← 2015 Paper 1

UPSC 2015 Maths Optional Paper 1 Q6a — Step-by-Step Solution

12 marks · Section B

Exact equations · ODEs · asked 9× in 13 yrs · Read the full method →

Question

Find the constant $a$ so that $(x+y)^a$ is the integrating factor of $(4x^2+2xy+6y)\,dx+(2x^2+9y+3x)\,dy=0$, and hence solve the differential equation.

Technique

Find $a$ by equating $M_y=N_x$ after multiplying by $\mu=(x+y)^a$ — the structural identity $(4x^2+2xy+6y)-(2x^2+9y+3x)=(2x-3)(x+y)$ and $(2x+6)-(4x+3)=-(2x-3)$ make the equation collapse to a single condition $a=1$. Then standard exact-equation procedure.

Solution

Let $M_0=4x^2+2xy+6y$, $N_0=2x^2+9y+3x$. Multiply by $\mu=(x+y)^a$:

Exactness requires $M_y=N_x$.

Step 1 — Compute $M_y$

Step 2 — Compute $N_x$

Step 3 — Equate $M_y=N_x$

Simplify the bracket on the first term: $(4x^2+2xy+6y)-(2x^2+9y+3x)=2x^2+2xy-3y-3x=2x(x+y)-3(x+y)=(2x-3)(x+y)$.

Simplify the bracket on the second term: $(2x+6)-(4x+3)=-2x+3=-(2x-3)$.

So

Wait: $(x+y)^{a-1}\cdot(x+y)=(x+y)^a$ — yes. So $a(2x-3)(x+y)^a-(2x-3)(x+y)^a=(a-1)(2x-3)(x+y)^a=0$.

For this to hold identically (for all $x,y$), $a-1=0$, so $\boxed{a=1}$.

Step 4 — Solve with $\mu=x+y$

$M=(x+y)(4x^2+2xy+6y)$ and $N=(x+y)(2x^2+9y+3x)$.

Expand $M$: $(x+y)(4x^2+2xy+6y)=4x^3+2x^2 y+6xy+4x^2 y+2xy^2+6y^2$ $=4x^3+6x^2 y+6xy+2xy^2+6y^2$.

Expand $N$: $(x+y)(2x^2+9y+3x)=2x^3+9xy+3x^2+2x^2 y+9y^2+3xy$ $=2x^3+3x^2+2x^2 y+12xy+9y^2$.

Check exactness:

Let me redo $M_y$ term by term. $M=4x^3+6x^2y+6xy+2xy^2+6y^2$: $M_y=0+6x^2+6x+4xy+12y$ ✓.

$N=2x^3+3x^2+2x^2y+12xy+9y^2$: $N_x=6x^2+6x+4xy+12y$ ✓.

$M_y=N_x$ ✓ (confirms $a=1$).

Step 5 — Find potential $F$ with $F_x=M$, $F_y=N$

Integrate $F_x=M=4x^3+6x^2y+6xy+2xy^2+6y^2$ w.r.t. $x$:

Wait, let me integrate term by term carefully:

• $\int 4x^3\,dx=x^4$.
• $\int 6x^2 y\,dx=2x^3 y$.
• $\int 6xy\,dx=3x^2 y$.
• $\int 2xy^2\,dx=x^2 y^2$.
• $\int 6y^2\,dx=6xy^2$.

So $F=x^4+2x^3 y+3x^2 y+x^2 y^2+6xy^2+g(y)$.

Step 6 — Differentiate w.r.t. $y$ and compare to $N$

$F_y=2x^3+3x^2+2x^2 y+12xy+g'(y)$.

Compare with $N=2x^3+3x^2+2x^2y+12xy+9y^2$: $g'(y)=9y^2$, so $g(y)=3y^3$.

Step 7 — General solution

Answer