Solving first order, first degree differential equations

What are the three methods?

In plain words: the first step is to check whether the equation is one of these three types. Not every first order equation is; some need another substitution, as Q13 shows. Each method ends in a general solution with one arbitrary constant C; a given condition, such as y = 1 when x = 0, may fix C and give a particular solution. Check that the condition is compatible with the equation and lies in the solution’s domain.

Variables separable

If F(x, y) can be expressed as a product g (x) h(y), where, g(x) is a function of x and h(y) is a function of y, then the differential equation (1) is said to be of variable separable type.

NCERT Mathematics Part II (Class 12), Section 9.4.1, page 307

Here H and G are anti derivatives of 1/h and g (NCERT page 307). The constant C goes in when you integrate, once, on one side.

NCERT’s examples (pages 308–310):

• dy/dx = (1 + y²)/(1 + x²): separating gives dy/(1 + y²) = dx/(1 + x²), so tan⁻¹ y = tan⁻¹ x + C (NCERT Example 5).
• dy/dx = −4xy² with y = 1 when x = 0: −1/y = −2x² + C, the condition gives C = −1, and y = 1/(2x² + 1) (NCERT Example 6).
• Money growing continuously at 5% a year: dP/dt = P/20, so P = 1000e^t/20 for Rs 1000 at t = 0. It doubles when e^t/20 = 2, that is, after 20 log_e 2 ≈ 13.86 years (NCERT Example 9).

Homogeneous equations

Homogeneous functions. A function F(x, y) is homogeneous of degree n if F(λx, λy) = λⁿF(x, y) for any nonzero constant λ (NCERT Section 9.4.2, page 312). Strictly, λ must be such that F is defined at (λx, λy) and λⁿ is defined, for example λ > 0 when n is a fraction such as ½ (our reasoning, beyond the NCERT definition). NCERT’s examples: y² + 2xy has degree 2, 2x − 3y degree 1, and cos (y/x) degree 0, while sin x + cos y is not homogeneous. A homogeneous function of degree n can be written xⁿg(y/x) (page 313).

Homogeneous equations. dy/dx = F(x, y) is homogeneous when F is homogeneous of degree zero, that is, F(x, y) = g(y/x) (page 313). The substitution y = vx turns it into a separable equation in v and x:

Work on an interval with x ≠ 0. Before dividing by g(v) − v, check its roots: each permissible constant root v = k gives a straight solution y = kx, which division can lose. At the end, put v = y/x back. NCERT’s Note (page 314): if the equation is dx/dy = h(x/y), put x = vy instead.

NCERT Example 13 (pages 319–320): curves whose tangent slope at (x, y) is (x² + y²)/(2xy). With y = vx, x dv/dx = (1 − v²)/(2v). Separating and integrating gives (1 − v²)x = C, that is, x² − y² = Cx.

Linear equations

dy/dx + Py = Q, where P and Q are constants or functions of x only, is a first order linear equation (NCERT Section 9.4.3, page 322). Examples: dy/dx + y = sin x, and dy/dx + y/x = e^x.

Use an interval where P and Q are continuous; singular points split the domain.

The integrating factor. Multiply the equation by g(x) = e^{∫P dx}. Then g′ = Pg, so the left side becomes g dy/dx + g′y, which is the derivative of y × g(x) by the product rule. This g is the integrating factor, I.F. (page 323).

NCERT slip (page 323): page 323 opens with “Choose g(x) in such a way that R.H.S. becomes a derivative of y . g (x)”. It is the left side, g(x) dy/dx + Pg(x)y, that becomes this derivative, as the next line and page 323’s own later sentence (“the L.H.S. becomes the derivative”) show. Read “L.H.S.”.

NCERT Example 14 (pages 324–325): for dy/dx − y = cos x, P = −1, so I.F. = e^−x. Then ye^−x = ∫e^−x cos x dx, which comes to e^−x(sin x − cos x)/2 + C. So y = (sin x − cos x)/2 + Ce^x.

Linear in x

Sometimes the equation is linear in x, with y as the independent variable: dx/dy + P₁x = Q₁, with P₁ and Q₁ constants or functions of y only (page 322). The same steps work with the letters swapped:

NCERT slip (page 324): in the 2026–27 PDF the integrating factor for dx/dy + P₁x = Q₁ shows as e to the power P₁ dy, with no integral sign; the file does not show whether the sign is missing or fails to display. It is e^{∫P₁ dy}, as in step (ii) for P and in NCERT Example 16 (page 326). In an exam, write the integral sign.

NCERT Example 16 (page 326): y dx − (x + 2y²) dy = 0 is not linear in y, but divided by y dy it is dx/dy − x/y = 2y. Here P₁ = −1/y, so I.F. = 1/y, and x/y = ∫2 dy = 2y + C. So x = 2y² + Cy on intervals with y ≠ 0. In the original differential-form equation, the horizontal solution y = 0 must also be checked separately; it cannot be expressed as a function x(y).

