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MHT-CET Maths · Differential Equations

Order, Degree, Formation, and Verification

The order is the highest derivative present; the degree is the power of that highest derivative once the equation is made polynomial in its derivatives; n independent arbitrary constants force an order-n differential equation, which you build by differentiating and eliminating the constants — or verify by substituting a proposed solution back.

Why this matters

This is the entire MHT-CET differential-equations subtopic and it is a mark-bank: 31 PYQs sit here, spanning EASY definitional order/degree right up to HARD elimination of circle and parabola families. Two mechanical skills carry almost every question — read order/degree only AFTER clearing radicals and fractional powers, and form a family's equation by differentiating once per independent constant and eliminating. The recurring traps are exactly three: the degree is undefined when a derivative sits inside a log/trig, redundant constants (like C₃e^{x+C₄}) must be collapsed before you count the order, and only INDEPENDENT constants count.

Concept 1 of 9: Differential Equation Terminology

Before classifying anything, fix the vocabulary. A differential equation relates a function to its derivatives. Its order and degree are two independent labels; its solution comes in two flavours — a general solution carrying arbitrary constants, and a particular solution with those constants pinned down by conditions.

Definition

The vocabulary you must have cold:

  • Differential equation: an equation involving derivatives of an unknown function, e.g. dydx=3x\dfrac{dy}{dx} = 3x or d2ydx2+4y=0\dfrac{d^2y}{dx^2} + 4y = 0.
  • Order: the order of the highest derivative present.
  • Degree: the power of the highest-order derivative once the equation is polynomial in its derivatives.
  • Arbitrary constants: free parameters (a,b,c,C1,…a, b, c, C_1, \dots) in a solution family.
  • General solution: contains as many independent arbitrary constants as the order.
  • Particular solution: a general solution with its constants fixed by given conditions.

The master link

order of the ODE  =  number of independent arbitrary constants in its general solution\text{order of the ODE} \;=\; \text{number of independent arbitrary constants in its general solution}
  • orderorder of the highest derivative appearing
  • arbitrary constantsindependent free parameters in the solution family

Worked example

For y=c1e2x+c2e−3xy = c_1 e^{2x} + c_2 e^{-3x}, name the order of the differential equation it solves and the type of solution it is.
Practice this conceptself-check · 4 quick reps

Order and degree are separate labels

Order is about WHICH derivative is highest; degree is about the POWER on it. (d2ydx2)3=x\big(\tfrac{d^2y}{dx^2}\big)^3 = x is order 2 but degree 3. Don't conflate the two.

"Number of constants" means INDEPENDENT constants

Two constants that always merge into one (like c1+c3c_1 + c_3) count as a single arbitrary constant. Collapse the family first, then count — the order equals the number of constants that survive.

Concept 2 of 9: Order = Order of the Highest Derivative Present

Order is the easiest classifier to read: scan the equation for derivatives and pick the one differentiated the most times. A cubed second derivative is still order 2 — the power never touches the order.

Definition

Order of a differential equation == the order of the highest-order derivative that appears in it.

  • d2ydx2\dfrac{d^2y}{dx^2} present but no higher derivative ⇒\Rightarrow order 2, regardless of any power on it.
  • A high power on a LOW derivative does not raise the order: (dydx)5+d3ydx3=0\big(\tfrac{dy}{dx}\big)^{5} + \tfrac{d^3y}{dx^3} = 0 is order 3 (because d3ydx3\tfrac{d^3y}{dx^3} is present), not order 5.
  • Mixed powers of the same top derivative also leave the order alone.

