← 2019 Paper 1
UPSC 2019 Maths Optional Paper 1 Q6c-ii — Step-by-Step Solution 10 marks · Section B
Laplace transform · ODEs · asked 2× in 14 yrs · Read the full method →
Question
Find the Laplace transforms of t − 1 / 2 t^{-1/2} t − 1/2 t 1 / 2 t^{1/2} t 1/2 t n + 1 2 t^{n+\frac12} t n + 2 1 n ∈ N n\in\mathbb N n ∈ N
Γ ( n + 1 + 1 2 ) s n + 1 + 1 2 . \frac{\Gamma\!\left(n+1+\dfrac12\right)}{s^{\,n+1+\frac12}}. s n + 1 + 2 1 Γ ( n + 1 + 2 1 ) . Technique
One master integral L { t p } = Γ ( p + 1 ) / s p + 1 \mathcal L\{t^p\}=\Gamma(p+1)/s^{p+1} L { t p } = Γ ( p + 1 ) / s p + 1 u = s t u=st u = s t p = − 1 2 , 1 2 , n + 1 2 p=-\tfrac12,\ \tfrac12,\ n+\tfrac12 p = − 2 1 , 2 1 , n + 2 1
Solution
For s > 0 s>0 s > 0 p > − 1 p>-1 p > − 1
L { t p } = ∫ 0 ∞ e − s t t p d t . \mathcal L\{t^p\}=\int_0^\infty e^{-st}t^p\,dt. L { t p } = ∫ 0 ∞ e − s t t p d t . Substitute u = s t ( t = u / s , d t = d u / s ) u=st\ (t=u/s,\ dt=du/s) u = s t ( t = u / s , d t = d u / s )
L { t p } = ∫ 0 ∞ e − u ( u s ) p d u s = 1 s p + 1 ∫ 0 ∞ e − u u p d u = Γ ( p + 1 ) s p + 1 . \mathcal L\{t^p\}=\int_0^\infty e^{-u}\Big(\frac us\Big)^p\frac{du}{s}=\frac{1}{s^{p+1}}\int_0^\infty e^{-u}u^p\,du=\frac{\Gamma(p+1)}{s^{p+1}}. L { t p } = ∫ 0 ∞ e − u ( s u ) p s d u = s p + 1 1 ∫ 0 ∞ e − u u p d u = s p + 1 Γ ( p + 1 ) . This is the key identity; everything follows by choosing p p p
Take p = − 1 2 p=-\tfrac12 p = − 2 1 p + 1 = 1 2 p+1=\tfrac12 p + 1 = 2 1 Γ ( 1 2 ) = π \Gamma(\tfrac12)=\sqrt\pi Γ ( 2 1 ) = π
L { t − 1 / 2 } = Γ ( 1 2 ) s 1 / 2 = π s . \boxed{\;\mathcal L\{t^{-1/2}\}=\frac{\Gamma(\tfrac12)}{s^{1/2}}=\sqrt{\frac{\pi}{s}}.\;} L { t − 1/2 } = s 1/2 Γ ( 2 1 ) = s π . Take p = 1 2 p=\tfrac12 p = 2 1 p + 1 = 3 2 p+1=\tfrac32 p + 1 = 2 3 Γ ( 3 2 ) = 1 2 Γ ( 1 2 ) = π 2 \Gamma(\tfrac32)=\tfrac12\Gamma(\tfrac12)=\tfrac{\sqrt\pi}{2} Γ ( 2 3 ) = 2 1 Γ ( 2 1 ) = 2 π
L { t 1 / 2 } = Γ ( 3 2 ) s 3 / 2 = π 2 s 3 / 2 . \boxed{\;\mathcal L\{t^{1/2}\}=\frac{\Gamma(\tfrac32)}{s^{3/2}}=\frac{\sqrt\pi}{2\,s^{3/2}}.\;} L { t 1/2 } = s 3/2 Γ ( 2 3 ) = 2 s 3/2 π . Take p = n + 1 2 p=n+\tfrac12 p = n + 2 1 n ∈ N n\in\mathbb N n ∈ N p + 1 = n + 3 2 = n + 1 + 1 2 p+1=n+\tfrac32=n+1+\tfrac12 p + 1 = n + 2 3 = n + 1 + 2 1
Answer
L { t n + 1 / 2 } = Γ ( n + 1 + 1 2 ) s n + 1 + 1 2 = Γ ( n + 3 2 ) s n + 3 2 . \boxed{\;\mathcal L\{t^{n+1/2}\}=\frac{\Gamma\!\left(n+1+\tfrac12\right)}{s^{\,n+1+\frac12}}=\frac{\Gamma\!\left(n+\tfrac32\right)}{s^{\,n+\frac32}}.\;} L { t n + 1/2 } = s n + 1 + 2 1 Γ ( n + 1 + 2 1 ) = s n + 2 3 Γ ( n + 2 3 ) .