What is the solution of the differential equation.
$\frac{{ydx - xdy}}{{{y^2}}} = 0$ ?
Where c is an arbitrary constant.

Question

What is the solution of the differential equation.$\frac{{ydx - xdy}}{{{y^2}}} = 0$ ?Where c is an arbitrary constant.

Prepp Team · Accepted Answer

Understanding the Differential Equation Problem

The question asks us to find the solution to a given differential equation. A differential equation is an equation that relates a function with its derivatives. Solving it means finding the function that satisfies the equation. The equation provided is:

$\frac{{ydx - xdy}}{{{y^2}}} = 0$

Here, $ydx - xdy$ involves differentials of $x$ and $y$, and it's divided by $y^2$.

Solving the Differential Equation

Let's analyze the structure of the given differential equation. The left side of the equation, $\frac{{ydx - xdy}}{{{y^2}}}$, looks very similar to the differential of a quotient. Recall the differential rule for a quotient $\frac{u}{v}$:

$d\left(\frac{u}{v}\right) = \frac{{vdu - udv}}{{{v^2}}}$

Comparing this formula with the term $\frac{{ydx - xdy}}{{{y^2}}}$, we can see a direct correspondence:

Let $u = x$. Then $du = dx$.
Let $v = y$. Then $dv = dy$.
The denominator is $v^2 = y^2$.

So, the expression $\frac{{ydx - xdy}}{{{y^2}}}$ is exactly the differential of $\frac{x}{y}$.

Therefore, the given differential equation can be rewritten as:

$d\left(\frac{x}{y}\right) = 0$

To find the solution, we need to integrate both sides of this equation. Integrating the differential of a function simply gives the function back, plus an integration constant.

$\int d\left(\frac{x}{y}\right) = \int 0$

Integrating the left side gives $\frac{x}{y}$. Integrating the right side (0) gives a constant. Let's call this constant $C$.

$\frac{x}{y} = C$

Where $C$ is an arbitrary constant of integration.

Now, we need to rearrange this equation to match the format of the options provided. We can multiply both sides by $y$ (assuming $y \neq 0$):

$x = Cy$

Or, we can solve for $y$:

$y = \frac{x}{C}$

Since $C$ is an arbitrary constant, $\frac{1}{C}$ is also an arbitrary constant. Let's denote $\frac{1}{C}$ as $c$.

$y = cx$

This is the general solution to the differential equation, where $c$ is an arbitrary constant.

Comparing with Options

Let's compare our derived solution $y = cx$ with the given options:

Option 1: $xy = c$. This is not our solution.
Option 2: $y = cx$. This exactly matches our derived solution.
Option 3: $x + y = c$. This is not our solution.
Option 4: $x - y = c$. This is not our solution.

Thus, the solution to the differential equation $\frac{{ydx - xdy}}{{{y^2}}} = 0$ is $y = cx$.

Revision Table: Key Concepts for Differential Equations

Concept	Description	Example
Differential Equation	An equation involving an unknown function and its derivatives.	$\frac{dy}{dx} = 2x$
Solution of DE	A function that satisfies the differential equation when substituted.	$y = x^2 + C$ is a solution for $\frac{dy}{dx} = 2x$.
Exact Differential	A differential expression $P(x,y)dx + Q(x,y)dy$ that can be written as the total differential $dF$ of some function $F(x,y)$.	$2xydx + x^2dy = d(x^2y)$
Differential of Quotient	The rule $d\left(\frac{u}{v}\right) = \frac{{vdu - udv}}{{{v^2}}}$.	$d\left(\frac{x}{y}\right) = \frac{{ydx - xdy}}{{{y^2}}}$

Additional Information on Solving Differential Equations

Differential equations are classified based on their type, order, and linearity. The method used to solve them depends on these characteristics.

Separable Equations: Can be written in the form $f(y)dy = g(x)dx$ and integrated directly.
Exact Equations: Written as $P(x,y)dx + Q(x,y)dy = 0$ where $\frac{{\partial P}}{{\partial y}} = \frac{{\partial Q}}{{\partial x}}$. The solution is found by integrating $P$ and $Q$ partially.
Linear Equations: Can be written in the form $\frac{dy}{dx} + P(x)y = Q(x)$ and solved using an integrating factor.
Homogeneous Equations: Where $f(tx, ty) = t^n f(x,y)$. Often solved using the substitution $y = vx$.

The equation $\frac{{ydx - xdy}}{{{y^2}}} = 0$ is a simple example that can be solved by recognizing it as an exact differential, specifically the differential of the quotient $\frac{x}{y}$. This is a common technique in solving differential equations.

Prepp Team · Answer

xy = c

Prepp Team · Answer

x + y = c

Prepp Team · Answer

x &ndash; y = c

	Expenditure (in Rs.)
Items	Family A	Family B
Food	3,500	2,700
Clothing	500	800
Rent	1,500	1,000
Education	2,000	1,800
Miscellaneous	2,500	1,800

Age	Frequency
15 – 25	2
25 – 35	4
35 – 45	6
45 – 55	5
55 - 65	3

Understanding the Differential Equation Problem

Solving the Differential Equation

Comparing with Options

Revision Table: Key Concepts for Differential Equations

Additional Information on Solving Differential Equations

Similar Questions

Important Questions from Solution of Differential Equations

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Concept	Description	Example
Differential Equation	An equation involving an unknown function and its derivatives.	\(\frac{dy}{dx} = 2x\)
Solution of DE	A function that satisfies the differential equation when substituted.	\(y = x^2 + C\) is a solution for \(\frac{dy}{dx} = 2x\).
Exact Differential	A differential expression \(P(x,y)dx + Q(x,y)dy\) that can be written as the total differential \(dF\) of some function \(F(x,y)\).	\(2xydx + x^2dy = d(x^2y)\)
Differential of Quotient	The rule \(d\left(\frac{u}{v}\right) = \frac{{vdu - udv}}{{{v^2}}}\).	\(d\left(\frac{x}{y}\right) = \frac{{ydx - xdy}}{{{y^2}}}\)