Order

order=the order of the highest derivative appearing in the equation\text{order} = \text{the order of the highest derivative appearing in the equation}

Worked example

Find the order of x2d3ydx3+(dydx)4−y=0x^2\dfrac{d^3y}{dx^3} + \big(\dfrac{dy}{dx}\big)^{4} - y = 0.
Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

MHT-CET · 2023 · 3rd May Shift 2 · Q139Moderate

Example 2 · Differential Equations · Order, Degree, Formation of ODE, and Verification of Solutions

If order and degree of the differential equation (d2ydx2)5+4(d2ydx2)5⋅d3ydx3+d3ydx3=sin⁡x\left(\frac{d^{2}y}{dx^{2}}\right)^{5}+4\left(\frac{d^{2}y}{dx^{2}}\right)^{5}\cdot\frac{d^{3}y}{dx^{3}}+\frac{d^{3}y}{dx^{3}}=\sin x, are mm and nn respectively, then the value of m2+n2m^{2}+n^{2} is equal to

A power on the top derivative is DEGREE, never order

(d2ydx2)5\big(\tfrac{d^2y}{dx^2}\big)^{5} reads as "order 2, degree 5", not "order 5". The exponent belongs to degree; the order only counts how many times you differentiated.

Concept 3 of 9: Degree = Power of the Highest Derivative After Clearing Radicals

Degree is only meaningful once the equation is polynomial in its derivatives. So the first move is always to rationalize: raise both sides to a power that clears every root and fractional exponent. Read the degree only from the CLEAN equation, never the raw one.

Definition

To find the degree: 1. Clear all radicals and fractional powers on the derivatives (raise to a suitable power). 2. Once the equation is polynomial in the derivatives, the degree is the power on the highest-order derivative.

  • Example shape: y′′=y′−55\sqrt{y''} = \sqrt[5]{y' - 5} becomes (y′′)5=(y′−5)2(y'')^5 = (y'-5)^2 after raising to the 10th power ⇒\Rightarrow degree 5.
  • The LCM of the fractional exponents tells you the power to raise both sides to.

Degree

degree=power of the highest-order derivative, once the equation is polynomial in its derivatives\text{degree} = \text{power of the highest-order derivative, once the equation is polynomial in its derivatives}

Worked example

Find the order and degree of (d2ydx2)2/3=(1+dydx)1/2\big(\dfrac{d^2y}{dx^2}\big)^{2/3} = \big(1 + \dfrac{dy}{dx}\big)^{1/2}, and their sum.
Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

MHT-CET · 2025 · 20 April Shift I · Q149Moderate

Example 3 · Differential Equations · Order, Degree, Formation of ODE, and Verification of Solutions

The sum of the degree and order of the differential equation d2y dx2=dy dx−55\sqrt{\frac{d^{2}y}{\text{ }dx^{2}}}=\sqrt[5]{\frac{dy}{\text{ }dx}- 5} is

Clear radicals BEFORE you read the degree

The degree is NOT the fractional exponent you first see. For (d2ydx2)0.6=y′\big(\tfrac{d^2y}{dx^2}\big)^{0.6} = y', raise to the 5th power to get (d2ydx2)3=(y′)5\big(\tfrac{d^2y}{dx^2}\big)^{3} = (y')^5: degree =3= 3, not 0.60.6. Make it polynomial first.

Raise to the LCM of the fractional exponents

With a   \sqrt{\;} (power 12\tfrac12) and a   5\sqrt[5]{\;} (power 15\tfrac15), raise both sides to the 10th power in one shot — squaring alone leaves the 5th root, and 5th-powering alone leaves the square root.

Concept 4 of 9: When Degree Is Undefined (Derivative Inside a Transcendental)

You can only clear radicals and fractional powers by algebra. If a derivative is trapped inside a log, a trig, or an exponential, no amount of raising to powers makes the equation polynomial in its derivatives — so the degree simply does not exist. The order still does.

Definition

Degree is undefined when the equation cannot be made polynomial in its derivatives:

  • A derivative appears inside a transcendental function: log⁡ ⁣(d2ydx2)\log\!\big(\tfrac{d^2y}{dx^2}\big), sin⁡ ⁣(dydx)\sin\!\big(\tfrac{dy}{dx}\big), e y′′e^{\,y''}, etc.
  • Order is still well-defined in these cases — read it as usual (the highest derivative present).
  • Only radicals/fractional powers can be cleared; a derivative inside log⁡\log/sin⁡\sin/cos⁡\cos/e(⋅)e^{(\cdot)} is permanent.

