UPSC 2024 Maths Optional Paper 2 Q8a — Step-by-Step Solution

15 marks · Section B

Classification and reduction to canonical form · PDEs · asked 8× in 13 yrs · Read the full method →

Question

Solve the PDE $\dfrac{\partial}{\partial y}\!\left(\dfrac{\partial\phi}{\partial x}+\phi\right)+2x^2 y\!\left(\dfrac{\partial\phi}{\partial x}+\phi\right)=0$ by transforming it to canonical form.

Technique

Substitute $\psi=\phi_x+\phi$ to reduce to a separable first-order ODE in $y$ ; then recover $\phi$ via an integrating-factor ODE in $x$ .

Solution

Step 1 — Canonical substitution.

Let $\psi=\phi_x+\phi$ . The PDE becomes:

\psi_y+2x^2 y\,\psi=0.

Step 2 — Solve for $\psi$ (separable ODE in $y$ ).

\frac{d\psi}{\psi}=-2x^2 y\,dy\;\Rightarrow\;\ln|\psi|=-x^2 y^2+g(x)\;\Rightarrow\;\psi(x,y)=h(x)\,e^{-x^2 y^2},

where $h(x)$ is an arbitrary function of $x$ .

Step 3 — Recover $\phi$ (linear ODE in $x$ ).

\phi_x+\phi=h(x)\,e^{-x^2 y^2}.

Integrating factor $e^x$ : $\partial_x(e^x\phi)=h(x)\,e^{x-x^2 y^2}$ .

Integrating:

e^x\phi(x,y)=k(y)+\int h(x)\,e^{x-x^2 y^2}\,dx,

where $k(y)$ is a second arbitrary function.

Step 4 — General solution.

\phi(x,y)=e^{-x}\!\left[k(y)+\int h(x)\,e^{x-x^2 y^2}\,dx\right].

The two arbitrary functions $h(x)$ and $k(y)$ encode the general solution.

Answer

\boxed{\;\phi(x,y)=e^{-x}\!\left[k(y)+\int h(x)\,e^{x-x^2 y^2}\,dx\right],\quad h,k\text{ arbitrary.}\;}