UPSC Maths 2025 Paper 1 Q7c-i — Solution

10 marks · Section B

Question

Find the general solution and singular solution of the differential equation $\left(1 + \dfrac{dy}{dx}\right)^3 = \dfrac{27}{8a}(x + y)\left(1 - \dfrac{dy}{dx}\right)^3$ .

Technique

The substitution $v = x + y$ reduces the equation to separable form, giving the general solution by integration; the singular solution comes from the $p$ -discriminant ( $\partial F/\partial p = 0$ ).

Solution

Let $p = \dfrac{dy}{dx}$ and $v = x + y$ , so $\dfrac{dv}{dx} = 1 + p$ , hence $1 + p = v'$ and $1 - p = 2 - v'$ .

The equation is $(1+p)^3 = \dfrac{27}{8a}v\,(1-p)^3$ . Taking cube roots (with $a>0$ ): $1 + p = \left(\frac{27}{8a}\right)^{1/3} v^{1/3}\,(1-p) = \frac{3}{2a^{1/3}}\,v^{1/3}\,(1-p).$

In terms of $v$ : $v' = \dfrac{3}{2a^{1/3}}v^{1/3}\,(2 - v')$ . Solving for $v'$ : $v'\left(1 + \tfrac{3}{2a^{1/3}}v^{1/3}\right) = \frac{3}{a^{1/3}}v^{1/3} \;\Longrightarrow\; v' = \frac{3a^{-1/3}v^{1/3}}{1 + \tfrac{3}{2}a^{-1/3}v^{1/3}}.$

Separate variables ( $dx = dv/v'$ ): $dx = \left(\frac{1}{2}\cdot\frac{2a^{1/3}}{3}\,v^{-1/3} + \frac{1}{2}\right)dv = \left(\frac{a^{1/3}}{3}v^{-1/3} + \frac{1}{2}\right)dv.$ Integrate: $x = \frac{a^{1/3}}{3}\cdot\frac{v^{2/3}}{2/3} + \frac{v}{2} + \text{const} = \frac{a^{1/3}}{2}v^{2/3} + \frac{v}{2} + C_1.$

Multiply by $2$ and substitute $v = x+y$ : $2x - (x+y) = a^{1/3}(x+y)^{2/3} + 2C_1 \;\Longrightarrow\; (x - y) - C = a^{1/3}(x+y)^{2/3},$ with $C = -2C_1$ an arbitrary constant. Cubing gives the general solution $\bigl(x - y + c\bigr)^3 = a\,(x + y)^2,\qquad c\ \text{arbitrary}.$

Singular solution. Write the ODE as $F(x,y,p) = (1+p)^3 - \dfrac{27}{8a}(x+y)(1-p)^3 = 0$ . Differentiating partially in $p$ : $\frac{\partial F}{\partial p} = 3(1+p)^2 + \frac{81}{8a}(x+y)(1-p)^2 = 0 \;\Longrightarrow\; (x+y) = -\frac{8a(1+p)^2}{27(1-p)^2}.$ Substituting back into $F=0$ gives $F = 2(1+p)^2 = 0$ , so $p = -1$ , and then $x+y = 0$ . Direct check: on $x+y=0$ , $p=-1$ , both sides of the ODE are $0$ . Hence the singular solution is $x + y = 0.$

Answer

$\boxed{\;\text{General: } (x - y + c)^3 = a(x+y)^2;\qquad \text{Singular: } x + y = 0.\;}$