Degree-undefined criterion

degree undefined  ⟺  a derivative sits inside a transcendental (log⁡, sin⁡, cos⁡, e(⋅))\text{degree undefined} \iff \text{a derivative sits inside a transcendental (}\log,\ \sin,\ \cos,\ e^{(\cdot)}\text{)}

Worked example

State the order and degree of d2ydx2=e dy/dx+x\dfrac{d^2y}{dx^2} = e^{\,dy/dx} + x.
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The same idea in a real exam question:

MHT-CET · 2025 · 22 April Shift I · Q114Easy

Example 4 · Differential Equations · Order, Degree, Formation of ODE, and Verification of Solutions

The degree of the differential equation d2y dx2+3( dy dx)2=x2log⁡( d2y dx2)\frac{d^{2}y}{\text{ }dx^{2}}+ 3\left( \frac{\text{ }dy}{\text{ }dx} \right)^{2}=x^{2}\log\left( \frac{{\text{ }d}^{2}y}{\text{ }dx^{2}} \right) is

Seeing a first power does NOT mean degree 1

For d2ydx2+sin⁡ ⁣(dydx)=0\tfrac{d^2y}{dx^2} + \sin\!\big(\tfrac{dy}{dx}\big) = 0, writing "degree 1" because d2ydx2\tfrac{d^2y}{dx^2} appears once is the trap. A derivative inside sin⁡\sin, cos⁡\cos, log⁡\log, or e(⋅)e^{(\cdot)} makes the degree UNDEFINED regardless of the visible power.

Order survives; only degree dies

"Degree undefined" never means "order undefined". Always still report the order — it is just the highest derivative present.

Concept 5 of 9: Collapse Redundant Arbitrary Constants Before Counting Order

The order of a family equals its number of INDEPENDENT arbitrary constants — but families are often written with fake extra constants that secretly merge. Simplify first: combine sums, absorb exponentials, and see how many truly-free constants remain. That count is the order.

Definition

Constants merge in predictable ways — spot and collapse them:

  • Sums merge: C1+C2→AC_1 + C_2 \to A (one constant), and C1+C3→AC_1 + C_3 \to A.
  • Exponential shifts absorb: C3ex+C4=(C3eC4)ex=BexC_3 e^{x + C_4} = (C_3 e^{C_4})e^x = B e^x — the C4C_4 vanishes into a single BB.
  • Same-form terms merge: (C1+C2)ex=Aex(C_1 + C_2)e^x = A e^x; two constants become one.
  • After collapsing, order = number of surviving independent constants.

Constant-absorption identity

C3 e x+C4=(C3eC4)ex=B exC_3\,e^{\,x + C_4} = \big(C_3 e^{C_4}\big)e^{x} = B\,e^{x}
  • Bthe single surviving constant after absorbing C3,C4C_3, C_4

Worked example

Find the order of the ODE whose general solution is y=C1+C2ex+C3ex+C4y = C_1 + C_2 e^x + C_3 e^{x + C_4}.
Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

MHT-CET · 2023 · 15th May Shift 2 · Q110Hard

Example 5 · Differential Equations · Order, Degree, Formation of ODE, and Verification of Solutions

The order of the differential equation, whose solution is y=C1+C2ex+C3ex+C4y = C_1 + C_2 e^x + C_3 e^{x+C_4}, is

ex+Ce^{x + C} hides a constant, it does not add one

C3ex+C4C_3 e^{x + C_4} LOOKS like two constants but is just BexB e^x — one constant. Counting C4C_4 separately over-states the order. Absorb every exponential shift before you count.

Only INDEPENDENT constants count

A sum like C1+C2C_1 + C_2 is a single free parameter. Two constants that can only ever appear as their sum contribute one to the order, not two.

Concept 6 of 9: Formation: n Independent Constants ⇒ Order-n Differential Equation

To build the differential equation of a family, you must get rid of every arbitrary constant. Each differentiation gives you one more equation to eliminate one constant — so a family with n independent constants needs n differentiations, producing an order-n equation. Count the constants first; that fixes the order before you compute anything.

