← 2026 Paper 1

UPSC 2026 Maths Optional Paper 1 Q7c-i — Step-by-Step Solution

10 marks · Section B

Exact equations · ODEs · asked 10× in 14 yrs · Read the full method →

Question

Show that (x+y+1)−4(x + y + 1)^{-4} is an integrating factor of

(2x−y−1)y dx+(2y−x−1)x dy=0.(2x - y - 1)y\,dx + (2y - x - 1)x\,dy = 0.

Hence solve it.

Technique

Multiply through by the proposed factor and verify exactness (∂M~/∂y=∂N~/∂x\partial \tilde M/\partial y = \partial \tilde N/\partial x) directly; then recover the potential uu from ux=M~u_x = \tilde M, fixing the yy-dependent constant by uy=N~u_y = \tilde N, and write u=u = const.

Solution

The equation is not exact as given

Write M=(2x−y−1)yM = (2x-y-1)y, N=(2y−x−1)xN = (2y-x-1)x. Then

∂M∂y=2x−2y−1,∂N∂x=2y−2x−1,\frac{\partial M}{\partial y} = 2x - 2y - 1,\qquad \frac{\partial N}{\partial x} = 2y - 2x - 1,

which differ (by 4x−4y4x-4y), so the equation is not exact.

(x+y+1)−4(x+y+1)^{-4} is an integrating factor

Put s=x+y+1s = x + y + 1 (so sx=sy=1s_x = s_y = 1) and multiply the equation by s−4s^{-4}:

M~=y(2x−y−1)s4,N~=x(2y−x−1)s4.\tilde M = \frac{y(2x-y-1)}{s^4},\qquad \tilde N = \frac{x(2y-x-1)}{s^4}.

Compute ∂M~/∂y\partial\tilde M/\partial y. By the quotient rule,

∂M~∂y=(2x−2y−1)s4−y(2x−y−1)⋅4s3s8=(2x−2y−1)s−4y(2x−y−1)s5.\frac{\partial \tilde M}{\partial y} = \frac{(2x-2y-1)s^4 - y(2x-y-1)\cdot 4s^3}{s^8} = \frac{(2x-2y-1)s - 4y(2x-y-1)}{s^{5}} .

Expand the numerator:

(2x−2y−1)(x+y+1)=2x2−2y2+x−3y−1,(2x-2y-1)(x+y+1) = 2x^2 - 2y^2 + x - 3y - 1, 4y(2x−y−1)=8xy−4y2−4y,4y(2x-y-1) = 8xy - 4y^2 - 4y, ⟹ numerator=2x2−2y2+x−3y−1−8xy+4y2+4y=2x2+2y2−8xy+x+y−1.\Longrightarrow\ \text{numerator} = 2x^2 - 2y^2 + x - 3y - 1 - 8xy + 4y^2 + 4y = 2x^2 + 2y^2 - 8xy + x + y - 1 .

Hence

∂M~∂y=2x2+2y2−8xy+x+y−1(x+y+1)5.(1)\frac{\partial \tilde M}{\partial y} = \frac{2x^2 + 2y^2 - 8xy + x + y - 1}{(x+y+1)^{5}} .\tag{1}

Compute ∂N~/∂x\partial\tilde N/\partial x. The interchange x↔yx \leftrightarrow y sends M~↦N~\tilde M \mapsto \tilde N and ∂y↦∂x\partial_y \mapsto \partial_x (note ss is symmetric in x,yx,y). Applying that swap to (1):

∂N~∂x=2y2+2x2−8xy+y+x−1(x+y+1)5,(2)\frac{\partial \tilde N}{\partial x} = \frac{2y^2 + 2x^2 - 8xy + y + x - 1}{(x+y+1)^{5}} ,\tag{2}

which is identical to (1), the numerator being symmetric in xx and yy.

