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UPSC 2017 Maths Optional Paper 1 Q8b — Step-by-Step Solution 17 marks · Section B
Laplace transform applied to IVP for second-order linear ODE with constant coefficients · ODEs · asked 11× in 14 yrs · Read the full method →
Question
Solve the following initial value problem using Laplace transform:
d 2 y d x 2 + 9 y = r ( x ) , y ( 0 ) = 0 , y ′ ( 0 ) = 4 , \frac{d^2y}{dx^2}+9y=r(x),\qquad y(0)=0,\ y'(0)=4, d x 2 d 2 y + 9 y = r ( x ) , y ( 0 ) = 0 , y ′ ( 0 ) = 4 , where r ( x ) = { 8 sin x if 0 < x < π 0 if x ≥ π . r(x)=\begin{cases}8\sin x & \text{if } 0<x<\pi\\ 0 & \text{if } x\ge\pi.\end{cases} r ( x ) = { 8 sin x 0 if 0 < x < π if x ≥ π .
Technique
Laplace transform; write forcing via Heaviside, use sin x = − sin ( x − π ) \sin x=-\sin(x-\pi) sin x = − sin ( x − π ) 1 ( s 2 + 1 ) ( s 2 + 9 ) = 1 8 ( 1 s 2 + 1 − 1 s 2 + 9 ) \tfrac1{(s^2+1)(s^2+9)}=\tfrac18(\tfrac1{s^2+1}-\tfrac1{s^2+9}) ( s 2 + 1 ) ( s 2 + 9 ) 1 = 8 1 ( s 2 + 1 1 − s 2 + 9 1 )
Solution
Step 1 — Write the forcing with a Heaviside step
r ( x ) = 8 sin x [ 1 − H ( x − π ) ] = 8 sin x − 8 sin x H ( x − π ) . r(x)=8\sin x\,\big[1-H(x-\pi)\big]=8\sin x-8\sin x\,H(x-\pi). r ( x ) = 8 sin x [ 1 − H ( x − π ) ] = 8 sin x − 8 sin x H ( x − π ) . For the second-shift theorem we need a function of ( x − π ) (x-\pi) ( x − π ) sin x = sin ( ( x − π ) + π ) = − sin ( x − π ) \sin x=\sin\big((x-\pi)+\pi\big)=-\sin(x-\pi) sin x = sin ( ( x − π ) + π ) = − sin ( x − π )
r ( x ) = 8 sin x + 8 sin ( x − π ) H ( x − π ) . r(x)=8\sin x+8\sin(x-\pi)\,H(x-\pi). r ( x ) = 8 sin x + 8 sin ( x − π ) H ( x − π ) . Let Y ( s ) = L { y } Y(s)=\mathcal L\{y\} Y ( s ) = L { y } y ( 0 ) = 0 , y ′ ( 0 ) = 4 y(0)=0,\ y'(0)=4 y ( 0 ) = 0 , y ′ ( 0 ) = 4
L { y ′ ′ } = s 2 Y − s y ( 0 ) − y ′ ( 0 ) = s 2 Y − 4. \mathcal L\{y''\}=s^2Y-sy(0)-y'(0)=s^2Y-4. L { y ′′ } = s 2 Y − sy ( 0 ) − y ′ ( 0 ) = s 2 Y − 4. Using L { sin x } = 1 s 2 + 1 \mathcal L\{\sin x\}=\dfrac{1}{s^2+1} L { sin x } = s 2 + 1 1 L { sin ( x − π ) H ( x − π ) } = e − π s s 2 + 1 \mathcal L\{\sin(x-\pi)H(x-\pi)\}=\dfrac{e^{-\pi s}}{s^2+1} L { sin ( x − π ) H ( x − π )} = s 2 + 1 e − π s
