← 2022 Paper 1
UPSC 2022 Maths Optional Paper 1 Q6b — Step-by-Step Solution 15 marks · Section B
Method of variation of parameters · ODEs · asked 11× in 13 yrs · Read the full method →
Question
Solve ( x 2 − 1 ) y ′ ′ − 2 x y ′ + 2 y = ( x 2 − 1 ) 2 (x^2-1)y''-2xy'+2y=(x^2-1)^2 ( x 2 − 1 ) y ′′ − 2 x y ′ + 2 y = ( x 2 − 1 ) 2 y 1 = x y_1=x y 1 = x
Technique
Reduction of order to find second homogeneous solution (y 2 = x 2 + 1 y_2=x^2+1 y 2 = x 2 + 1 u 1 ′ y 1 + u 2 ′ y 2 = 0 , u 1 ′ y 1 ′ + u 2 ′ y 2 ′ = ( x 2 − 1 ) / 1 u_1'y_1+u_2'y_2=0,\;u_1'y_1'+u_2'y_2'=(x^2-1)/1 u 1 ′ y 1 + u 2 ′ y 2 = 0 , u 1 ′ y 1 ′ + u 2 ′ y 2 ′ = ( x 2 − 1 ) /1 u 1 , u 2 u_1,u_2 u 1 , u 2
Solution
Standard form. Divide by ( x 2 − 1 ) (x^2-1) ( x 2 − 1 )
y ′ ′ − 2 x x 2 − 1 y ′ + 2 x 2 − 1 y = x 2 − 1. y''-\dfrac{2x}{x^2-1}y'+\dfrac{2}{x^2-1}y=x^2-1. y ′′ − x 2 − 1 2 x y ′ + x 2 − 1 2 y = x 2 − 1. Step 1 — Find second homogeneous solution via reduction of order
Given y 1 = x y_1=x y 1 = x y 2 = v ( x ) ⋅ x y_2=v(x)\cdot x y 2 = v ( x ) ⋅ x
y 2 ′ = v ′ x + v y_2'=v'x+v y 2 ′ = v ′ x + v y 2 ′ ′ = v ′ ′ x + 2 v ′ y_2''=v''x+2v' y 2 ′′ = v ′′ x + 2 v ′
Substitute into homogeneous equation ( x 2 − 1 ) y ′ ′ − 2 x y ′ + 2 y = 0 (x^2-1)y''-2xy'+2y=0 ( x 2 − 1 ) y ′′ − 2 x y ′ + 2 y = 0 ( x 2 − 1 ) ( v ′ ′ x + 2 v ′ ) − 2 x ( v ′ x + v ) + 2 v x = 0 (x^2-1)(v''x+2v')-2x(v'x+v)+2vx=0 ( x 2 − 1 ) ( v ′′ x + 2 v ′ ) − 2 x ( v ′ x + v ) + 2 v x = 0 x ( x 2 − 1 ) v ′ ′ + 2 ( x 2 − 1 ) v ′ − 2 x 2 v ′ − 2 x v + 2 x v = 0 x(x^2-1)v''+2(x^2-1)v'-2x^2 v'-2xv+2xv=0 x ( x 2 − 1 ) v ′′ + 2 ( x 2 − 1 ) v ′ − 2 x 2 v ′ − 2 xv + 2 xv = 0 x ( x 2 − 1 ) v ′ ′ + [ 2 ( x 2 − 1 ) − 2 x 2 ] v ′ = 0 x(x^2-1)v''+[2(x^2-1)-2x^2]v'=0 x ( x 2 − 1 ) v ′′ + [ 2 ( x 2 − 1 ) − 2 x 2 ] v ′ = 0 x ( x 2 − 1 ) v ′ ′ − 2 v ′ = 0 x(x^2-1)v''-2v'=0 x ( x 2 − 1 ) v ′′ − 2 v ′ = 0
Let u = v ′ u=v' u = v ′ x ( x 2 − 1 ) u ′ = 2 u x(x^2-1)u'=2u x ( x 2 − 1 ) u ′ = 2 u
d u u = 2 d x x ( x 2 − 1 ) . \dfrac{du}{u}=\dfrac{2\,dx}{x(x^2-1)}. u d u = x ( x 2 − 1 ) 2 d x . Partial fractions: 2 x ( x 2 − 1 ) = 2 x ( x − 1 ) ( x + 1 ) = A x + B x − 1 + C x + 1 \dfrac{2}{x(x^2-1)}=\dfrac{2}{x(x-1)(x+1)}=\dfrac{A}{x}+\dfrac{B}{x-1}+\dfrac{C}{x+1} x ( x 2 − 1 ) 2 = x ( x − 1 ) ( x + 1 ) 2 = x A + x − 1 B + x + 1 C
2 = A ( x − 1 ) ( x + 1 ) + B x ( x + 1 ) + C x ( x − 1 ) 2=A(x-1)(x+1)+Bx(x+1)+Cx(x-1) 2 = A ( x − 1 ) ( x + 1 ) + B x ( x + 1 ) + C x ( x − 1 )