∴∂M~∂y=∂N~∂x⟹(x+y+1)−4 is an integrating factor.\therefore\quad \frac{\partial \tilde M}{\partial y} = \frac{\partial \tilde N}{\partial x} \quad\Longrightarrow\quad (x+y+1)^{-4}\ \text{is an integrating factor.}

Solving the exact equation

Seek u(x,y)u(x,y) with ux=M~u_x = \tilde M, uy=N~u_y = \tilde N; the solution is then u=u = const.

Integrate ux=M~u_x = \tilde M with respect to xx (holding yy fixed). Split the numerator against ss:

2x−y−1=2(x+y+1)−3(y+1)=2s−3(y+1),2x - y - 1 = 2(x+y+1) - 3(y+1) = 2s - 3(y+1),

so

M~=y[2s3−3(y+1)s4].\tilde M = y\left[\frac{2}{s^{3}} - \frac{3(y+1)}{s^{4}}\right].

Since ∂s/∂x=1\partial s/\partial x = 1,

∫2s3 dx=−1s2,∫3(y+1)s4 dx=−y+1s3,\int \frac{2}{s^{3}}\,dx = -\frac{1}{s^{2}},\qquad \int \frac{3(y+1)}{s^{4}}\,dx = -\frac{y+1}{s^{3}},

giving

u=y[−1s2+y+1s3]+ϕ(y)=−ys+y(y+1)s3+ϕ(y).u = y\left[-\frac{1}{s^{2}} + \frac{y+1}{s^{3}}\right] + \phi(y) = \frac{-ys + y(y+1)}{s^{3}} + \phi(y).

Now −ys+y2+y=−y(x+y+1)+y2+y=−xy-ys + y^2 + y = -y(x+y+1) + y^2 + y = -xy, so

u=−xy(x+y+1)3+ϕ(y).u = -\frac{xy}{(x+y+1)^{3}} + \phi(y).

Fix ϕ\phi from uy=N~u_y = \tilde N:

∂∂y(−xys3)=−x s3−xy⋅3s2s6=−xs+3xys4=x(−(x+y+1)+3y)s4=x(2y−x−1)s4=N~.\frac{\partial}{\partial y}\left(-\frac{xy}{s^{3}}\right) = -\frac{x\,s^{3} - xy\cdot 3s^{2}}{s^{6}} = \frac{-x s + 3xy}{s^{4}} = \frac{x\bigl(-(x+y+1) + 3y\bigr)}{s^{4}} = \frac{x(2y - x - 1)}{s^{4}} = \tilde N .

So ϕ′(y)=0\phi'(y) = 0, i.e. ϕ\phi is a constant, which we absorb.

General solution

u=−xy(x+y+1)3=const⟺xy(x+y+1)3=c.u = -\frac{xy}{(x+y+1)^{3}} = \text{const} \qquad\Longleftrightarrow\qquad \frac{xy}{(x+y+1)^{3}} = c .

Check by differentiating back. From xy=c (x+y+1)3xy = c\,(x+y+1)^3,

y dx+x dy=3c(x+y+1)2(dx+dy)=3xyx+y+1(dx+dy)(using c=xys3).y\,dx + x\,dy = 3c(x+y+1)^2(dx+dy) = \frac{3xy}{x+y+1}(dx + dy)\qquad\left(\text{using } c = \frac{xy}{s^3}\right).

Multiplying by s=x+y+1s = x+y+1:

y(s−3x) dx+x(s−3y) dy=0 ⟹ y(y+1−2x) dx+x(x+1−2y) dy=0,y(s - 3x)\,dx + x(s - 3y)\,dy = 0 \ \Longrightarrow\ y(y+1-2x)\,dx + x(x+1-2y)\,dy = 0,

which is −1-1 times the given equation. ✓

Answer

  ∂∂y ⁣[M(x+y+1)4]=∂∂x ⁣[N(x+y+1)4],and the general solution isxy(x+y+1)3=c.  \boxed{\;\frac{\partial}{\partial y}\!\left[\frac{M}{(x+y+1)^{4}}\right] = \frac{\partial}{\partial x}\!\left[\frac{N}{(x+y+1)^{4}}\right],\qquad\text{and the general solution is}\quad \frac{xy}{(x+y+1)^{3}} = c .\;}
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