UPSC 2023 Maths Optional Paper 1 Q7a-i — Step-by-Step Solution

10 marks · Section B

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

Question

Find the solution of the differential equation $\dfrac{dy}{dx}=-\dfrac{2xy^3+2}{3x^2 y^2+8e^{4y}}$ .

Technique

Test exactness via $\partial M/\partial y=\partial N/\partial x$ ; if exact, find potential function $F$ by integrating $M$ in $x$ and matching $g(y)$ .

Solution

Step 1 — Recognise as exact.

Rewrite: $(2xy^3+2)\,dx+(3x^2y^2+8e^{4y})\,dy=0$ .

Let $M=2xy^3+2,\,N=3x^2y^2+8e^{4y}$ .

Test exactness: $\dfrac{\partial M}{\partial y}=6xy^2,\quad\dfrac{\partial N}{\partial x}=6xy^2.$

Equal — the equation is exact.

Step 2 — Find $F(x,y)$ such that $F_x=M,\,F_y=N$ .

$F_x=2xy^3+2\Rightarrow F=x^2y^3+2x+g(y)$ for some $g$ .

Then $F_y=3x^2y^2+g'(y)$ . Equate to $N$ : $3x^2y^2+g'(y)=3x^2y^2+8e^{4y}\Rightarrow g'(y)=8e^{4y}\Rightarrow g(y)=2e^{4y}.$

So $F(x,y)=x^2y^3+2x+2e^{4y}$ .

Step 3 — Solution.

Answer

\boxed{\;x^2y^3+2x+2e^{4y}=C.\;}