x = 0 x=0 x = 0 2 = A ( − 1 ) ( 1 ) = − A ⇒ A = − 2 2=A(-1)(1)=-A\Rightarrow A=-2 2 = A ( − 1 ) ( 1 ) = − A ⇒ A = − 2 x = 1 x=1 x = 1 2 = B ( 1 ) ( 2 ) = 2 B ⇒ B = 1 2=B(1)(2)=2B\Rightarrow B=1 2 = B ( 1 ) ( 2 ) = 2 B ⇒ B = 1 x = − 1 x=-1 x = − 1 2 = C ( − 1 ) ( − 2 ) = 2 C ⇒ C = 1 2=C(-1)(-2)=2C\Rightarrow C=1 2 = C ( − 1 ) ( − 2 ) = 2 C ⇒ C = 1
So 2 x ( x 2 − 1 ) = − 2 x + 1 x − 1 + 1 x + 1 \dfrac{2}{x(x^2-1)}=-\dfrac{2}{x}+\dfrac{1}{x-1}+\dfrac{1}{x+1} x ( x 2 − 1 ) 2 = − x 2 + x − 1 1 + x + 1 1
∫ d u u = ∫ ( − 2 x + 1 x − 1 + 1 x + 1 ) d x \int\dfrac{du}{u}=\int\!\left(-\dfrac{2}{x}+\dfrac{1}{x-1}+\dfrac{1}{x+1}\right)dx ∫ u d u = ∫ ( − x 2 + x − 1 1 + x + 1 1 ) d x ln ∣ u ∣ = − 2 ln ∣ x ∣ + ln ∣ x − 1 ∣ + ln ∣ x + 1 ∣ + C 1 = ln ∣ ( x − 1 ) ( x + 1 ) x 2 ∣ + C 1 = ln ∣ x 2 − 1 x 2 ∣ + C 1 \ln|u|=-2\ln|x|+\ln|x-1|+\ln|x+1|+C_1=\ln\!\left|\dfrac{(x-1)(x+1)}{x^2}\right|+C_1=\ln\!\left|\dfrac{x^2-1}{x^2}\right|+C_1 ln ∣ u ∣ = − 2 ln ∣ x ∣ + ln ∣ x − 1∣ + ln ∣ x + 1∣ + C 1 = ln x 2 ( x − 1 ) ( x + 1 ) + C 1 = ln x 2 x 2 − 1 + C 1
So u = v ′ = x 2 − 1 x 2 = 1 − 1 x 2 u=v'=\dfrac{x^2-1}{x^2}=1-\dfrac{1}{x^2} u = v ′ = x 2 x 2 − 1 = 1 − x 2 1
Integrate: v = x + 1 x + C 2 v=x+\dfrac{1}{x}+C_2 v = x + x 1 + C 2 C 2 = 0 C_2=0 C 2 = 0 v = x + 1 / x v=x+1/x v = x + 1/ x
So y 2 = v x = x ( x + 1 / x ) = x 2 + 1 y_2=vx=x(x+1/x)=x^2+1 y 2 = v x = x ( x + 1/ x ) = x 2 + 1
Step 2 — Verify y 2 y_2 y 2
Test y 2 = x 2 + 1 y_2=x^2+1 y 2 = x 2 + 1 y 2 ′ = 2 x y_2'=2x y 2 ′ = 2 x y 2 ′ ′ = 2 y_2''=2 y 2 ′′ = 2 ( x 2 − 1 ) ( 2 ) − 2 x ( 2 x ) + 2 ( x 2 + 1 ) = 2 x 2 − 2 − 4 x 2 + 2 x 2 + 2 = 0 (x^2-1)(2)-2x(2x)+2(x^2+1)=2x^2-2-4x^2+2x^2+2=0 ( x 2 − 1 ) ( 2 ) − 2 x ( 2 x ) + 2 ( x 2 + 1 ) = 2 x 2 − 2 − 4 x 2 + 2 x 2 + 2 = 0
Step 3 — Variation of parameters
General solution: y p = u 1 ( x ) y 1 + u 2 ( x ) y 2 y_p=u_1(x)y_1+u_2(x)y_2 y p = u 1 ( x ) y 1 + u 2 ( x ) y 2
u 1 ′ y 1 + u 2 ′ y 2 = 0 , u_1'y_1+u_2'y_2=0, u 1 ′ y 1 + u 2 ′ y 2 = 0 , u 1 ′ y 1 ′ + u 2 ′ y 2 ′ = R H S leading coefficient = x 2 − 1. u_1'y_1'+u_2'y_2'=\dfrac{RHS}{\text{leading coefficient}}=x^2-1. u 1 ′ y 1 ′ + u 2 ′ y 2 ′ = leading coefficient R H S = x 2 − 1. With y 1 = x , y 1 ′ = 1 , y 2 = x 2 + 1 , y 2 ′ = 2 x y_1=x,y_1'=1,y_2=x^2+1,y_2'=2x y 1 = x , y 1 ′ = 1 , y 2 = x 2 + 1 , y 2 ′ = 2 x
