Sample 2022-23 Class 12 Maths Q15 Solution: General solution of y dx − x dy = 0

The question

The general solution of the differential equation $y\,dx - x\,dy = 0$ is

(a) $xy = C$

(b) $x = Cy^2$

(c) $y = Cx$

(d) $y = Cx^2$

Exam answer

\begin{aligned} &y\,dx - x\,dy = 0 \\ \Rightarrow\ &x\,dy = y\,dx \\ \Rightarrow\ &\frac{dy}{y} = \frac{dx}{x} \end{aligned}

Why?

The variables separate: move $x\,dy$ to the other side, then divide both sides by $xy$ (for $x, y \ne 0$ ⁠). Now each side has only one variable.

\begin{aligned} y\,dx - x\,dy &= 0 \\ y\,dx &= x\,dy \\ \frac{y\,dx}{xy} &= \frac{x\,dy}{xy} \\ \frac{dx}{x} &= \frac{dy}{y} \end{aligned}

Integrating both sides, with $K > 0$ ⁠:

\begin{aligned} \log|y| &= \log|x| + \log K \\ \Rightarrow |y| &= K|x| \\ \Rightarrow y &= \pm Kx \end{aligned}

Why?

The standard integral is $\int\frac{dx}{x} = \log|x|$ ⁠. Writing the arbitrary constant as $\log K$ lets the logs combine:

\begin{aligned} \int\frac{dy}{y} &= \int\frac{dx}{x} \\ \log|y| &= \log|x| + \log K \end{aligned}

Then the law $\log m + \log n = \log(mn)$ ⁠:

\begin{aligned} \log|x| + \log K &= \log(K|x|) \end{aligned}

Equal logs mean equal numbers, so $|y| = K|x|$ ⁠. Removing the modulus gives $y = Kx$ or $y = -Kx$ ⁠.

$\therefore\ y = Cx$ ⁠, where $C = \pm K$ ⁠.

Why?

Check: if $y = Cx$ ⁠, then $dy = C\,dx$ and

\begin{aligned} &y\,dx - x\,dy \\ &= Cx\,dx - Cx\,dx = 0 \end{aligned}

None of the other options passes this check. For example, $y = Cx^2$ gives $Cx^2\,dx - 2Cx^2\,dx$ ⁠, which is not 0 for $C \ne 0$ ⁠.

Answer: (c) $y = Cx$

Where this comes from

Chapter 9, Differential Equations: Variable separable differential equation.

Question type: Variables separable

How to spot it: click to see

Type 3. Variables separable: $y$ on one side, $x$ on the other

Spot it: $\frac{dy}{dx}$ equals a part in $x$ times a part in $y$ ⁠, like $(1 + x^2)(1 + y^2)$ ⁠.

Here: Rearrange to $x\,dy = y\,dx$ and divide by $xy$ ⁠: $\frac{dy}{y} = \frac{dx}{x}$ ⁠. Both sides give logs, so $y = Cx$ ⁠.

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Q15: General solution of y dx−x dy=0y\,dx - x\,dy = 0ydx−xdy=0