Definition

The formation recipe:

  • Count the independent arbitrary constants nn in the family (collapse redundant ones first).
  • Differentiate the family nn times, then eliminate all nn constants using the original equation plus the derived equations.
  • The result is a differential equation of order nn, free of arbitrary constants.
  • The degree of that equation is read afterwards (clear radicals first).

Formation order

n independent arbitrary constants  ⟹  differential equation of order nn \text{ independent arbitrary constants} \;\Longrightarrow\; \text{differential equation of order } n

Worked example

What order and degree of differential equation represents the family of tangent lines to x2=4yx^2 = 4y?
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The same idea in a real exam question:

MHT-CET · 2025 · 19 April Shift I · Q114Moderate

Example 6 · Differential Equations · Order, Degree, Formation of ODE, and Verification of Solutions

The order and degree of differential equation of all tangent lines to the parabola x2=4yx^{2}= 4y is respectively.

Collapse constants BEFORE fixing the order

For all parabolas with axis parallel to Y, (x−h)2=4a(y−k)(x-h)^2 = 4a(y-k) has THREE independent constants h,a,kh, a, k — so its ODE is order 3 (d3ydx3=0\tfrac{d^3y}{dx^3} = 0). Miscounting the constants sets the wrong order from the start.

A fixed point removes a constant

All lines through a fixed point have only the slope free (order 1), while all lines in the plane have slope AND intercept free (order 2). Read what is fixed before counting.

Concept 7 of 9: Forming the Differential Equation of a Curve Family

Once you know the order equals the number of constants, the mechanics are pure elimination: differentiate the family, solve for a constant, and substitute back. Known functions like e^x stay in the equation — only the ARBITRARY constants must go. The visual: one equation with a free constant is a whole family of curves; the differential equation is the single rule they all obey.

Definition

For a family with constants, differentiate as many times as there are constants, then eliminate:

  • One constant: differentiate once, solve for the constant, substitute back.
  • Two constants (e.g. Ax2+By2=1Ax^2 + By^2 = 1): differentiate twice and eliminate A,BA, B, giving a second-order equation.
  • Keep known functions: in x2y=4ex+cx^2 y = 4e^x + c, the exe^x is a known function, NOT the arbitrary constant — only cc is eliminated, so exe^x survives in the answer.
  • For y=ex(a+bx+x2)y = e^x(a + bx + x^2): use y=exuy = e^x u, differentiate, and eliminate a,ba, b.

Elimination recipe

differentiate n times  →  solve for the constants  →  substitute back to eliminate them\text{differentiate } n \text{ times} \;\to\; \text{solve for the constants} \;\to\; \text{substitute back to eliminate them}
y = c·x²(one curve per c)eliminate c → x·y′ = 2y

Worked example

Form the differential equation of the family x2y=4ex+cx^2 y = 4e^x + c, where cc is arbitrary.
Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

MHT-CET · 2025 · 20 April Shift I · Q150Moderate

Example 7 · Differential Equations · Order, Degree, Formation of ODE, and Verification of Solutions

The differential equation whose solution represents the family x2y=4ex+cx^{2}y= 4e^{x}+ c, where c is an arbitrary constant, is

Eliminate the CONSTANT, not the known function

In x2y=4ex+cx^2 y = 4e^x + c only cc is arbitrary — the exe^x is a fixed function that survives differentiation. Dropping exe^x as if it were the constant gives the wrong equation. The correct ODE keeps the 4ex4e^x term.

Differentiate ONCE per constant — no more, no less

Ax2+By2=1Ax^2 + By^2 = 1 has two constants, so it needs TWO differentiations to eliminate both (giving xyy′′+x(y′)2−yy′=0xy y'' + x(y')^2 - yy' = 0). Stopping after one differentiation leaves a constant behind.

Concept 8 of 9: Forming the Differential Equation of Circles and Parabolas

Geometric families are just curve families with a geometric constraint that fixes some constants and frees others. The whole skill is translating the words ("centre on the X-axis", "touching the Y-axis", "vertex at origin, axis along +Y") into an equation with the RIGHT number of free constants, then eliminating them exactly as before.

