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UPSC 2014 Maths Optional Paper 1 Q5a — Step-by-Step Solution

10 marks · Section B

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

Question

Justify that a differential equation of the form

where $f(x^{2}+y^{2})$ is an arbitrary function of $(x^{2}+y^{2})$, is not an exact differential equation and $\dfrac{1}{x^{2}+y^{2}}$ is an integrating factor for it. Hence solve this differential equation for $f(x^{2}+y^{2})=(x^{2}+y^{2})^{2}$.

Technique

Show $M_y\ne N_x$ for non-exactness; verify exactness after multiplying by $\mu=1/r^{2}$; integrate potential function.

Solution

Write $r^{2}=x^{2}+y^{2}$, $f=f(r^{2})$, $f'=f'(r^{2})$ (derivative w.r.t. its argument).

Part 1 — Not exact

$M_y=1+x\,f'\cdot 2y=1+2xyf'.$

$N_x=y\,f'\cdot 2x-1=2xyf'-1.$

So $M_y-N_x=1-(-1)=2\ne 0$. Not exact.

Part 2 — $1/r^{2}$ is an integrating factor

After multiplying through by $\mu=1/r^{2}$:

Compute $M'_y$ and $N'_x$ using $\partial/\partial y[r^{2}]=2y$ and similar:

So $M'_y=N'_x$. The multiplied equation is exact, confirming $\mu=1/r^{2}$ is an integrating factor.

Part 3 — Solve for $f(r^{2})=r^{4}$

Then $f=r^{4}$, $f'(u)=2u$ where $u=r^{2}$. Multiplied equation:

i.e.

Find potential $F(x,y)$ with $F_x=M'$, $F_y=N'$. Integrate $F_x$ w.r.t. $x$:

Differentiate this w.r.t. $y$ and match $N'$:

$N'=yr^{2}-x/r^{2}=x^{2}y+y^{3}-x/(x^{2}+y^{2})$. Match:

Integrate: $g(y)=y^{4}/4$.

So

General solution: $F=$ const:

Answer