Solve the differential equation: ylog ydx-xdy=0

Worked Examples · Example 10

y\log y\,dx-x\,dy=0

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Concept understanding — Order and Degree of Differential Equations

Order and Degree of Differential Equations

The Intuition: What Are We Counting?

A differential equation is an equation that involves derivatives — rates of change. When you see something like

\frac{dy}{dx} = 3x^2

\frac{d^2y}{dx^2} + 5\frac{dy}{dx} + 6y = 0

you are looking at relationships between a function and its derivatives. The order and degree are two numbers that classify such equations. Think of them as the "size" and "shape" of the derivative information.

Order tells you the highest number of times the function has been differentiated. If the highest derivative is $\frac{dy}{dx}$ (first derivative), the order is 1. If it's $\frac{d^2y}{dx^2}$ (second derivative), the order is 2. Simple.

Degree is trickier. It tells you the power to which the highest-order derivative is raised — after the equation has been cleared of radicals and fractions involving derivatives. If the highest derivative appears squared, the degree is 2. If it appears inside a square root, you must first remove that root before declaring the degree.

The Precise Definitions

Important

Order of a differential equation is the order of the highest derivative present in the equation.

Important

Degree of a differential equation is the power of the highest-order derivative, provided the equation is a polynomial equation in all the derivatives (i.e., no fractional powers, no radicals, no trigonometric functions of derivatives).

The "provided" part is critical. You cannot read the degree directly from a messy equation — you must first rewrite it so that every derivative appears with a whole-number exponent.

Examples That Build Understanding

Example 1: $\frac{dy}{dx} + y = x$

Highest derivative is $\frac{dy}{dx}$ (first derivative). Order = 1. That derivative appears to the power 1. Degree = 1.

Example 2: $\left(\frac{d^2y}{dx^2}\right)^3 + 5\frac{dy}{dx} = \sin x$

Highest derivative is $\frac{d^2y}{dx^2}$ (second derivative). Order = 2. That derivative is raised to the power 3. Degree = 3.

Example 3: $\sqrt{1 + \left(\frac{dy}{dx}\right)^2} = \frac{d^2y}{dx^2}$

Here the highest derivative is $\frac{d^2y}{dx^2}$ (order 2). But the equation is not a polynomial in derivatives — there is a square root. To find the degree, square both sides:

1 + \left(\frac{dy}{dx}\right)^2 = \left(\frac{d^2y}{dx^2}\right)^2

Now the highest derivative $\frac{d^2y}{dx^2}$ appears with power 2. Degree = 2.

Watch out

Never read the degree from an equation that still has radicals, fractional powers, or trigonometric functions of derivatives. You must first make it a polynomial in the derivatives.

Example 4: $\frac{d^2y}{dx^2} = \sin\left(\frac{dy}{dx}\right)$

Order = 2. But the equation contains $\sin$ of a derivative — it is not a polynomial in derivatives at all. Degree is not defined for such equations. Many exam questions test exactly this: if the derivative appears inside a trigonometric, logarithmic, or exponential function, the degree is simply not defined.

A Quick Reference Table

Equation	Order	Degree	Notes
$\frac{dy}{dx} + y = 0$	1	1
$\left(\frac{d^2y}{dx^2}\right)^2 + y = x$	2	2
$\sqrt{\frac{dy}{dx}} + y = x$	1	1	Square both sides: $\frac{dy}{dx} = (x - y)^2$ , then degree = 1
$\frac{d^3y}{dx^3} + \left(\frac{dy}{dx}\right)^{1/2} = 0$	3	1	Clear the fractional power: raise to power 2
$\frac{d^2y}{dx^2} + \sin\left(\frac{dy}{dx}\right) = 0$	2	Not defined	Derivative inside a trigonometric function
$\left(\frac{d^2y}{dx^2}\right)^3 + \frac{dy}{dx} = e^x$	2	3

The One Rule That Saves You in Exams

Tip

When finding degree, always ask: "Is the equation a polynomial in the derivatives?" If yes, the degree is the exponent of the highest derivative. If no (because of roots, fractional powers, or trig/log of derivatives), the degree is not defined.

Order is almost always straightforward — just find the highest derivative. Degree is where students lose marks, and it is always because they forgot to clear radicals first.

Why This Matters

Order and degree are the first things you write when classifying a differential equation. They tell you what method of solution might work. A first-order equation is solved differently from a second-order one. A degree-1 equation is often linear; higher degrees usually mean nonlinearity. The classification is the starting point for everything that follows.

Separating gives $\dfrac{dy}{y\log y}=\dfrac{dx}{x}$ ; integrating yields $\log(\log y)=\log x+\log C$ , so $y=e^{Cx}$ .

u=\log y,\;du=\dfrac{dy}{y}

Given: $y\log y\,dx-x\,dy=0$ .

Rearrange: $y\log y\,dx=x\,dy\;\Rightarrow\;\dfrac{dx}{x}=\dfrac{dy}{y\log y}$ .
Integrate both sides: $\displaystyle\int\dfrac{dx}{x}=\int\dfrac{dy}{y\log y}$ .
LHS $=\log|x|$ . RHS: put $u=\log y\Rightarrow du=\dfrac{dy}{y}$ , so $\displaystyle\int\dfrac{du}{u}=\log|u|=\log|\log y|$ .
Thus $\log|\log y|=\log|x|+\log C=\log|Cx|$ .
Remove logs: $\log y=Cx$ .
Hence $y=e^{Cx}$ .

✓Final answer

\;\log y=Cx\;

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Q.Solve the differential equation: ylog⁡y dx−x dy=0y\log y\,dx-x\,dy=0ylogydx−xdy=0

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Q.Solve the differential equation:
$y\log y\,dx-x\,dy=0$

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