Definition

Set up the standard form from the geometric description, then eliminate:

  • Circles, centre on X-axis, through origin: (x−a)2+y2=a2⇒x2+y2=2ax(x-a)^2 + y^2 = a^2 \Rightarrow x^2 + y^2 = 2ax; eliminate aa →\to y2=x2+2xy y′y^2 = x^2 + 2xy\,y' (order 1, one constant).
  • Circles through origin, centre on Y-axis: x2+y2=2ky⇒(x2−y2)y′−2xy=0x^2 + y^2 = 2ky \Rightarrow (x^2 - y^2)y' - 2xy = 0.
  • Circles touching Y-axis at origin, centre on X-axis: x2+y2=2hx⇒x2−y2+2xy y′=0x^2 + y^2 = 2hx \Rightarrow x^2 - y^2 + 2xy\,y' = 0.
  • Parabolas, vertex origin, axis along +Y: x2=4ay⇒xdydx=2yx^2 = 4ay \Rightarrow x\dfrac{dy}{dx} = 2y (one constant aa, order 1).
  • All parabolas, axis parallel to Y: three constants ⇒\Rightarrow order 3, d3ydx3=0\dfrac{d^3y}{dx^3} = 0.

Two workhorses

x2+y2=2ax  (circle)x2=4ay  (parabola, axis +Y)x^2 + y^2 = 2ax \;(\text{circle}) \qquad x^2 = 4ay \;(\text{parabola, axis } +Y)
  • athe single geometric parameter to eliminate by one differentiation

Worked example

Form the differential equation of all circles passing through the origin with centres on the X-axis.
Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

MHT-CET · 2024 · 10th May Shift 2 · Q142Moderate

Example 8 · Differential Equations · Order, Degree, Formation of ODE, and Verification of Solutions

The differential equation of all circles, passing through the origin and having their centres on the X-axis, is

Translate the geometry into the RIGHT free constants

"Centre on the X-axis and touching the Y-axis" fixes the centre as (a,0)(a,0) with radius ∣a∣|a| — ONE free constant, giving an order-1 equation. Treating it as a general circle (two/three constants) inflates the order and the answer.

Mind the sign when substituting the eliminated constant

For circles through the origin centred on the X-axis, substituting a=x+yy′a = x + yy' yields y2=x2+2xy y′y^2 = x^2 + 2xy\,y' — a plus sign. Careless algebra flips it to y2=x2−2xy y′y^2 = x^2 - 2xy\,y', which is a different (wrong) option.

Concept 9 of 9: Verifying a Solution and Identifying Its Family

Sometimes you are handed a candidate solution and asked to check it, find a constant that makes it fit, or say what curve it represents. The move is the reverse of formation: substitute the function (and its derivatives) into the differential equation and simplify — matching both sides confirms it, or reveals the unknown constant.

Definition

Three verification tasks, all by substitution:

  • Confirm a solution: compute y′,y′′y', y'' from the given yy, plug into the ODE, and check the equation holds identically.
  • Find a constant kk: for a PARAMETRIC solution x=x(t), y=y(t)x = x(t),\ y = y(t), use dydx=dy/dtdx/dt\dfrac{dy}{dx} = \dfrac{dy/dt}{dx/dt} and d2ydx2=ddx ⁣(dydx)\dfrac{d^2y}{dx^2} = \dfrac{d}{dx}\!\big(\tfrac{dy}{dx}\big) to substitute, then solve for kk.
  • Identify the family: solve/simplify the given ODE to its solution curve and name it (circle, hyperbola, ellipse, pair of lines).

Parametric derivative

dydx=dy/dtdx/dtd2ydx2=1dx/dtddt ⁣(dydx)\dfrac{dy}{dx} = \dfrac{dy/dt}{dx/dt} \qquad \dfrac{d^2y}{dx^2} = \dfrac{1}{dx/dt}\dfrac{d}{dt}\!\left(\dfrac{dy}{dx}\right)
  • tthe parameter — differentiate x and y with respect to it, then divide

Worked example

For x=sin⁡tx = \sin t, y=aet2+be−t2y = a e^{t\sqrt{2}} + b e^{-t\sqrt{2}}, find kk so that (1−x2)y′′−xy′=ky(1 - x^2)y'' - x y' = k y.
Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

