The general solution of a first-order differential equation contains:

Exactly two arbitrary constants

As many constants as there are variables

Newton's law of cooling translates into:

dT/dt = −k(T − T_s), where T_s is the surrounding temperature

Differential Equations

A differential equation is a statement about a rate rather than a value, and solving one means recovering the function from information about how it changes. For JEE the scope is narrow and therefore very manageable: first-order first-degree equations only, in four recognisable forms. The whole skill is classification — decide which of the four you are looking at, and the method follows mechanically. Marks are lost almost entirely in two places: misidentifying the form, and forgetting that the general solution of a first-order equation carries exactly one arbitrary constant.

Order, Degree and What Counts as a Solution

The order of a differential equation is the order of the highest derivative it contains, and the degree is the power of that highest derivative once the equation has been made free of radicals and fractions in the derivatives. So (d²y/dx²)³ + dy/dx = 0 has order 2 and degree 3, while √(dy/dx) = x must first be squared to give dy/dx = x², order 1 and degree 1.

⚠️ Degree is undefined when a derivative sits inside a transcendental function

An equation such as sin(dy/dx) + y = 0 or log(dy/dx) = x has an order — one — but no degree at all, because the derivative cannot be brought to a polynomial power by any algebraic manipulation. The same applies to e^dy/dx. So a question asking for the degree of such an equation is asking you to say that it is undefined, and the option list will offer 1 as a plausible answer. The rule to hold is that degree exists only when the equation is a polynomial in its derivatives.

A general solution contains as many arbitrary constants as the order of the equation, so a first-order equation has exactly one and a second-order equation two. A particular solution fixes those constants using given conditions. Read in reverse, this tells you the order of the equation a given family of curves satisfies: a family with two parameters, such as all circles through the origin with centres on the x-axis, satisfies a first-order equation only if one parameter is already pinned down.

The Four Forms and Their Methods

Recognise the form, then apply the method
Form	How to recognise it	Method
Variable separable	dy/dx = f(x)g(y)	Separate and integrate both sides
Homogeneous	dy/dx = f(y/x); every term has the same total degree	Substitute y = vx
Linear	dy/dx + Py = Q, with P and Q functions of x only	Multiply by the integrating factor e^{∫P dx}
Reducible to linear	dx/dy + Px = Q, with P and Q functions of y	Treat x as the dependent variable and repeat

🎯 Why the integrating factor works

For dy/dx + Py = Q the left side is almost the derivative of a product but not quite. Multiplying through by I = e^{∫P dx} fixes that: the left side becomes exactly d(Iy)/dx, because differentiating Iy gives I dy/dx + yI′ and I′ = PI by construction. So the equation collapses to d(Iy)/dx = IQ, which integrates directly to give Iy = ∫IQ dx. Understanding it this way means you never have to remember the final formula — you only have to remember to multiply by e^{∫P dx} and then recognise the product rule running backwards. It also explains why the equation must be written with the coefficient of dy/dx equal to 1 before P is read off, which is the commonest source of a wrong integrating factor.

🧠 Check homogeneity by counting the total degree of every term

An equation is homogeneous when x and y appear only in combinations of the same total degree, so that dy/dx can be written purely as a function of y/x. In (x² + y²)dx = 2xy dy every term is of degree two — x², y² and xy each count as two — so the substitution y = vx will work. But (x² + y)dx = xy dy is not homogeneous, because y alone is of degree one while x² is of degree two. Counting degrees takes a few seconds and prevents a substitution that will not simplify anything, which is the usual failure mode when the form is guessed rather than tested.

Applications appear as short word problems and all reduce to the same move: translate the sentence into a rate. "The rate of growth is proportional to the population" is dP/dt = kP, which separates to give exponential growth. "The rate of cooling is proportional to the excess temperature" is Newton's law of cooling, dT/dt = −k(T − T_s), which gives exponential decay towards the surroundings. Radioactive decay, compound interest and the draining of a tank all take one of these two shapes.

Key takeaways

Order is the highest derivative; degree is its power once radicals and fractions are cleared.
Degree is undefined when a derivative sits inside sin, log or an exponential.
A general solution has as many arbitrary constants as the order of the equation.
Variable separable: split and integrate. Homogeneous: substitute y = vx.
Linear: multiply by e^{∫P dx} so the left side becomes d(Iy)/dx, then integrate.
Normalise the coefficient of dy/dx to 1 before reading off P — otherwise the factor is wrong.
Test homogeneity by counting the total degree of every term, rather than guessing.
Word problems reduce to dP/dt = kP for growth or dT/dt = −k(T − T_s) for cooling.

