← 2019 Paper 1
UPSC 2019 Maths Optional Paper 1 Q7a — Step-by-Step Solution 15 marks · Section B
Method of variation of parameters · ODEs · asked 12× in 14 yrs · Read the full method →
Question
Find the linearly independent solutions of the corresponding homogeneous differential equation of the equation x 2 y ′ ′ − 2 x y ′ + 2 y = x 3 sin x x^2y''-2xy'+2y=x^3\sin x x 2 y ′′ − 2 x y ′ + 2 y = x 3 sin x
Technique
Cauchy–Euler indicial equation for the CF; variation of parameters using the standard form (R = x sin x R=x\sin x R = x sin x x 2 x^2 x 2
Solution
Step 1 — Homogeneous (Cauchy–Euler) equation
x 2 y ′ ′ − 2 x y ′ + 2 y = 0. x^2y''-2xy'+2y=0. x 2 y ′′ − 2 x y ′ + 2 y = 0. Try y = x m y=x^m y = x m x m [ m ( m − 1 ) − 2 m + 2 ] = x m [ m 2 − 3 m + 2 ] = 0 x^m\big[m(m-1)-2m+2\big]=x^m\big[m^2-3m+2\big]=0 x m [ m ( m − 1 ) − 2 m + 2 ] = x m [ m 2 − 3 m + 2 ] = 0 m 2 − 3 m + 2 = ( m − 1 ) ( m − 2 ) = 0 m^2-3m+2=(m-1)(m-2)=0 m 2 − 3 m + 2 = ( m − 1 ) ( m − 2 ) = 0 m = 1 , 2 m=1,2 m = 1 , 2
y 1 = x , y 2 = x 2 . \boxed{\;y_1=x,\qquad y_2=x^2.\;} y 1 = x , y 2 = x 2 . (Their Wronskian W = ∣ x x 2 1 2 x ∣ = 2 x 2 − x 2 = x 2 ≠ 0 W=\begin{vmatrix}x&x^2\\1&2x\end{vmatrix}=2x^2-x^2=x^2\neq0 W = x 1 x 2 2 x = 2 x 2 − x 2 = x 2 = 0
Divide the full equation by x 2 x^2 x 2 1 1 1
y ′ ′ − 2 x y ′ + 2 x 2 y = x sin x , so R ( x ) = x sin x . y''-\frac2x y'+\frac{2}{x^2}y=x\sin x,\qquad\text{so}\quad R(x)=x\sin x. y ′′ − x 2 y ′ + x 2 2 y = x sin x , so R ( x ) = x sin x . With y p = u 1 y 1 + u 2 y 2 y_p=u_1y_1+u_2y_2 y p = u 1 y 1 + u 2 y 2
u 1 ′ = − y 2 R W = − x 2 ⋅ x sin x x 2 = − x sin x , u 2 ′ = y 1 R W = x ⋅ x sin x x 2 = sin x . u_1'=-\frac{y_2R}{W}=-\frac{x^2\cdot x\sin x}{x^2}=-x\sin x,\qquad
u_2'=\frac{y_1R}{W}=\frac{x\cdot x\sin x}{x^2}=\sin x. u 1 ′ = − W y 2 R = − x 2 x 2 ⋅ x sin x = − x sin x , u 2 ′ = W y 1 R = x 2 x ⋅ x sin x = sin x . Step 4 — Integrate
u 1 = ∫ ( − x sin x ) d x = x cos x − sin x , u 2 = ∫ sin x d x = − cos x . u_1=\int(-x\sin x)\,dx=x\cos x-\sin x,\qquad u_2=\int\sin x\,dx=-\cos x. u 1 = ∫ ( − x sin x ) d x = x cos x − sin x , u 2 = ∫ sin x d x = − cos x . (∫ x sin x d x = − x cos x + sin x \int x\sin x\,dx=-x\cos x+\sin x ∫ x sin x d x = − x cos x + sin x u 1 = − ( − x cos x + sin x ) = x cos x − sin x u_1=-(-x\cos x+\sin x)=x\cos x-\sin x u 1 = − ( − x cos x + sin x ) = x cos x − sin x
Step 5 — Particular integral
y p = u 1 y 1 + u 2 y 2 = ( x cos x − sin x ) x + ( − cos x ) x 2 = x 2 cos x − x sin x − x 2 cos x = − x sin x . y_p=u_1y_1+u_2y_2=(x\cos x-\sin x)\,x+(-\cos x)\,x^2=x^2\cos x-x\sin x-x^2\cos x=-x\sin x. y p = u 1 y 1 + u 2 y 2 = ( x cos x − sin x ) x + ( − cos x ) x 2 = x 2 cos x − x sin x − x 2 cos x = − x sin x . Step 6 — General solution
Answer
y = C 1 x + C 2 x 2 − x sin x . \boxed{\;y=C_1x+C_2x^2-x\sin x.\;} y = C 1 x + C 2 x 2 − x sin x .