CBSE Sample Paper 2023-24 · Section A · 1 mark

Q15: General solution of $y\,dx - x\,dy = 0$ ( $x, y \gt 0$ ⁠)

Sample 2023-24Ch 9 · Differential Equations

The question

The general solution of the differential equation $y\,dx - x\,dy = 0$ ⁠; (Given $x, y > 0$ ⁠), is of the form

(a) $xy = c$

(b) $x = cy^2$

(d) $y = cx^2$ ⁠;

(Where ' $c$ ' is an arbitrary positive constant of integration)

Exam answer

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

Why?

First add $x\,dy$ to both sides:

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

The variables can be separated: divide both sides of $x\,dy = y\,dx$ by $xy$ ⁠, which is not 0 since $x, y > 0$ ⁠.

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

Then every $y$ is with $dy$ and every $x$ is with $dx$ ⁠, so each side can be integrated on its own.

Integrating both sides ( $x, y, c > 0$ ⁠):

\begin{aligned} \log y &= \log x + \log c \\ &= \log cx \end{aligned}

Why?

Standard integral: $\int\frac{1}{y}\,dy = \log\lvert y\rvert$ ⁠, and $\lvert y\rvert = y$ since $y > 0$ ⁠. The same holds for $x$ ⁠.

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

The constant is written as $\log c$ so that the logs combine, using $\log m + \log n = \log mn$ ⁠.

$\therefore\ y = cx$

Why?

$\log$ is one-one, so equal logs mean equal numbers.

Check: $y = cx$ gives $dy = c\,dx$ ⁠, and then $y\,dx - x\,dy = cx\,dx - cx\,dx = 0$ ⁠.

Answer: (c) $y = cx$

Where this comes from

Chapter 9, Differential Equations: Variables separable: general solution.

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: From $x\,dy = y\,dx$ ⁠, divide by $xy$ ⁠: $\frac{dy}{y} = \frac{dx}{x}$ ⁠. With $x$ ⁠, $y$ and $c$ positive, $\log y = \log cx$ ⁠, so $y = cx$ ⁠.

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Differential Equations in past papers: question 22 of 31 · see them all

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

Q15: General solution of $y\,dx - x\,dy = 0$ ( $x, y \gt 0$ ⁠)