UPSC 2018 Maths Optional Paper 1 Q6a — Step-by-Step Solution

13 marks · Section B

First-order higher-degree ODEs · ODEs · asked 5× in 13 yrs · Read the full method →

Question

Solve: $\left(\dfrac{dy}{dx}\right)^2 y+2\dfrac{dy}{dx}\,x-y=0$ .

Equation linear in $x$ → solve for $x$ , differentiate w.r.t. $y$ using $dx/dy=1/p$ , separate to get $yp=c$ , eliminate $p$ .

Write $p=\dfrac{dy}{dx}$ , so the equation is $y\,p^2+2x\,p-y=0$ .

Although it is quadratic in $p$ , it is linear in $x$ , which is the cleaner route:

2x\,p=y-y\,p^2\;\Rightarrow\; x=\frac{y(1-p^2)}{2p}.

Use $\dfrac{dx}{dy}=\dfrac1p$ . Differentiate $x=\dfrac{y(1-p^2)}{2p}=\dfrac{y}{2p}-\dfrac{yp}{2}$ :

\frac{dx}{dy}=\frac{1}{2p}-\frac{y}{2p^2}\frac{dp}{dy}-\frac{p}{2}-\frac{y}{2}\frac{dp}{dy}.

Set this equal to $\dfrac1p$ :

\frac1p=\frac{1}{2p}-\frac{p}{2}-\frac{y}{2}\left(\frac{1}{p^2}+1\right)\frac{dp}{dy}.

Move terms:

\frac1p-\frac{1}{2p}+\frac{p}{2}=-\frac{y}{2}\,\frac{1+p^2}{p^2}\,\frac{dp}{dy} \;\Rightarrow\;\frac{1}{2p}+\frac{p}{2}=\frac{1+p^2}{2p}=-\frac{y}{2}\,\frac{1+p^2}{p^2}\,\frac{dp}{dy}.

Cancel $\dfrac{1+p^2}{2}$ from both sides (it is non-zero):

\frac1p=-\frac{y}{p^2}\,\frac{dp}{dy}\;\Rightarrow\;\frac{dy}{y}=-\frac{dp}{p}.

Integrate: $\ln y=-\ln p+\text{const}$ , i.e. $yp=c$ (a constant). Thus $p=\dfrac{c}{y}$ .

Substitute $p=c/y$ into $x=\dfrac{y(1-p^2)}{2p}$ :

x=\frac{y\big(1-c^2/y^2\big)}{2c/y}=\frac{y^2-c^2}{2c}\;\Rightarrow\;2cx=y^2-c^2.

\boxed{\;y^2=2cx+c^2\;}\qquad(c\ \text{an arbitrary constant}).

Geometrically this is a one-parameter family of parabolas with axis along the $x$ -axis.

$ python3 -c "import sympy as sp; x,c=sp.symbols('x c'); y=sp.sqrt(c*(2*x+c)); p=sp.diff(y,x); print(sp.simplify(p**2*y**2+2*p*x*y-y**2))"
# 0

Substituting $y^2=2cx+c^2$ (so $y\,y'=c$ , $p=c/y$ ) into $yp^2+2xp-y$ gives $0$ identically. ✓