Exact Differential Equations
Differential Equations
12
JEE Qs
8%
Hard
60
min
Always verify the exactness condition ∂M/∂y = ∂N/∂x before proceeding and be meticulous with partial differentiation and integration steps, paying close attention to which variables are treated as constants.
🧮 Key Formulas
✅ Key Points for JEE
- 1The fundamental step is to correctly identify M(x, y) and N(x, y) from the given differential equation in the form M dx + N dy = 0.
- 2The exactness condition, ∂M/∂y = ∂N/∂x, must be verified before applying the solution method. If this condition is not met, the equation is not exact (unless an integrating factor is used, which is a related but separate topic).
- 3When integrating M(x,y) with respect to x, treat y as a constant. When integrating N(x,y) with respect to y, treat x as a constant.
- 4When using the solution formula, ensure you correctly identify 'terms of N that do not contain x' (or 'terms of M that do not contain y') to avoid double counting or missing terms.
- 5Always remember to add the arbitrary constant 'C' to the general solution.
⚠️ Common Mistakes
- ✕Incorrectly identifying M and N, especially if the equation is not explicitly given in the M dx + N dy = 0 form.
- ✕Making errors in partial differentiation when checking the exactness condition (∂M/∂y and ∂N/∂x).
- ✕Misapplying the integration step, such as incorrectly treating variables as constants or making mistakes in selecting which terms from N (or M) to integrate.
- ✕Forgetting to include the arbitrary constant of integration 'C' in the final solution.
📝 Practice Questions
See allQ12.Let f : R →R be a twice differentiable function such that f(x + y) = f(x)f(y) for all x, y ∈R. If f ′(0) = 4a and f satisfies f ′′(x) −3af ′(x) −f(x) = 0, a > 0, then the area of the region R = {(x, y) ∣0 ≤y ≤f(ax), 0 ≤x ≤2} is: (1) e2 −1 (2) e2 + 1 (3) e4 + 1 (4) e4 −1
Q24.Let y = f(x) be the solution of the differential equation dydx + x2−1xy = √1−x2x6+4x f(0) = 0. If 6 ∫1/2−1/2 f(x)dx = 2π −α then α2 is equal to _______ .
Q2. Let x = x(y) be the solution of the differential equation y2 dx + (x −1y )dy (1) 1 2 + e (2) 3 + e (3) 3 −e (4) 32 + e
Q9. Let f(x) be a real differentiable function such that f(0) = 1 and f(x + y) = f(x)f ′(y) + f ′(x)f(y) for all x, y ∈R. Then ∑100n=1 loge f(n) is equal to : (1) 2525 (2) 5220 (3) 2384 (4) 2406
Q16.If x = f(y) is the solution of the differential equation (1 + y2) + (x −2etan−1 y) dydx is equal to : f(0) = 1, then f ( √31 ) (1) eπ/12 (2) eπ/4 (3) eπ/3 (4) eπ/6
Q6. Let a curve y = f(x) pass through the points (0, 5) and (loge 2, k). If the curve satisfies the differential equation 2(3 + y)e2xdx −(7 + e2x)dy = 0, then k is equal to (1) 4 (2) 32 (3) 8 (4) 16
NCERT Chapters
- Class 12 Maths Ch 9: Differential Equations