← 2019 Paper 1
UPSC 2019 Maths Optional Paper 1 Q6c-i — Step-by-Step Solution 10 marks · Section B
Reduction of order with one solution known · ODEs · asked 3× in 14 yrs · Read the full method →
Question
Solve the differential equation
d 2 y d x 2 + ( 3 sin x − cot x ) d y d x + 2 y sin 2 x = e − cos x sin 2 x . \frac{d^2y}{dx^2}+(3\sin x-\cot x)\frac{dy}{dx}+2y\sin^2 x=e^{-\cos x}\sin^2 x. d x 2 d 2 y + ( 3 sin x − cot x ) d x d y + 2 y sin 2 x = e − c o s x sin 2 x . Technique
Change of independent variable t = − cos x t=-\cos x t = − cos x d t / d x = sin x dt/dx=\sin x d t / d x = sin x y ′ y' y ′ t t t
Solution
Step 1 — Choose the substitution
The combination 3 sin x − cot x 3\sin x-\cot x 3 sin x − cot x sin 2 x \sin^2 x sin 2 x
t = − cos x , d t d x = sin x , d 2 t d x 2 = cos x . t=-\cos x,\qquad \frac{dt}{dx}=\sin x,\qquad \frac{d^2t}{dx^2}=\cos x. t = − cos x , d x d t = sin x , d x 2 d 2 t = cos x . With y = Y ( t ) y=Y(t) y = Y ( t )
d y d x = Y ′ ( t ) sin x , d 2 y d x 2 = Y ′ ′ ( t ) sin 2 x + Y ′ ( t ) cos x . \frac{dy}{dx}=Y'(t)\sin x,\qquad
\frac{d^2y}{dx^2}=Y''(t)\sin^2 x+Y'(t)\cos x. d x d y = Y ′ ( t ) sin x , d x 2 d 2 y = Y ′′ ( t ) sin 2 x + Y ′ ( t ) cos x . Step 3 — Substitute into the equation
Y ′ ′ sin 2 x + Y ′ cos x ⏟ y ′ ′ + ( 3 sin x − cot x ) Y ′ sin x ⏟ y ′ + 2 Y sin 2 x = e − cos x sin 2 x . \underbrace{Y''\sin^2 x+Y'\cos x}_{y''}+(3\sin x-\cot x)\underbrace{Y'\sin x}_{y'}+2Y\sin^2 x=e^{-\cos x}\sin^2 x. y ′′ Y ′′ sin 2 x + Y ′ cos x + ( 3 sin x − cot x ) y ′ Y ′ sin x + 2 Y sin 2 x = e − c o s x sin 2 x . Expand the middle term: ( 3 sin x − cot x ) Y ′ sin x = 3 Y ′ sin 2 x − Y ′ cos x . (3\sin x-\cot x)Y'\sin x=3Y'\sin^2 x-Y'\cos x. ( 3 sin x − cot x ) Y ′ sin x = 3 Y ′ sin 2 x − Y ′ cos x . ± Y ′ cos x \pm Y'\cos x ± Y ′ cos x
Y ′ ′ sin 2 x + 3 Y ′ sin 2 x + 2 Y sin 2 x = e − cos x sin 2 x . Y''\sin^2 x+3Y'\sin^2 x+2Y\sin^2 x=e^{-\cos x}\sin^2 x. Y ′′ sin 2 x + 3 Y ′ sin 2 x + 2 Y sin 2 x = e − c o s x sin 2 x . Divide by sin 2 x \sin^2 x sin 2 x e − cos x = e t e^{-\cos x}=e^{t} e − c o s x = e t
Y ′ ′ + 3 Y ′ + 2 Y = e t . Y''+3Y'+2Y=e^{t}. Y ′′ + 3 Y ′ + 2 Y = e t . Step 4 — Solve the constant-coefficient equation in t t t
Auxiliary: m 2 + 3 m + 2 = ( m + 1 ) ( m + 2 ) = 0 ⇒ m = − 1 , − 2 m^2+3m+2=(m+1)(m+2)=0\Rightarrow m=-1,-2 m 2 + 3 m + 2 = ( m + 1 ) ( m + 2 ) = 0 ⇒ m = − 1 , − 2 = C 1 e − t + C 2 e − 2 t =C_1e^{-t}+C_2e^{-2t} = C 1 e − t + C 2 e − 2 t 1 1 1 Y p = A e t Y_p=Ae^{t} Y p = A e t ( 1 + 3 + 2 ) A = 1 ⇒ A = 1 6 (1+3+2)A=1\Rightarrow A=\tfrac16 ( 1 + 3 + 2 ) A = 1 ⇒ A = 6 1
Y ( t ) = C 1 e − t + C 2 e − 2 t + 1 6 e t . Y(t)=C_1e^{-t}+C_2e^{-2t}+\tfrac16 e^{t}. Y ( t ) = C 1 e − t + C 2 e − 2 t + 6 1 e t . Step 5 — Return to x x x t = − cos x t=-\cos x t = − cos x
e − t = e cos x e^{-t}=e^{\cos x} e − t = e c o s x e − 2 t = e 2 cos x e^{-2t}=e^{2\cos x} e − 2 t = e 2 c o s x e t = e − cos x e^{t}=e^{-\cos x} e t = e − c o s x
Answer
y = C 1 e cos x + C 2 e 2 cos x + 1 6 e − cos x . \boxed{\;y=C_1e^{\cos x}+C_2e^{2\cos x}+\frac16e^{-\cos x}.\;} y = C 1 e c o s x + C 2 e 2 c o s x + 6 1 e − c o s x .