JEE Main 2005

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Animated Solution for Mathematics - Differential Equations: If $x \frac{d y}{d x} = y (lo g y - lo g x + 1)$ , then the solution of the equation is

Q: How do I know for sure if it is a homogeneous differential equation?

A differential equation $\frac{dy}{dx} = F(x, y)$ is homogeneous if the function $F(x, y)$ can be written as a function of the ratio $\frac{y}{x}$ (or $\frac{x}{y}$). In our case, after using the log property, we got $\frac{y}{x}(\log(\frac{y}{x}) + 1)$. Since every term is a function of $\frac{y}{x}$, it is perfectly homogeneous. It is a beautiful way to reduce a complex equation into a separable one!

Select Answer:

Visualized Solution

Identify the Differential Equation

We are given the first-order differential equation: $x \frac{d y}{d x} = y (lo g y - lo g x + 1)$
Let's visualize the family of solution curves on the coordinate plane.
Our goal is to find the general solution that represents these curves.

Apply Logarithm Properties

Recall the quotient rule of logarithms: $lo g a - lo g b = lo g (\frac{a}{b})$
Applying this to $lo g y - lo g x$ , we get: $lo g (\frac{y}{x})$
Substitute this back into the original equation: $x \frac{d y}{d x} = y (lo g (\frac{y}{x}) + 1)$

Isolate the Derivative $\frac{d y}{d x}$

Divide both sides of the equation by $x$ to isolate the derivative term:
$\frac{d y}{d x} = \frac{y}{x} (lo g (\frac{y}{x}) + 1)$
Observe that the entire right-hand side is now a function of the single ratio $\frac{y}{x}$ .

Recognize Homogeneous Equation

A differential equation of the form $\frac{d y}{d x} = f (\frac{y}{x})$ is called a Homogeneous Differential Equation.
This suggests a standard substitution: let $v = \frac{y}{x}$ , which means $y = v x$ .

Differentiate $y = v x$

We substitute $y = v x$ .
Differentiating both sides with respect to $x$ using the Product Rule:
$\frac{d y}{d x} = v \cdot \frac{d}{d x} (x) + x \cdot \frac{d v}{d x}$
$\frac{d y}{d x} = v + x \frac{d v}{d x}$

Substitute into the Differential Equation

Substitute $\frac{d y}{d x} = v + x \frac{d v}{d x}$ and $\frac{y}{x} = v$ into our isolated equation:
$v + x \frac{d v}{d x} = v (lo g v + 1)$

Expand and Simplify

Expand the right-hand side:
$v + x \frac{d v}{d x} = v lo g v + v$
Subtract $v$ from both sides:
$x \frac{d v}{d x} = v lo g v$

Variable Separation Method

Rearrange the terms to group all $v$ terms on one side and $x$ terms on the other:
$\frac{1}{v l o g v} d v = \frac{1}{x} d x$

Set Up the Integrals

Integrate both sides of the separated equation:
$\int \frac{1}{v l o g v} d v = \int \frac{1}{x} d x$

Integration by Substitution

To evaluate $\int \frac{1}{v l o g v} d v$ , use substitution:
Let $t = lo g v ⟹ d t = \frac{1}{v} d v$
The integral becomes:
$\int \frac{1}{t} d t = lo g ∣ t ∣ = lo g ∣ lo g v ∣$

Integrate and Add Constant

The integral of the right-hand side is:
$\int \frac{1}{x} d x = lo g ∣ x ∣$
Add the constant of integration in logarithmic form, $lo g c$ , to make simplification easier:
$lo g (lo g v) = lo g x + lo g c$

Simplify Using Log Properties

Use the product rule of logarithms: $lo g a + lo g b = lo g (ab)$
$lo g (lo g v) = lo g (c x)$
Taking the antilog (exponentiating) both sides:
$lo g v = c x$

Substitute Back $v = \frac{y}{x}$

Recall our original substitution: $v = \frac{y}{x}$
Substitute this back into the simplified equation:
$lo g (\frac{y}{x}) = c x$
This matches Option 3.

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The Sigma Insight: Homogeneous Differential Equations

Solution Diagram

Analyzing the Setup

The given differential equation is:

x \frac{d y}{d x} = y (lo g y - lo g x + 1)

To simplify the expression, we apply the logarithmic quotient rule,

lo g a - lo g b = lo g (\frac{a}{b})

. This transforms the equation into:

x \frac{d y}{d x} = y (lo g (\frac{y}{x}) + 1)

Dividing both sides by

x

, we isolate the derivative:

\frac{d y}{d x} = \frac{y}{x} (lo g (\frac{y}{x}) + 1)

The structure of the right-hand side, being a function solely of the ratio

\frac{y}{x}

, confirms that this is a Homogeneous Differential Equation.