( s 2 + 9 ) Y − 4 = 8 s 2 + 1 + 8 e − π s s 2 + 1 . (s^2+9)Y-4=\frac{8}{s^2+1}+\frac{8e^{-\pi s}}{s^2+1}. ( s 2 + 9 ) Y − 4 = s 2 + 1 8 + s 2 + 1 8 e − π s . Y = 4 s 2 + 9 + 8 ( s 2 + 1 ) ( s 2 + 9 ) + 8 e − π s ( s 2 + 1 ) ( s 2 + 9 ) . Y=\frac{4}{s^2+9}+\frac{8}{(s^2+1)(s^2+9)}+\frac{8e^{-\pi s}}{(s^2+1)(s^2+9)}. Y = s 2 + 9 4 + ( s 2 + 1 ) ( s 2 + 9 ) 8 + ( s 2 + 1 ) ( s 2 + 9 ) 8 e − π s . Step 3 — Partial fractions
8 ( s 2 + 1 ) ( s 2 + 9 ) = 8 ⋅ 1 8 ( 1 s 2 + 1 − 1 s 2 + 9 ) = 1 s 2 + 1 − 1 s 2 + 9 . \frac{8}{(s^2+1)(s^2+9)}=8\cdot\frac1{8}\Big(\frac{1}{s^2+1}-\frac{1}{s^2+9}\Big)=\frac{1}{s^2+1}-\frac{1}{s^2+9}. ( s 2 + 1 ) ( s 2 + 9 ) 8 = 8 ⋅ 8 1 ( s 2 + 1 1 − s 2 + 9 1 ) = s 2 + 1 1 − s 2 + 9 1 . Step 4 — Invert term by term
First two terms (no shift):
L − 1 { 4 s 2 + 9 } = 4 3 sin 3 x , L − 1 { 1 s 2 + 1 − 1 s 2 + 9 } = sin x − 1 3 sin 3 x . \mathcal L^{-1}\!\Big\{\frac{4}{s^2+9}\Big\}=\frac43\sin 3x,\qquad
\mathcal L^{-1}\!\Big\{\frac1{s^2+1}-\frac1{s^2+9}\Big\}=\sin x-\frac13\sin 3x. L − 1 { s 2 + 9 4 } = 3 4 sin 3 x , L − 1 { s 2 + 1 1 − s 2 + 9 1 } = sin x − 3 1 sin 3 x . Sum of the un-shifted part:
y 0 ( x ) = 4 3 sin 3 x + sin x − 1 3 sin 3 x = sin x + sin 3 x . y_0(x)=\frac43\sin3x+\sin x-\frac13\sin3x=\sin x+\sin 3x. y 0 ( x ) = 3 4 sin 3 x + sin x − 3 1 sin 3 x = sin x + sin 3 x . Shifted term: L − 1 { 8 e − π s ( s 2 + 1 ) ( s 2 + 9 ) } = g ( x − π ) H ( x − π ) \mathcal L^{-1}\Big\{\dfrac{8e^{-\pi s}}{(s^2+1)(s^2+9)}\Big\}=g(x-\pi)\,H(x-\pi) L − 1 { ( s 2 + 1 ) ( s 2 + 9 ) 8 e − π s } = g ( x − π ) H ( x − π ) g ( x ) = sin x − 1 3 sin 3 x g(x)=\sin x-\tfrac13\sin 3x g ( x ) = sin x − 3 1 sin 3 x g ( x − π ) = sin ( x − π ) − 1 3 sin ( 3 x − 3 π ) = − sin x + 1 3 sin 3 x g(x-\pi)=\sin(x-\pi)-\tfrac13\sin(3x-3\pi)=-\sin x+\tfrac13\sin3x g ( x − π ) = sin ( x − π ) − 3 1 sin ( 3 x − 3 π ) = − sin x + 3 1 sin 3 x x > π x>\pi x > π − sin x + 1 3 sin 3 x -\sin x+\tfrac13\sin 3x − sin x + 3 1 sin 3 x
Step 5 — Assemble (piecewise)
y ( x ) = ( sin x + sin 3 x ) + ( − sin x + 1 3 sin 3 x ) H ( x − π ) . y(x)=\big(\sin x+\sin 3x\big)+\big(-\sin x+\tfrac13\sin 3x\big)H(x-\pi). y ( x ) = ( sin x + sin 3 x ) + ( − sin x + 3 1 sin 3 x ) H ( x − π ) . Answer
y ( x ) = { sin x + sin 3 x , 0 ≤ x ≤ π , 4 3 sin 3 x , x ≥ π . \boxed{\;y(x)=\begin{cases}\sin x+\sin 3x, & 0\le x\le \pi,\\[4pt]\dfrac43\sin 3x, & x\ge \pi.\end{cases}\;} y ( x ) = ⎩ ⎨ ⎧ sin x + sin 3 x , 3 4 sin 3 x , 0 ≤ x ≤ π , x ≥ π .