← 2015 Paper 1

UPSC 2015 Maths Optional Paper 1 Q8d — Step-by-Step Solution

13 marks · Section B

Euler-Cauchy equation · ODEs · asked 9× in 14 yrs · Read the full method →

Question

Solve $x^4 y^{(4)}+6x^3 y^{(3)}+4x^2 y''-2x y'-4y=x^2+2\cos(\log_e x).$

Technique

Euler equation ⇒ substitute $x=e^t$ to get constant-coefficient operator $f(D)=D^4-3D^2-4=(D-2)(D+2)(D^2+1)$; CF from the roots; particular integrals handle resonance ($e^{2t}$ and $\cos t$ are both in the kernel) via $t$-multiplication; convert back via $t=\log x$.

Solution

Strategy. The LHS is an Euler equation. Substitute $x=e^t$ (so $t=\log x$) and use $xD_x=D_t$, $x^2 D_x^2=D_t(D_t-1)$, etc.

Step 1 — Convert to constant coefficients

Let $D=d/dt$. Standard formulae:

• $xy'=Dy$.
• $x^2 y''=D(D-1)y$.
• $x^3 y'''=D(D-1)(D-2)y$.
• $x^4 y^{(4)}=D(D-1)(D-2)(D-3)y$.

Substitute:

(RHS: $x^2=e^{2t}$, $\cos(\log x)=\cos t$.)

Step 2 — Simplify the LHS polynomial in $D$

Let me compute the operator coefficient by coefficient.

$D(D-1)(D-2)(D-3)$: expand step by step.

$D(D-1)=D^2-D$.

$(D^2-D)(D-2)=D^3-2D^2-D^2+2D=D^3-3D^2+2D$.

$(D^3-3D^2+2D)(D-3)=D^4-3D^3-3D^3+9D^2+2D^2-6D=D^4-6D^3+11D^2-6D$.

$6D(D-1)(D-2)=6(D^3-3D^2+2D)=6D^3-18D^2+12D$.

$4D(D-1)=4D^2-4D$.

$-2D=-2D$.

$-4=-4$.

Sum:

• $D^4$: $1$.
• $D^3$: $-6+6=0$.
• $D^2$: $11-18+4=-3$.
• $D^1$: $-6+12-4-2=0$.
• $D^0$: $-4$.

LHS operator: $D^4-3D^2-4$.

Step 3 — Factor the operator

$D^4-3D^2-4=(D^2-4)(D^2+1)=(D-2)(D+2)(D^2+1)$.

Step 4 — Auxiliary equation; complementary function

Roots: $D=2,-2,\pm i$.

Complementary function in $t$:

In $x$ (using $t=\log x$, $e^t=x$):

Step 5 — Particular integral for $e^{2t}$

RHS contains $e^{2t}$. Standard formula: $y_p=\dfrac{e^{2t}}{f(D)}\bigg|_{\text{at }D=2}$, where $f(D)=D^4-3D^2-4$. But $f(2)=16-12-4=0$ — resonance.

Use $f(D)=(D-2)(D+2)(D^2+1)$. With $D=2$: $f(2)=0$, single root. Apply $\dfrac{1}{D-2}$ specially:

For a simple-pole resonance, $y_p=\dfrac{e^{2t}}{f'(2)}\cdot t$ — standard “shift” formula.

$f(D)=D^4-3D^2-4$, $f'(D)=4D^3-6D=2D(2D^2-3)$. $f'(2)=2\cdot 2\cdot(8-3)=20$.

So $y_{p,1}=\dfrac{te^{2t}}{20}$.

Step 6 — Particular integral for $2\cos t$

RHS contains $2\cos t$. Operator $f(D)$ evaluated at $D=\pm i$: $f(i)=i^4-3i^2-4=1+3-4=0$ — resonance again!

Resonance with $D^2+1$ factor. Use the shift method.

For $\cos t=\Re e^{it}$, $y_p$ for $e^{it}$ part: $f(i)=0$, simple root of $D^2+1=(D-i)(D+i)$.

$f'(D)=4D^3-6D$. $f'(i)=4i^3-6i=-4i-6i=-10i$.

Particular for $e^{it}$: $y_{p,e^{it}}=\dfrac{te^{it}}{f'(i)}=\dfrac{te^{it}}{-10i}=\dfrac{it e^{it}}{10}$.

For $2\cos t=2\Re(e^{it})$: $y_{p,2}=2\Re\bigl(\dfrac{it e^{it}}{10}\bigr)=\dfrac{t}{5}\Re(ie^{it})=\dfrac{t}{5}\Re(i\cos t-\sin t)=\dfrac{t}{5}(-\sin t)=-\dfrac{t\sin t}{5}$.

Hmm wait: $\Re(i\cos t-\sin t)=\Re(i\cos t)-\Re(\sin t)=0-\sin t=-\sin t$. So $y_{p,2}=2\cdot t\cdot(-\sin t)/10=-t\sin t/5$.