Practice questions (8)

Attempt each one before opening the answer. Every explanation names the tempting wrong option as well as the right one, because that is where marks are lost.

The degree of the differential equation sin(dy/dx) + y = 0 is:
1. 1
2. 2
3. 0
4. Not defined
Show answer
Answer: D — Not defined
Degree is defined only when the equation is a polynomial in its derivatives, and no algebraic manipulation can bring sin(dy/dx) to a polynomial power. The order is 1, since the highest derivative is the first, but the degree simply does not exist. The option 1 is offered because the order is 1 and the two are easily conflated.
The integrating factor of dy/dx + 2y = eˣ is:
1. e^2x
2. e^−2x
3. eˣ
4. 2x
Show answer
Answer: A — e^{2x}
Here P = 2, so the integrating factor is e^∫2dx = e^2x. Multiplying through makes the left side d(ye^2x)/dx, which integrates directly. The sign matters: e^−2x would be the factor for dy/dx − 2y = eˣ, and mixing the two up is the most frequent error, closely followed by reading P off before normalising the coefficient of dy/dx to 1.
The substitution used to solve a homogeneous differential equation is:
1. y = vx
2. y = eˣ
3. x = vy only
4. No substitution is needed
Show answer
Answer: A — y = vx
A homogeneous equation can be written as dy/dx = f(y/x), so putting y = vx makes the right side a function of v alone and the equation becomes variable separable in v and x. The alternative x = vy also works and is sometimes cleaner, but y = vx is the standard first attempt. The essential prior step is to confirm homogeneity by checking that all terms share the same total degree.
The general solution of a first-order differential equation contains:
1. No arbitrary constant
2. Exactly one arbitrary constant
3. Exactly two arbitrary constants
4. As many constants as there are variables
Show answer
Answer: B — Exactly one arbitrary constant
The number of arbitrary constants in a general solution equals the order of the equation, because each integration introduces exactly one. A solution with no constant is a particular solution, obtained by imposing an initial condition. Read backwards, this also tells you the order of the equation satisfied by a given family of curves — a two-parameter family needs a second-order equation.
The order and degree of (d²y/dx²)³ + (dy/dx)⁵ = 0 are:
1. Order 2, degree 5
2. Order 2, degree 3
3. Order 3, degree 2
4. Order 1, degree 5
Show answer
Answer: B — Order 2, degree 3
The highest derivative present is the second, so the order is 2, and the degree is the power of that highest derivative, namely 3. The exponent 5 belongs to the first derivative and is irrelevant to the degree — taking the largest exponent in the equation regardless of which derivative it sits on is the standard error, and it is what the first option is there to catch.
Newton's law of cooling translates into:
1. dT/dt = kT
2. dT/dt = −k(T − T_s), where T_s is the surrounding temperature
3. dT/dt = −kt
4. T = kt + c
Show answer
Answer: B — dT/dt = −k(T − T_s), where T_s is the surrounding temperature
The rate of cooling is proportional to the excess of the body temperature over the surroundings, not to the temperature itself, so the driving quantity is the difference T − T_s and the minus sign makes the body cool towards its environment. Writing dT/dt = kT would describe a body whose cooling accelerated as it got colder, which is the opposite of what is observed.
Which equation is homogeneous?
1. (x² + y)dx = xy dy
2. (x² + y²)dx = 2xy dy
3. dy/dx = x + 3
4. dy/dx + y = eˣ
Show answer
Answer: B — (x² + y²)dx = 2xy dy
In the second option every term — x², y² and xy — has total degree two, so dy/dx can be written as a function of y/x alone and y = vx will separate the variables. The first option fails because y has degree one while x² has degree two. The third is directly separable and the fourth is linear, so each has its own method rather than the homogeneous substitution.
For 2 dy/dx + 4y = x, the integrating factor is:
1. e^4x
2. e^2x
3. e^x/2
4. e^−2x
Show answer
Answer: B — e^{2x}
The equation must first be divided through by 2 to give dy/dx + 2y = x/2, so P = 2 and the factor is e^2x. Reading P as 4 from the original form gives e^4x, which is the trap — the standard form requires the coefficient of dy/dx to be exactly 1 before P is identified, and skipping that normalisation is the single most common way to get an integrating factor wrong.