Linear and Bernoulli equations are the other two standard first-order types in the ordinary differential equations module of VTU Engineering Mathematics 1 (BMATS101 under the 2022 scheme, 18MAT11 under 2018). Each has a fixed structure and a fixed method, which makes them reliable marks once the steps are practised. This guide shows how to recognise each type, uses the integrating factor method for linear equations, reduces Bernoulli equations to linear form, and works through examples with the checks included.
The Linear First-Order Equation
A first-order equation is linear if it can be written as:
dy/dx + P(x) y = Q(x)
The unknown function y and its derivative appear only to the first power, and no product of y with its own derivative occurs. P(x) and Q(x) are functions of x alone.
The Integrating Factor Method
Multiply the equation by an integrating factor:
IF = e^(∫P(x) dx)
With this factor, the left side becomes the derivative of a product:
d/dx (y · IF) = Q(x) · IF
Integrate both sides with respect to x, then divide by IF to get y. Write the constant of integration when you integrate the right side.
The method has four steps:
- Write the equation in standard form dy/dx + P y = Q.
- Compute P(x), then IF = e^(∫P dx). Do not add a constant when computing this integral, because it cancels.
- Multiply through by IF and recognise the left side as d/dx(y · IF).
- Integrate and solve for y.
Worked Example 1: dy/dx + y = e^(−x)
Here P(x) = 1 and Q(x) = e^(−x).
IF = e^(∫1 dx) = eˣ.
Multiply: eˣ dy/dx + eˣ y = eˣ e^(−x) = 1.
The left side is d/dx (y eˣ), so d/dx(y eˣ) = 1.
Integrate: y eˣ = x + C.
y = (x + C) e^(−x)
Check: dy/dx = e^(−x) − (x + C) e^(−x) = e^(−x) − y. Rearranged, dy/dx + y = e^(−x) ✓.
Worked Example 2: dy/dx + 2xy = x
Here P(x) = 2x and Q(x) = x.
IF = e^(∫2x dx) = e^(x²).
The equation becomes d/dx (y e^(x²)) = x e^(x²).
Integrate the right side: ∫x e^(x²) dx = (1/2) e^(x²) + C (substitute t = x², so dt = 2x dx).
So y e^(x²) = (1/2) e^(x²) + C, and dividing by e^(x²):
y = 1/2 + C e^(−x²)
Check: the particular part 1/2 gives 2x · (1/2) = x, which matches the right side. The term C e^(−x²) satisfies the homogeneous equation dy/dx + 2xy = 0 ✓.
Worked Example 3: Initial Value Problem
Solve dy/dx + y tan x = sin x with y(0) = 1.
P(x) = tan x. Then ∫tan x dx = −ln|cos x|, so IF = e^(−ln|cos x|) = sec x.
Multiply: sec x · dy/dx + sec x tan x · y = sec x · sin x = tan x.
The left side is d/dx (y sec x). Integrate the right side: ∫tan x dx = −ln|cos x| + C.
So y sec x = −ln|cos x| + C, giving y = cos x (C − ln|cos x|).
Apply y(0) = 1: 1 = cos 0 (C − ln 1) = C, so C = 1.
y = cos x (1 − ln|cos x|)
Check at x = 0: y = 1 · (1 − 0) = 1 ✓.
The Bernoulli Equation
A Bernoulli equation has the form:
dy/dx + P(x) y = Q(x) yⁿ
It is not linear unless n = 0 or n = 1, because of the yⁿ term. The method is to make a substitution that turns it into a linear equation.
Divide by yⁿ and set v = y^(1−n). Then dv/dx = (1 − n) y^(−n) dy/dx. Substituting gives the linear equation:
dv/dx + (1 − n) P(x) v = (1 − n) Q(x)
Solve this linear equation for v using the integrating factor method, then substitute back y = v^(1/(1−n)).
Worked Example 4: dy/dx + y = x y²
Here P = 1, Q = x, and n = 2. The substitution is v = y^(1−2) = y^(−1) = 1/y.
Then dv/dx = −y^(−2) dy/dx. Multiply the equation by −y^(−2):
−y^(−2) dy/dx − y^(−1) = −x
Substitute dv/dx = −y^(−2) dy/dx and v = 1/y: dv/dx − v = −x.
This is linear in v. Here P = −1, so IF = e^(−x).
d/dx (v e^(−x)) = −x e^(−x).
Integrate the right side by parts: ∫ −x e^(−x) dx = (x + 1) e^(−x) + C.
So v e^(−x) = (x + 1) e^(−x) + C, giving v = x + 1 + C eˣ.