Actually let me redo more carefully. $2\cos t=\Re(2e^{it})$. Particular for $2e^{it}$: $2\cdot \dfrac{te^{it}}{f'(i)}=\dfrac{2t e^{it}}{-10i}=\dfrac{2t e^{it}\cdot i}{-10i\cdot i}=\dfrac{2it e^{it}}{10}=\dfrac{it e^{it}}{5}$.

Take real part: $\Re(it e^{it}/5)=(t/5)\Re(ie^{it})=(t/5)\Re(i\cos t-\sin t)=(t/5)(-\sin t)=-t\sin t/5$.

Hmm: but my standard sign-tracking suggests $\Re(\cos t+i\sin t)\cdot i=i\cos t-\sin t$, so $\Re=-\sin t$. Confirmed.

$y_{p,2}=-\dfrac{t\sin t}{5}$.

Wait — let me double-check by direct substitution. Try $y=At\sin t+Bt\cos t$, compute $(D^4-3D^2-4)y$ and match to $2\cos t$.

Actually easier: $y_p=t\,g(t)$ where $g(t)=A\cos t+B\sin t$. By the operator factoring $f(D)=(D-2)(D+2)(D^2+1)$, the resonance is with the $(D^2+1)$ factor. Reduce: let $L_1=D^2+1$ and $L_2=(D-2)(D+2)=D^2-4$. Then $f(D)=L_1 L_2=L_2 L_1$.

For RHS $=2\cos t$, $L_1(2\cos t)=(D^2+1)(2\cos t)=2(-\cos t+\cos t)=0$. So $2\cos t$ is in the kernel of $L_1$.

Apply $f^{-1}$: $y_p=L_2^{-1}L_1^{-1}(2\cos t)$. We have $L_1^{-1}(2\cos t)=$ particular solution of $L_1 z=2\cos t$, i.e. $z''+z=2\cos t$. Standard: $z=t\sin t$. (Check: $z'=\sin t+t\cos t$, $z''=\cos t+\cos t-t\sin t=2\cos t-t\sin t$. $z''+z=2\cos t-t\sin t+t\sin t=2\cos t$ ✓.)

Now apply $L_2^{-1}=1/(D^2-4)$: $y_p=\dfrac{t\sin t}{D^2-4}$. Plug $D^2-4$ acting on $t\sin t$ — find $w$ such that $(D^2-4)w=t\sin t$.

Try $w=(At+B)\sin t+(Ct+E)\cos t$. Then $w'=A\sin t+(At+B)\cos t+C\cos t-(Ct+E)\sin t$ $=(A-Ct-E)\sin t+(At+B+C)\cos t$.

$w''=-C\sin t+(A-Ct-E)\cos t+A\cos t-(At+B+C)\sin t$ $=(-C-At-B-C)\sin t+(2A-Ct-E)\cos t$ $=(-At-B-2C)\sin t+(2A-Ct-E)\cos t$.

$w''-4w=(-At-B-2C-4At-4B)\sin t+(2A-Ct-E-4Ct-4E)\cos t$

Wait, $-4w=-4[(At+B)\sin t+(Ct+E)\cos t]=(-4At-4B)\sin t+(-4Ct-4E)\cos t$.

$w''-4w$: coefficient of $\sin t$: $-At-B-2C-4At-4B=-5At-5B-2C$.

Coefficient of $\cos t$: $2A-Ct-E-4Ct-4E=2A-5Ct-5E$.

Match to $t\sin t$: coefficient of $\sin t$ is $-5At-5B-2C=t$, giving $-5A=1$ and $-5B-2C=0$.

Coefficient of $\cos t$ is $2A-5Ct-5E=0$, giving $-5C=0$ and $2A-5E=0$.

So $A=-1/5$, $C=0$, $B$ from $-5B-2\cdot 0=0\Rightarrow B=0$. $E$ from $2(-1/5)-5E=0\Rightarrow -2/5=5E\Rightarrow E=-2/25$.

$w=-\dfrac{t}{5}\sin t-\dfrac{2}{25}\cos t$.

Check: derivative of $-\dfrac{2}{25}\cos t$ is in $L_2^{-1}$ kernel? No — it’s a particular contribution.

So $y_{p,2}=w=-\dfrac{t\sin t}{5}-\dfrac{2\cos t}{25}$.

The $-\dfrac{2\cos t}{25}$ term can be absorbed into $y_c$ (since $\cos t$ is in the kernel of $L_1$ and hence of $f$). So drop it:

Good — matches my earlier answer.

Step 7 — Total general solution in $t$

Step 8 — Back to $x$

$e^{2t}=x^2$, $e^{-2t}=1/x^2$, $\cos t=\cos(\log x)$, $\sin t=\sin(\log x)$, $te^{2t}=x^2\log x$, $t\sin t=\sin(\log x)\cdot\log x$.

Answer