UPSC 2014 Maths Optional Paper 1 Q6b — Step-by-Step Solution

20 marks · Section B

Euler-Cauchy equation · ODEs · asked 8× in 13 yrs · Read the full method →

Question

Solve the differential equation:

x^{3}\dfrac{d^{3}y}{dx^{3}}+3x^{2}\dfrac{d^{2}y}{dx^{2}}+x\dfrac{dy}{dx}+8y=65\cos(\log_e x).

Cauchy–Euler $x=e^{t}$ substitution → constant-coefficient ODE; CF via roots of auxiliary; PI via operator method.

Strategy. Cauchy–Euler equation: substitute $x=e^{t}$ to convert to constant-coefficient ODE. Solve CF + PI.

With $D=d/dt$ :

xy'=Dy,\quad x^{2}y''=D(D-1)y,\quad x^{3}y'''=D(D-1)(D-2)y.

LHS becomes

D(D-1)(D-2)y+3D(D-1)y+Dy+8y.

Expand:

Sum: $D^{3}-3D^{2}+2D+3D^{2}-3D+D+8=D^{3}+8$ .

So the ODE becomes

(D^{3}+8)y=65\cos t.

Auxiliary $D^{3}+8=0$ : $D^{3}=-8$ . The three cube roots of $-8=8e^{i\pi}$ are $2e^{i\pi/3},\,2e^{i\pi},\,2e^{i5\pi/3}$ , i.e.

D=1+i\sqrt 3,\;-2,\;1-i\sqrt 3.

(Check: $(D-(-2))(D-(1+i\sqrt 3))(D-(1-i\sqrt 3))=(D+2)((D-1)^{2}+3)=(D+2)(D^{2}-2D+4)=D^{3}+8$ ✓.)

CF:

y_c=c_1 e^{-2t}+e^{t}\bigl[c_2\cos\sqrt 3 t+c_3\sin\sqrt 3 t\bigr].

In $x$ -form (using $e^{t}=x$ , $e^{-2t}=x^{-2}$ , $\sqrt 3 t=\sqrt 3\log x$ ):

y_c=\dfrac{c_1}{x^{2}}+x\bigl[c_2\cos(\sqrt 3\log x)+c_3\sin(\sqrt 3\log x)\bigr].

For $65\cos t$ , operator-method:

y_p=\dfrac{65\cos t}{D^{3}+8}.

Acting on $\cos t$ : $D^{2}\to-1$ , so $D^{3}=D\cdot D^{2}\to D\cdot(-1)=-D$ . Therefore $D^{3}+8\to 8-D$ (when acting on $\cos t$ ):

y_p=\dfrac{65\cos t}{8-D}=65\cdot\dfrac{(8+D)\cos t}{(8-D)(8+D)}=\dfrac{65(8+D)\cos t}{64-D^{2}}=\dfrac{65(8+D)\cos t}{64-(-1)}=\dfrac{65(8\cos t-\sin t)}{65}=8\cos t-\sin t.

(Using $D\cos t=-\sin t$ and $D^{2}\cos t=-\cos t$ .)

In $x$ -form: $y_p=8\cos(\log x)-\sin(\log x)$ .

\boxed{\;y=\dfrac{c_1}{x^{2}}+x\bigl[c_2\cos(\sqrt 3\log x)+c_3\sin(\sqrt 3\log x)\bigr]+8\cos(\log x)-\sin(\log x).\;}