MHT-CET · 2023 · 16th May Shift 2 · Q124Hard

Example 9 · Differential Equations · Order, Degree, Formation of ODE, and Verification of Solutions

The function y(x)y(x) represented by x=sin⁡tx = \sin t, y=aet2+be−t2y = ae^{t\sqrt{2}} + be^{-t\sqrt{2}}, t∈(−π2,π2)t \in \left(-\frac{\pi}{2}, \frac{\pi}{2}\right) satisfies the equation (1−x2)y′′−xy′=ky(1-x^2)y'' - xy' = ky, then the value of kk is

Convert parametric derivatives correctly

dydx≠dydt\dfrac{dy}{dx} \ne \dfrac{dy}{dt} — you must divide by dxdt\dfrac{dx}{dt}. For x=sin⁡tx = \sin t, dxdt=cos⁡t\dfrac{dx}{dt} = \cos t; skipping this factor is the most common error in find-kk questions and gives the wrong constant.

Identify the conic from the SIMPLIFIED solution

x2=c(1+y2)x^2 = c(1 + y^2) only becomes x2−y2=1x^2 - y^2 = 1 (a hyperbola) AFTER applying the given point to fix cc. Reading the conic type off the un-simplified, constant-carrying form is unreliable.

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (9)

  • Differential Equation Terminology

    The master link

    order of the ODE  =  number of independent arbitrary constants in its general solution\text{order of the ODE} \;=\; \text{number of independent arbitrary constants in its general solution}
  • Order = Order of the Highest Derivative Present

    Order

    order=the order of the highest derivative appearing in the equation\text{order} = \text{the order of the highest derivative appearing in the equation}
  • Degree = Power of the Highest Derivative After Clearing Radicals

    Degree

    degree=power of the highest-order derivative, once the equation is polynomial in its derivatives\text{degree} = \text{power of the highest-order derivative, once the equation is polynomial in its derivatives}
  • When Degree Is Undefined (Derivative Inside a Transcendental)

    Degree-undefined criterion

    degree undefined  ⟺  a derivative sits inside a transcendental (log⁡, sin⁡, cos⁡, e(⋅))\text{degree undefined} \iff \text{a derivative sits inside a transcendental (}\log,\ \sin,\ \cos,\ e^{(\cdot)}\text{)}
  • Collapse Redundant Arbitrary Constants Before Counting Order

    Constant-absorption identity

    C3 e x+C4=(C3eC4)ex=B exC_3\,e^{\,x + C_4} = \big(C_3 e^{C_4}\big)e^{x} = B\,e^{x}
  • Formation: n Independent Constants ⇒ Order-n Differential Equation

    Formation order

    n independent arbitrary constants  ⟹  differential equation of order nn \text{ independent arbitrary constants} \;\Longrightarrow\; \text{differential equation of order } n
  • Forming the Differential Equation of a Curve Family

    Elimination recipe

    differentiate n times  →  solve for the constants  →  substitute back to eliminate them\text{differentiate } n \text{ times} \;\to\; \text{solve for the constants} \;\to\; \text{substitute back to eliminate them}
  • Forming the Differential Equation of Circles and Parabolas

    Two workhorses

    x2+y2=2ax  (circle)x2=4ay  (parabola, axis +Y)x^2 + y^2 = 2ax \;(\text{circle}) \qquad x^2 = 4ay \;(\text{parabola, axis } +Y)
  • Verifying a Solution and Identifying Its Family

    Parametric derivative

    dydx=dy/dtdx/dtd2ydx2=1dx/dtddt ⁣(dydx)\dfrac{dy}{dx} = \dfrac{dy/dt}{dx/dt} \qquad \dfrac{d^2y}{dx^2} = \dfrac{1}{dx/dt}\dfrac{d}{dt}\!\left(\dfrac{dy}{dx}\right)

Watch out for (17)

Test yourself on Differential Equations

20 past MHT-CET questions from this chapter, timed at 36 minutes and marked the way the exam marks it. You see your score and every answer the moment you finish. Free to start.