← 2014 Paper 1

UPSC 2014 Maths Optional Paper 1 Q8a — Step-by-Step Solution

15 marks · Section B

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

Question

Find the sufficient condition for the differential equation M(x,y) dx+N(x,y) dy=0M(x,y)\,dx+N(x,y)\,dy=0 to have an integrating factor as a function of (x+y)(x+y). What will be the integrating factor in that case? Hence find the integrating factor for the differential equation

(x2+xy) dx+(y2+xy) dy=0,(x^{2}+xy)\,dx+(y^{2}+xy)\,dy=0,

and solve it.

Technique

Standard derivation of the condition for μ=μ(x+y)\mu=\mu(x+y); apply to specific ODE; the IF simplifies the equation to a separable form.

Solution

Part 1 — Sufficient condition for IF as μ(x+y)\mu(x+y)

Strategy. After multiplying by μ(x+y)\mu(x+y), exactness requires ∂(μM)/∂y=∂(μN)/∂x\partial(\mu M)/\partial y=\partial(\mu N)/\partial x.

Compute:

∂(μM)/∂y=μ′M+μMy,∂(μN)/∂x=μ′N+μNx,\partial(\mu M)/\partial y=\mu' M+\mu M_y,\quad\partial(\mu N)/\partial x=\mu' N+\mu N_x,

where μ′=dμ/d(x+y)\mu'=d\mu/d(x+y).

Setting equal:

μ′(M−N)=μ(Nx−My)  ⟹  μ′μ=Nx−MyM−N.\mu'(M-N)=\mu(N_x-M_y)\;\Longrightarrow\;\dfrac{\mu'}{\mu}=\dfrac{N_x-M_y}{M-N}.

Sufficient condition: Nx−MyM−N\dfrac{N_x-M_y}{M-N} is a function of (x+y)(x+y) only.

If so, denote this function f(x+y)f(x+y). Then

μ(x+y)=exp⁡ ⁣∫f(x+y) d(x+y).\mu(x+y)=\exp\!\int f(x+y)\,d(x+y).

Part 2 — Apply to (x2+xy)dx+(y2+xy)dy=0(x^{2}+xy)dx+(y^{2}+xy)dy=0

M=x2+xy,  N=y2+xyM=x^{2}+xy,\;N=y^{2}+xy.

My=x,  Nx=yM_y=x,\;N_x=y.

Nx−MyM−N=y−x(x2+xy)−(y2+xy)=y−xx2−y2=−(x−y)(x−y)(x+y)=−1x+y.\dfrac{N_x-M_y}{M-N}=\dfrac{y-x}{(x^{2}+xy)-(y^{2}+xy)}=\dfrac{y-x}{x^{2}-y^{2}}=\dfrac{-(x-y)}{(x-y)(x+y)}=-\dfrac{1}{x+y}.

Function of (x+y)(x+y) only ✓. So

μ=exp⁡ ⁣∫ ⁣(−1x+y)d(x+y)=exp⁡(−ln⁡∣x+y∣)=1x+y.\mu=\exp\!\int\!\left(-\dfrac{1}{x+y}\right)d(x+y)=\exp(-\ln|x+y|)=\dfrac{1}{x+y}.

Part 3 — Solve the multiplied equation

Multiply the original equation by μ=1/(x+y)\mu=1/(x+y):

x2+xyx+y dx+y2+xyx+y dy=0.\dfrac{x^{2}+xy}{x+y}\,dx+\dfrac{y^{2}+xy}{x+y}\,dy=0.

Factor numerators: x2+xy=x(x+y)x^{2}+xy=x(x+y), y2+xy=y(y+x)y^{2}+xy=y(y+x). So

x dx+y dy=0.x\,dx+y\,dy=0.

This is trivially exact (both sides separable). Integrate:

x22+y22=const,  i.e.,    x2+y2=C.  \dfrac{x^{2}}{2}+\dfrac{y^{2}}{2}=\text{const},\;\text{i.e.,}\;\boxed{\;x^{2}+y^{2}=C.\;}

The solution is a family of concentric circles centred at the origin.

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