Substitute back v = 1/y:
y = 1 / (x + 1 + C eˣ)
Check: with v = x + 1 + C eˣ, dv/dx − v = (1 + C eˣ) − (x + 1 + C eˣ) = −x ✓.
How to Recognise Which Method to Use
- If the equation has the form dy/dx + P(x) y = Q(x), with no powers of y other than y¹, it is linear. Use the integrating factor.
- If it has the form dy/dx + P(x) y = Q(x) yⁿ with n ≠ 0 and n ≠ 1, it is Bernoulli. Use the substitution v = y^(1−n).
- If the variables separate, use separation, which is quicker. Separable equations are covered in our guide on separable and exact equations.
Common Mistakes That Cost Marks
Mistake 1: Forgetting to write the equation in standard form first. If the coefficient of dy/dx is not 1, divide through before finding P and Q. Otherwise the integrating factor is wrong.
Mistake 2: Adding the constant of integration too early. Find IF without a constant. Adding one at that stage changes the answer and is a frequent error.
Mistake 3: Using the wrong exponent in the Bernoulli substitution. The substitution is v = y^(1−n), not y^n or y^(n−1). Check the sign of 1 − n before substituting.
Mistake 4: Forgetting to substitute back. After solving the linear equation for v, convert back to y. The Bernoulli answer is in terms of y, not v.
Practice Questions
- Solve dy/dx + y/x = x². (Hint: P = 1/x, IF = x.)
- Solve dy/dx + y = y³ eˣ. (Hint: n = 3, v = y^(−2).)
- Solve dy/dx − y = eˣ with y(0) = 2.
Checks: (1) y = x³/4 + C/x. Multiply by x: d/dx(xy) = x³, so xy = x⁴/4 + C. (2) Using v = y^(−2): dv/dx − 2v = −2eˣ. IF = e^(−2x), so d/dx(v e^(−2x)) = −2e^(−x). Integrating gives v = 2eˣ + C e^(2x), that is y⁻² = 2eˣ + C e^(2x). Verify by substituting back. (3) IF = e^(−x), so d/dx(y e^(−x)) = 1, giving y = (x + C) eˣ. With y(0) = 2: C = 2, so y = (x + 2) eˣ.
An Applied Example: A Series RL Circuit
Linear first-order equations model many engineering systems. Consider a series circuit with resistance R, inductance L, and a constant voltage E. The current i(t) satisfies L di/dt + R i = E, with i(0) = 0.
Write the equation in standard form: di/dt + (R/L) i = E/L. Here P = R/L and Q = E/L, both constant. The integrating factor is e^(Rt/L).
Multiply through: d/dt (i e^(Rt/L)) = (E/L) e^(Rt/L).
Integrate: i e^(Rt/L) = (E/R) e^(Rt/L) + C. Divide by e^(Rt/L): i = E/R + C e^(−Rt/L).
Apply i(0) = 0: 0 = E/R + C, so C = −E/R. Therefore:
i(t) = (E/R)(1 − e^(−Rt/L))
Check: at t = 0, i = 0 ✓. As t grows, the exponential term vanishes and the current approaches E/R, which is the value predicted by Ohm’s law once the inductor has no further effect. The time constant L/R governs how quickly this happens, and it is a quantity engineers use directly.
Why the Method Works for Physical Problems
The integrating factor method is not just a classroom technique. It describes any system where a quantity changes at a rate proportional to a driving input minus a term proportional to itself, such as temperature change, drainage, and charging. Recognising the linear structure in a word problem is often the hardest step, so practise writing the equation from the physical description before solving it.
Connecting This Topic to the Full Module
Linear and Bernoulli equations form the second and third types of first-order equation in the ordinary differential equations module of VTU Engineering Mathematics 1. For the full module map, the topics most often tested, and the places where marks are lost, see our VTU Engineering Maths 1 Notes — Complete Module-Wise Guide (BMATS101).
Frequently Asked Questions
Q: Can I use the integrating factor method on a separable equation?
Yes, but separation is usually quicker. Use whichever method is simpler for the particular equation.
Q: Why does the integrating factor work?
Multiplying by e^(∫P dx) turns the left side into the derivative of a product. That is the only reason the method works, and it is worth understanding rather than memorising.
Q: Where does this fit in the module?
These are the second and third types of first-order equation after separable and exact. The full module outline is in our VTU Engineering Maths 1 notes guide.
For fully solved differential equation problems, see the Differential Equations Solved Problems page and the Engineering Math Survival Kit. To discuss batch options, use the contact page.