Q.Form a differential equation representing the family of parabolas having vertex at origin and axis along positive direction of y-axis.
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
dxdy=3x2
or
dx2d2y+5dxdy+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 dxdy (first derivative), the order is 1. If it's dx2d2y (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
Order of a differential equation is the order of the highest derivative present in the equation.
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: dxdy+y=x
Highest derivative is dxdy (first derivative). Order = 1. That derivative appears to the power 1. Degree = 1.
Example 2: (dx2d2y)3+5dxdy=sinx
Highest derivative is dx2d2y (second derivative). Order = 2. That derivative is raised to the power 3. Degree = 3.
Example 3: 1+(dxdy)2=dx2d2y
Here the highest derivative is dx2d2y (order 2). But the equation is not a polynomial in derivatives — there is a square root. To find the degree, square both sides:
1+(dxdy)2=(dx2d2y)2
Now the highest derivative dx2d2y appears with power 2. Degree = 2.
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: dx2d2y=sin(dxdy)
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 |
|---|---|---|---|
| dxdy+y=0 | 1 | 1 | |
| (dx2d2y)2+y=x | 2 | 2 | |
| dxdy+y=x | 1 | 1 | Square both sides: dxdy=(x−y)2, then degree = 1 |
| dx3d3y+(dxdy)1/2=0 | 3 | 1 | Clear the fractional power: raise to power 2 |
| dx2d2y+sin(dxdy)=0 | 2 | Not defined | Derivative inside a trigonometric function |
| (dx2d2y)3+dxdy=ex | 2 | 3 |
The One Rule That Saves You in Exams
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.
A parabola with vertex at the origin and axis along the positive y-axis has the one-parameter equation x2=4ay; differentiating once eliminates the constant a.
The family x2=4ay gives the differential equation xdxdy−2y=0 (i.e. xdxdy=2y).
A parabola with vertex at the origin and axis along +y is x2=4ay; eliminating a by differentiation gives xdxdy=2y.
A one-parameter family needs one differentiation to eliminate its single arbitrary constant, giving a first-order DE. Standard parabola about the y-axis: x2=4ay (a = parameter).
Set up the family: vertex (0,0), axis along the positive y-axis ⇒ x2=4ay, where a>0 is arbitrary. ...(1)
- Differentiate (1) w.r.t. x: 2x=4adxdy⇒4a=dy/dx2x.
- From (1), 4a=yx2.
- Equate the two expressions for 4a: yx2=dy/dx2x.
- Cross-multiply: x2dxdy=2xy.
- Divide by x (for x=0): xdxdy=2y.
xdxdy−2y=0 (order 1, degree 1).
- CBSE 2024Set 465/RQPS/41 markMCQQ.The order and the degree of the differential equation ydx+xlog(xy)dy−2xdy=0 are respectively : (A) 1,1 (B) 1,2 (C) 2,1 (D) 1, not defined
›Reveal solutionSolution
Only a first derivative appears, to the first power, so order =1 and degree =1.
Order = the highest-order derivative in the equation; Degree = the power of that highest derivative when the equation is a polynomial in the derivatives.
- Divide the equation ydx+xlog(xy)dy−2xdy=0 by dx: y+xlog(xy)dxdy−2xdxdy=0.
- The highest derivative is dxdy, so the order is 1.
- The term log(xy) contains no derivative — it is just a coefficient, so the equation stays polynomial in dxdy, which appears to the first power. Hence the degree is 1.
✓Final answer(A) 1,1
- CBSE 2024Set 465/S/RQPS/41 markMCQQ.Assertion (A) : The degree of the differential equation (dx2d2y)3+(dxdy)2+sin(dxdy)+1=0 is 3. Reason (R) : The highest power of the highest order derivative involved in a differential equation, when it is written as a polynomial in derivatives, is called its degree. (A) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A). (B) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A). (C) Assertion (A) is true, but Reason (R) is false. (D) Assertion (A) is false, but Reason (R) is true.
›Reveal solutionSolution
Treating the degree as the highest power (3) of the highest-order derivative, both statements are true and R correctly explains A — the answer the marking scheme accepts, with a noted caveat about the sin term.
Degree of a differential equation = the highest power of the highest-order derivative, after the equation is made a polynomial in its derivatives (it is undefined if that is impossible).
- The highest-order derivative here is dx2d2y (order 2), appearing with power 3 in (dx2d2y)3.
- Reading the degree directly as that power gives degree =3, so Assertion (A) is taken as true.
- Reason (R) states the standard definition of degree correctly, and it is exactly the rule used in step 1 — so R is the correct explanation of A.
- Honest caveat: strictly, the term sin(dxdy) is transcendental, so the equation is not a polynomial in its derivatives and its degree is technically not defined (which would make A false and point to option (D)). The CBSE marking scheme explicitly notes that option (A) "may be considered correct since students are not familiar with trigonometric functions," so the keyed answer is (A).
✓Final answer(A) Both A and R true and R is the correct explanation of A
- CBSE 2023Set 465/EF1GH/41 markMCQQ.Assertion (A) : The differential equation representing the family of parabolas y2=4ax, where 'a' is a parameter, is xdxdy−2y=0. Reason (R) : If the given family of curves has n parameters, then it is to be differentiated n times to eliminate the parameter and obtain the nth order differential equation. Select the correct answer from the codes (a), (b),(c) and(d) as given below.(a) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).(b) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).(c) Assertion (A) is true and Reason (R) is false.(d) Assertion (A) is false and Reason (R) is true.
›Reveal solutionSolution
Eliminating the single parameter a gives 2xy′−y=0, so the stated equation xy′−2y=0 is wrong — A false; R (n parameters ⇒ differentiate n times) true — answer (d).
To eliminate n arbitrary parameters, differentiate the family n times and eliminate; a one-parameter family yields a first-order differential equation.
- Family: y2=4ax (one parameter a). Differentiate once: 2ydxdy=4a, so 4a=2ydxdy.
- Substitute 4a back into y2=4ax: y2=(2ydxdy)x.
- Divide by y: y=2xdxdy, i.e. 2xdxdy−y=0.
- The Assertion's equation xdxdy−2y=0 does not match, so A is FALSE; the Reason correctly describes the elimination rule, so R is TRUE ⇒ code (d).
✓Final answer(d) Assertion (A) is false and Reason (R) is true
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