UPSC 2018 Maths Optional Paper 1 Q7d — Step-by-Step Solution

12 marks · Section B

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

Question

Find $\alpha$ and $\beta$ such that $x^\alpha y^\beta$ is an integrating factor of $(4y^2+3xy)\,dx-(3xy+2x^2)\,dy=0$ and solve the equation.

Technique

Multiply by $x^\alpha y^\beta$ , impose $M^\ast_y=N^\ast_x$ , match monomial coefficients to get a $2\times2$ linear system for $\alpha,\beta$ ; then integrate the now-exact equation.

Solution

Write $M\,dx+N\,dy=0$ with $M=4y^2+3xy$ and $N=-(3xy+2x^2)$ .

Step 1 — Multiply by the trial factor and impose exactness

Let $\mu=x^\alpha y^\beta$ . Set $M^\ast=x^\alpha y^\beta(4y^2+3xy)$ and $N^\ast=-x^\alpha y^\beta(3xy+2x^2)$ . Expand:

M^\ast=4x^\alpha y^{\beta+2}+3x^{\alpha+1}y^{\beta+1},\qquad N^\ast=-3x^{\alpha+1}y^{\beta+1}-2x^{\alpha+2}y^{\beta}.

Exactness requires $M^\ast_y=N^\ast_x$ :

M^\ast_y=4(\beta+2)x^\alpha y^{\beta+1}+3(\beta+1)x^{\alpha+1}y^{\beta},

N^\ast_x=-3(\alpha+1)x^{\alpha}y^{\beta+1}-2(\alpha+2)x^{\alpha+1}y^{\beta}.

Step 2 — Match coefficients

Equate the coefficients of like monomials.

Coefficient of $x^\alpha y^{\beta+1}$ : $\;4(\beta+2)=-3(\alpha+1)\Rightarrow 3\alpha+4\beta+11=0.$

Coefficient of $x^{\alpha+1}y^{\beta}$ : $\;3(\beta+1)=-2(\alpha+2)\Rightarrow 2\alpha+3\beta+7=0.$

Solve the linear system:

\begin{cases}3\alpha+4\beta=-11\\ 2\alpha+3\beta=-7\end{cases}\Rightarrow \alpha=-5,\ \beta=1.

\boxed{\;\alpha=-5,\qquad \beta=1.\;}\qquad \mu=x^{-5}y.

Step 3 — Make exact and integrate

With $\mu=x^{-5}y$ :

M^\ast=4x^{-5}y^{3}+3x^{-4}y^{2},\qquad N^\ast=-3x^{-4}y^{2}-2x^{-3}y.

Find $F$ with $F_x=M^\ast$ . Integrate $M^\ast$ in $x$ :

F=\int\big(4x^{-5}y^{3}+3x^{-4}y^{2}\big)\,dx=-x^{-4}y^{3}-x^{-3}y^{2}+g(y).

Then $F_y=-3x^{-4}y^{2}-2x^{-3}y+g'(y)$ , which must equal $N^\ast=-3x^{-4}y^{2}-2x^{-3}y$ , forcing $g'(y)=0$ .

Step 4 — General solution

F=-\frac{y^{3}}{x^{4}}-\frac{y^{2}}{x^{3}}=\text{const},

\boxed{\;\frac{y^{2}}{x^{3}}+\frac{y^{3}}{x^{4}}=C\;}\qquad\text{equivalently}\qquad x\,y^{2}+y^{3}=C\,x^{4}.

Verification

$ python3 -c "import sympy as sp; x,y=sp.symbols('x y'); M=x**-5*y*(4*y**2+3*x*y); N=x**-5*y*(-(3*x*y+2*x**2)); print(sp.simplify(sp.diff(M,y)-sp.diff(N,x))); F=sp.integrate(M,x); print(sp.simplify(F), sp.simplify(sp.diff(F,y)-N))"
# 0
# y**2*(-x - y)/x**4    0

With $\alpha=-5,\beta=1$ the equation is exact ( $M^\ast_y-N^\ast_x=0$ ) and the potential is $F=-\dfrac{y^2(x+y)}{x^4}$ , giving $\dfrac{y^2}{x^3}+\dfrac{y^3}{x^4}=C$ . ✓