First-order differential equations are the core of the ordinary differential equations module in VTU Engineering Mathematics 1 (BMATS101 under the 2022 scheme, 18MAT11 under 2018). Two methods cover a large share of the questions: variable separable equations and exact equations. This guide explains when each method applies, shows how to recognise the type of equation, and works through examples in full, including the exactness test and the common errors students make.
Recognising a First-Order Equation
A first-order differential equation involves the first derivative of the unknown function and no higher derivatives. The general form is dy/dx = f(x, y), or equivalently M(x, y) dx + N(x, y) dy = 0. Before choosing a method, rewrite the equation in one of these forms and check which structure it has.
Method 1: Variable Separable Equations
An equation is variable separable if it can be written as:
dy/dx = f(x) g(y)
Separate the variables by moving all y terms to one side and all x terms to the other:
dy / g(y) = f(x) dx
Integrate both sides and add a constant of integration once, on one side:
∫ dy/g(y) = ∫ f(x) dx + C
Worked Example 1: dy/dx = xy
Separate the variables: dy/y = x dx.
Integrate: ln|y| = x²/2 + C.
Exponentiate: y = A e^(x²/2), where A = ±e^C is an arbitrary non-zero constant.
You can check the result by differentiating. dy/dx = A · x · e^(x²/2) = x y ✓.
Notice that the constant is written as A, not as a sum. Because y = 0 is also a solution, you can allow A to take the value zero to include it.
Worked Example 2: Finding a Particular Solution
Solve dy/dx = (1 + y²)·x, given that y = 0 when x = 0.
Separate: dy/(1 + y²) = x dx.
Integrate: arctan y = x²/2 + C.
Apply the condition y(0) = 0: arctan 0 = 0 + C, so C = 0.
The particular solution is y = tan(x²/2).
Check: the derivative of tan(x²/2) is (x) sec²(x²/2) = x(1 + tan²(x²/2)) = x(1 + y²) ✓.
Method 2: Exact Differential Equations
Write the equation as M(x, y) dx + N(x, y) dy = 0. The equation is exact if:
∂M/∂y = ∂N/∂x
When this condition holds, there is a function F(x, y) with ∂F/∂x = M and ∂F/∂y = N. The solution is F(x, y) = C.
The method for finding F is:
- Integrate M with respect to x, treating y as a constant. Add an arbitrary function g(y).
- Differentiate the result with respect to y and set it equal to N. Solve for g′(y).
- Integrate g′(y) to find g(y), then write the solution F(x, y) = C.
Worked Example 3: Testing and Solving an Exact Equation
Solve (2xy) dx + (x² + 1) dy = 0.
Exactness test. M = 2xy, so ∂M/∂y = 2x. N = x² + 1, so ∂N/∂x = 2x. The two are equal, so the equation is exact.
Integrate M with respect to x. ∫ 2xy dx = x² y + g(y).
Differentiate with respect to y. ∂F/∂y = x² + g′(y). This must equal N = x² + 1, so g′(y) = 1, and g(y) = y.
Write the solution. F(x, y) = x² y + y = C.
Check: differentiating x² y + y gives (2xy) dx + (x² + 1) dy ✓.
Worked Example 4: An Equation That Needs an Integrating Factor
Test (y + x) dx + (x + 2y) dy = 0… Here ∂M/∂y = 1 and ∂N/∂x = 1, so this one is exact. Solve it using the method above: F = xy + x²/2 + y² = C.
Now consider 2y dx + x dy = 0. Here ∂M/∂y = 2 and ∂N/∂x = 1, so it is not exact. Multiplying by an integrating factor x gives 2xy dx + x² dy = 0, which is exact (∂M/∂y = 2x = ∂N/∂x). The solution is x² y = C.
Finding integrating factors is beyond the basic exact method, and your course may treat it as a separate topic. The example shows why the exactness test matters: a non-exact equation cannot be solved by the direct method.
How to Choose Between the Two Methods
- If the equation separates into a product of a function of x and a function of y, use separation.
- If it is written as M dx + N dy = 0 and the exactness test passes, use the exact method.
- If neither applies, check whether the equation is linear or Bernoulli, which are covered in a separate guide.
Common Mistakes That Cost Marks
Mistake 1: Losing solutions when dividing. Dividing by g(y) assumes g(y) is not zero. Check separately whether the values where g(y) = 0 give solutions.
Mistake 2: Adding a constant on both sides. Only one arbitrary constant is needed. Adding one on each side gives an equivalent but untidy answer.
Mistake 3: Skipping the exactness test. Applying the exact method without checking ∂M/∂y = ∂N/∂x gives a wrong answer. Always write the test before starting.
Mistake 4: Forgetting g(y) in the exact method. When you integrate M with respect to x, the constant of integration is a function of y, not a number. Omitting it loses the terms that depend only on y.
Practice Questions
- Solve dy/dx = y/x.
- Solve (3x² + 2xy) dx + (x² + y²) dy = 0.
- Solve dy/dx = e^(x+y) with y(0) = 0.
Checks: (1) Separate: dy/y = dx/x, so y = Ax. (2) M = 3x² + 2xy, ∂M/∂y = 2x; N = x² + y², ∂N/∂x = 2x. Exact. F = x³ + x²y + y³/3 = C. (3) Separate: e^(−y) dy = e^x dx, so −e^(−y) = e^x + C. With y(0) = 0: −1 = 1 + C, so C = −2. The solution is e^(−y) = 2 − e^x, or y = −ln(2 − e^x).
A Separable Example with a Rational Right Side
Solve dy/dx = x y / (x² + 1), with y(0) = 2.
Separate the variables: dy/y = x/(x² + 1) dx.
Integrate both sides. The left side gives ln|y|. For the right side, substitute t = x² + 1, so dt = 2x dx and x dx = dt/2. Then ∫ x/(x² + 1) dx = (1/2) ln(x² + 1) + C.
So ln|y| = (1/2) ln(x² + 1) + C, which gives y = A √(x² + 1), where A = ±e^C is a constant.
Apply y(0) = 2: 2 = A √1, so A = 2. The particular solution is y = 2√(x² + 1).
Check: dy/dx = 2 · x / √(x² + 1). Compare with x y / (x² + 1) = x · 2√(x² + 1) / (x² + 1) = 2x / √(x² + 1) ✓.
Why the Constant Must Be Carried Through
Students often drop the arbitrary constant after integrating the logarithm, or they write ln|y| = (1/2) ln(x² + 1) + C and then forget to exponentiate both sides properly. The exponential of a sum is a product, so e^C becomes a multiplicative constant A. Writing the constant as A from that point keeps the algebra clean and makes the initial condition easy to apply. Always apply the condition after you have written the general solution, not before.
Connecting This Topic to the Full Module
Variable separable and exact equations form the first topic of the ordinary differential equations module in VTU Engineering Mathematics 1. Mastering them makes the later linear and Bernoulli types easier to follow. The module outline, and the places where students most often lose marks, are in our VTU Engineering Maths 1 Notes — Complete Module-Wise Guide (BMATS101).
Frequently Asked Questions
Q: Is every first-order equation solvable by one of these methods?
No. Linear and Bernoulli equations need a separate method, and many equations need numerical methods in practice.
Q: How do I know if an equation is exact before I write the test?
Rewrite it as M dx + N dy = 0 first. Only then can you read off M and N and test them.
Q: Where does this fit in the module?
It is the first topic in the ordinary differential equations module. 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 batches, use the contact page.