Solved examples

Common mistakes to avoid

Quick revision summary

• First order, first degree: dy/dx = F(x, y). Decide first whether it is separable, homogeneous or linear.
• Separable: write dy/h(y) = g(x) dx and integrate both sides, adding one C.
• Homogeneous (F a function of y/x): put y = vx, so dy/dx = v + x dv/dx, and separate v and x; for dx/dy = h(x/y), put x = vy.
• Linear, dy/dx + Py = Q: the integrating factor is I.F. = e^{∫P dx}, and then y × I.F. = ∫(Q × I.F.) dx + C. For dx/dy + P₁x = Q₁, swap the roles of x and y.

Questions students ask

Why is no constant of integration added when finding the integrating factor?

No constant is added in the integrating factor because any nonzero constant multiple of an integrating factor is one too: e^{∫P dx + k} is e^k × e^∫P dx, and the factor e^k, multiplying both y × I.F. and ∫(Q × I.F.) dx, divides out, leaving the same family with C/e^k as the arbitrary constant. For the same reason the modulus of a logarithm can go: for P = 3/x, e^3 log |x| = |x|³ is −x³ for x < 0, so I.F. = x³ serves on either side of 0, each solution holding on an interval without 0.

What is the difference between a homogeneous and a non-homogeneous differential equation?

The word “homogeneous” has two meanings for differential equations: in NCERT, dy/dx = F(x, y) is homogeneous when F is a homogeneous function of degree zero, a function of y/x alone, while beyond the NCERT text a linear equation dy/dx + Py = Q is called homogeneous when Q = 0 and non-homogeneous when Q ≠ 0. The two tests differ: dy/dx + 2y = 0 is a homogeneous linear equation but not homogeneous in NCERT’s sense, since F = −2y has degree 1, while dy/dx = 2y/x + 3 is homogeneous in NCERT’s sense but, written dy/dx − 2y/x = 3, a non-homogeneous linear equation. NCERT uses only the first meaning, the type solved by putting y = vx.

Can a differential equation be both separable and linear?

Yes, a first order equation can be both variable separable and linear: dy/dx + Py = Q separates exactly when one of P and Q is a constant multiple of the other, since Q = kP (k = 0 included) gives dy/dx = P(k − y) and P = 0 gives dy/dx = Q. For dy/dx + 4y = 8, the integrating factor e^4x gives ye^4x = 2e^4x + C, that is y = 2 + Ce^−4x; separating as dy/(8 − 4y) = dx, for y ≠ 2, gives the same curves, and the constant solution y = 2 lost by dividing is the case C = 0. Otherwise, as in dy/dx + y = x, the variables cannot be separated and the integrating factor is needed.

What is the difference between a linear and a nonlinear differential equation?

A first order differential equation is linear when it can be written dy/dx + Py = Q with P and Q constants or functions of x only (NCERT page 322): y and dy/dx occur only to the first power, never multiplied together or inside another function. Linearity concerns y, not x: dy/dx = x²y + sin x is linear, while y dy/dx = 2x, dy/dx = y² + x and dy/dx = sin y are nonlinear, though each has degree 1. Some equations nonlinear in y, such as NCERT Example 16, are linear in x.

What is a Bernoulli differential equation, and how is it made linear?

Beyond the NCERT text, a Bernoulli equation is dy/dx + Py = Qyⁿ, with P and Q functions of x and n ≠ 0, 1; dividing by yⁿ (for y ≠ 0) and putting z = y¹⁻ⁿ turns it into a linear equation in z. For dy/dx + y/x = xy², n = 2 and z = 1/y gives dz/dx − z/x = −x, whose integrating factor is 1/x for x > 0, so z/x = −x + C and y = 1/(Cx − x²), wherever Cx − x² ≠ 0. The solution y = 0, lost on dividing, has to be added separately.

Practice questions

Try each question first, then open the answer to check your working.

1. Q1. Solve sec² x tan y dx + sec² y tan x dy = 0 (NCERT Exercise 9.3, Q4, page 311).

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   Divide by tan x tan y: (sec² x/tan x) dx = −(sec² y/tan y) dy. Integrating, log |tan x| = −log |tan y| + log |C|, so tan x tan y = C. Keep cos x cos y ≠ 0. Direct substitution restores the constant solutions y = kπ, represented by C = 0; a vertical line is not a function y(x).

2. Q2. Solve y′ = (x + y)/x (NCERT Exercise 9.4, Q2, page 321).

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   With y = vx: v + x dv/dx = 1 + v, so dv = dx/x and v = log |x| + C. So y = x log |x| + Cx.

3. Q3. Solve dy/dx + y/x = x² (NCERT Exercise 9.5, Q3, page 328).

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   P = 1/x, so I.F. = e^{log x} = x (for x > 0). Then xy = ∫x³ dx, which is x⁴/4 + C.

4. Q4. The general solution of dy/dx = e^x+y is (A) e^x + e^−y = C; (B) e^x + e^y = C; (C) e^−x + e^y = C; (D) e^−x + e^−y = C (NCERT Exercise 9.3, Q23, page 312).

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   e^x+y = e^xe^y, so e^−y dy = e^x dx. Integrating, −e^−y = e^x + K, that is, e^x + e^−y = C: (A), on intervals where C − e^ˣ > 0 so that y = −log (C − e^ˣ) is real.