frac{3\pi}{80}$$
b) $$\frac{3\pi}{240}$$
c) $$\frac{3\pi}{16}$$
d) $$\frac{\pi}{240}$$

Explanation: $$\int_0^∞ (sech x)^5 dx = \frac{2^5}{4} \beta(\frac{5}{2}, \frac{5}{2})$$
$$\displaystyle = \frac{8 \Gamma(\frac{5}{2})\Gamma(\frac{5}{2})}{\Gamma(5)} = \frac{(8*\frac{3}{2}*\frac{1}{2}*\sqrt{\pi}*\frac{3}{2}*\frac{1}{2}*\sqrt{\pi})}{24} = \frac{9π}{48}=\frac{3π}{16}.$$

13. What is the value of $$\beta(\frac{9}{2},3)$$?
a) $$\frac{16}{1287}$$
b) $$\frac{16}{1278}$$
c) $$\frac{14}{1287}$$
d) $$\frac{16}{127}$$

Explanation: $$\beta(\frac{9}{2},3) = \frac{\Gamma(\frac{9}{2})\Gamma(3)}{\Gamma(\frac{9}{2}+3)}$$
$$= \frac{\Gamma(\frac{9}{2})2}{\frac{13}{2}*\frac{11}{2}*\frac{9}{2}*\Gamma(\frac{9}{2})} = \frac{16}{1287}.$$

14. What is the value of $$\int_0^1 x^5 (1-x)^6 dx$$?
a) $$\frac{1}{12*11*10*9*8*7}$$
b) $$\frac{1}{12*11*10*9*8}$$
c) $$\frac{1}{12*11*10*9*8*7*6}$$
d) $$\frac{1}{12*11*10*9*8*7*6*5}$$

Explanation: Here, m-1 = 5 which implies m=6 and n-1 = 6 which implies n=7.
$$\beta(6,7)= \frac{\Gamma(6)\Gamma(7)}{\Gamma(13)} = \frac{1}{12*11*10*9*8*7*6}.$$

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