System:
u 1 ′ ⋅ x + u 2 ′ ( x 2 + 1 ) = 0 , u_1'\cdot x+u_2'(x^2+1)=0, u 1 ′ ⋅ x + u 2 ′ ( x 2 + 1 ) = 0 , u 1 ′ + u 2 ′ ⋅ 2 x = x 2 − 1. u_1'+u_2'\cdot 2x=x^2-1. u 1 ′ + u 2 ′ ⋅ 2 x = x 2 − 1. Solve. From first: u 1 ′ = − ( x 2 + 1 ) x u 2 ′ u_1'=-\dfrac{(x^2+1)}{x}u_2' u 1 ′ = − x ( x 2 + 1 ) u 2 ′
Substitute into second: − ( x 2 + 1 ) x u 2 ′ + 2 x u 2 ′ = x 2 − 1 -\dfrac{(x^2+1)}{x}u_2'+2x u_2'=x^2-1 − x ( x 2 + 1 ) u 2 ′ + 2 x u 2 ′ = x 2 − 1
u 2 ′ [ − x 2 + 1 x + 2 x ] = x 2 − 1 u_2'\!\left[-\dfrac{x^2+1}{x}+2x\right]=x^2-1 u 2 ′ [ − x x 2 + 1 + 2 x ] = x 2 − 1
u 2 ′ [ − x 2 − 1 + 2 x 2 x ] = x 2 − 1 u_2'\!\left[\dfrac{-x^2-1+2x^2}{x}\right]=x^2-1 u 2 ′ [ x − x 2 − 1 + 2 x 2 ] = x 2 − 1
u 2 ′ ⋅ x 2 − 1 x = x 2 − 1 u_2'\cdot\dfrac{x^2-1}{x}=x^2-1 u 2 ′ ⋅ x x 2 − 1 = x 2 − 1
u 2 ′ = x u_2'=x u 2 ′ = x
Then u 1 ′ = − ( x 2 + 1 ) / x ⋅ x = − ( x 2 + 1 ) u_1'=-(x^2+1)/x\cdot x=-(x^2+1) u 1 ′ = − ( x 2 + 1 ) / x ⋅ x = − ( x 2 + 1 )
Step 4 — Integrate
u 1 = − ∫ ( x 2 + 1 ) d x = − x 3 / 3 − x u_1=-\int(x^2+1)\,dx=-x^3/3-x u 1 = − ∫ ( x 2 + 1 ) d x = − x 3 /3 − x u 2 = ∫ x d x = x 2 / 2 u_2=\int x\,dx=x^2/2 u 2 = ∫ x d x = x 2 /2
Step 5 — Particular solution
y p = u 1 y 1 + u 2 y 2 = ( − x 3 / 3 − x ) ( x ) + ( x 2 / 2 ) ( x 2 + 1 ) y_p=u_1 y_1+u_2 y_2=(-x^3/3-x)(x)+(x^2/2)(x^2+1) y p = u 1 y 1 + u 2 y 2 = ( − x 3 /3 − x ) ( x ) + ( x 2 /2 ) ( x 2 + 1 ) = − x 4 / 3 − x 2 + x 4 / 2 + x 2 / 2 =-x^4/3-x^2+x^4/2+x^2/2 = − x 4 /3 − x 2 + x 4 /2 + x 2 /2 = x 4 ( − 1 / 3 + 1 / 2 ) + x 2 ( − 1 + 1 / 2 ) =x^4(-1/3+1/2)+x^2(-1+1/2) = x 4 ( − 1/3 + 1/2 ) + x 2 ( − 1 + 1/2 ) = x 4 ⋅ 1 / 6 + x 2 ⋅ ( − 1 / 2 ) =x^4\cdot 1/6+x^2\cdot(-1/2) = x 4 ⋅ 1/6 + x 2 ⋅ ( − 1/2 ) = x 4 6 − x 2 2 =\dfrac{x^4}{6}-\dfrac{x^2}{2} = 6 x 4 − 2 x 2
Step 6 — General solution
y = C 1 x + C 2 ( x 2 + 1 ) + x 4 6 − x 2 2 y=C_1 x+C_2(x^2+1)+\dfrac{x^4}{6}-\dfrac{x^2}{2} y = C 1 x + C 2 ( x 2 + 1 ) + 6 x 4 − 2 x 2
The x 2 x^2 x 2 C 2 C_2 C 2 C 2 ( x 2 + 1 ) − x 2 / 2 = ( C 2 − 1 / 2 ) x 2 + C 2 C_2(x^2+1)-x^2/2=(C_2-1/2)x^2+C_2 C 2 ( x 2 + 1 ) − x 2 /2 = ( C 2 − 1/2 ) x 2 + C 2 C 2 ′ = C 2 − 1 / 2 C_2'=C_2-1/2 C 2 ′ = C 2 − 1/2 − x 2 / 2 -x^2/2 − x 2 /2
Or just keep as is:
Answer
y = C 1 x + C 2 ( x 2 + 1 ) + x 4 6 − x 2 2 . \boxed{\;y=C_1 x+C_2(x^2+1)+\dfrac{x^4}{6}-\dfrac{x^2}{2}.\;} y = C 1 x + C 2 ( x 2 + 1 ) + 6 x 4 − 2 x 2 .