The Substitution Dance

To solve this, we use the standard substitution

v = \frac{y}{x}

, which implies

y = v x

. Differentiating with respect to

x

using the product rule, we obtain:

\frac{d y}{d x} = v + x \frac{d v}{d x}

Substituting these into our differential equation yields:

v + x \frac{d v}{d x} = v (lo g v + 1)

Expanding the right side gives

v + x \frac{d v}{d x} = v lo g v + v

. The

v

terms cancel out perfectly, leaving us with the separable form:

x \frac{d v}{d x} = v lo g v

The Integration Finale

We now separate the variables by grouping the

v

terms and

x

terms:

\frac{1}{v lo g v} d v = \frac{1}{x} d x

Integrating both sides:

\int \frac{1}{v lo g v} d v = \int \frac{1}{x} d x

For the left integral, we use the substitution

t = lo g v

, which implies

d t = \frac{1}{v} d v

. The integral becomes:

\int \frac{1}{t} d t = lo g ∣ t ∣ = lo g ∣ lo g v ∣

Equating this to the integral of the right side plus a constant

lo g c

lo g (lo g v) = lo g x + lo g c = lo g (c x)

Taking the antilog of both sides, we find

lo g v = c x

. Substituting

v = \frac{y}{x}

back into the equation, we reach the final solution:

$lo g (\frac{y}{x}) = c x$

Common Traps & FAQs

Wait, why do we use the substitution

v = \frac{y}{x}

instead of

v = \frac{x}{y}

Great question! We look at the structure of the derivative. Our equation is

\frac{d y}{d x} = f (\frac{y}{x})

. Since the derivative is

\frac{d y}{d x}

, it is mathematically standard and much easier to express

y

in terms of

v

and

x

(i.e.,

y = v x

) so that

\frac{d y}{d x}

becomes

v + x \frac{d v}{d x}

. If you used

v = \frac{x}{y}

, you would be dealing with

\frac{d x}{d y}

, which would unnecessarily complicate the algebra.

How do I know for sure if it is a homogeneous differential equation?

A differential equation

\frac{d y}{d x} = F (x, y)

is homogeneous if the function

F (x, y)

can be written as a function of the ratio

\frac{y}{x}

(or

\frac{x}{y}

). In our case, after using the log property, we got

\frac{y}{x} (lo g (\frac{y}{x}) + 1)

. Since every term is a function of

\frac{y}{x}

, it is perfectly homogeneous. It is a beautiful way to reduce a complex equation into a separable one!

Why did we write the constant of integration as

lo g c

instead of just

c

This is a pro-tip for JEE! When both sides of your equation involve logarithms (like

lo g (lo g v) = lo g x + \dots

), adding a constant

lo g c

allows you to use the property

lo g A + lo g B = lo g (A B)

. It makes the final step of removing the logarithms much cleaner and avoids messy exponential terms like

e^{c}

Animated Solution for Mathematics - Differential Equations: If $x \frac{d y}{d x} = y (lo g y - lo g x + 1)$ , then the solution of the equation is

Select Answer:

Visualized Solution

Identify the Differential Equation

Apply Logarithm Properties

Isolate the Derivative $\frac{d y}{d x}$

Recognize Homogeneous Equation

Differentiate $y = v x$

Substitute into the Differential Equation

Expand and Simplify

Variable Separation Method

Set Up the Integrals

Integration by Substitution

Integrate and Add Constant

Simplify Using Log Properties

Substitute Back $v = \frac{y}{x}$

The Sigma Insight: Homogeneous Differential Equations

Analyzing the Setup

The Substitution Dance

The Integration Finale

Similar Questions

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Animated Solution for Mathematics - Differential Equations: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If xdxdy​=y(logy−logx+1), then the solution of the equation is

Select Answer:

Visualized Solution

Identify the Differential Equation

Apply Logarithm Properties

Isolate the Derivative .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } dxdy​

Recognize Homogeneous Equation

Differentiate .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } y=vx

Substitute into the Differential Equation

Expand and Simplify

Variable Separation Method

Set Up the Integrals

Integration by Substitution

Integrate and Add Constant

Simplify Using Log Properties

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The Sigma Insight: Homogeneous Differential Equations

Analyzing the Setup

The Substitution Dance

The Integration Finale

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Animated Solution for Mathematics - Differential Equations: If $x \frac{d y}{d x} = y (lo g y - lo g x + 1)$ , then the solution of the equation is

Isolate the Derivative $\frac{d y}{d x}$

Differentiate $y = v x$

Substitute Back $v = \frac{y}{x}$

The solution curve of the differential equation $y \frac{d x}{d y} = x (lo g_{e} x - lo g_{e} y + 1), x > 0, y > 0$ passing through the point $(e, 1)$ is

The solution of the differential equation $\frac{d y}{d x} = \frac{x + y}{x}$ satisfying the condition $y (1) = 1$ is

The solution of the differential equation $\frac{d y}{d x} = - (\frac{x ^{2} + 3 y ^{2}}{3 x ^{2} + y ^{2}}), y (1) = 0$ is

If a curve $y = f (x)$ , passing through the point $(1, 2)$ , is the solution of the differential equation, $2 x^{2} d y = (2 x y + y^{2}) d x$ , then $f (\frac{1}{2})$ is equal to :

The solution of the differential equation $(x^{2} + y^{2}) d x - 5 x y d y = 0, y (1) = 0$ , is :

If $y = y (x)$ is the solution of the differential equation $2 x^{2} \frac{d y}{d x} - 2 x y + 3 y^{2} = 0$ such that $y (e) = \frac{e}{3}$ , then $y (1)$ is equal to

Let the solution curve of the differential equation $x d y - y d x = x^{2} + y^{2} d x, x > 0, y (1) = 0$ , be $y = y (x)$ . Then $y (3)$ is equal to

If $\frac{d y}{d x} = \frac{x y}{x ^{2} + y ^{2}}$ ; $y (1) = 1$ : then a value of $x$ satisfying $y (x) = e$ is: