Exercise 9.5 of Class 12 Maths Chapter 9 Differential Equations covers linear differential equations and the integrating factor method. These NCERT Solutions show every step in full. The free PDF download is available on this page.
Question count: 19 problems, where Q1 to Q12 ask for the general solution, Q13 to Q15 for a particular solution, Q16 to Q17 are curve word problems, and Q18 to Q19 are integrating-factor MCQs.
Exercise 9.5 contains the only place in Chapter 9 where the sister form dxdy + P1x = Q1 appears, used in Q10, Q11 and Q12, and tested directly by the Q19 MCQ.
Every solved problem in this Collegedunia PDF rewrites the equation in standard linear form, computes the integrating factor μ = e∫ P dx, multiplies through so the left-hand side becomes ddx(μ y) , and integrates once. The Collegedunia editorial team has matched every general and particular solution against the official NCERT answer key and the 2026-27 textbook.
How Collegedunia's NCERT Solutions Help You Clear Exercise 9.5
Exercise 9.5 marks are lost in two places: misidentifying P when the equation is not yet in standard form, and integrating the right-hand side after multiplying by the integrating factor. The Collegedunia solutions isolate both steps.
Standard-form rewrite first, so P and Q are read off correctly.
Integrating factor computed explicitly as μ = e∫ P dx before any multiplication.
Left side collapsed to ddx(μ y) , the step that justifies a single integration.
Sister form handled for Q10 to Q12 using μ = e∫ P1 dy on dxdy + P1x = Q1.
Initial condition applied last for Q13 to Q15 to fix the constant.
The table lists the headline answer for representative problems across the four blocks of Exercise 9.5. Use it to verify your own working.
Q No.
Equation / condition
Answer
1
dydx + 2y = sin x
y = 2sin x - cos x5 + Ce-2x
2
dydx + 3y = e-2x
y = e-2x + Ce-3x
3
dydx + yx = x2
y = x34 + Cx
5
cos2x dydx + y = tan x
y = tan x - 1 + Ce-tan x
8
(1+x2) dy + 2xy dx = cot x dx
y(1+x2) = log|sin x| + C
10
(x+y)dydx = 1
x + y + 1 = Cey
13
dydx + 2ytan x = sin x, y(π/3)=0
y = cos x - 2cos2x
15
dydx - 3ycot x = sin 2x , y(π/2)=2
y = 4sin3x - 2sin2x
16
Slope equals sum of coordinates, through origin
y + x + 1 = ex
18
MCQ: IF of xdydx - y = 2x2
(C) 1x
19
MCQ: IF of (1-y2)dxdy + yx = ay
(D) 1√1-y2
The particular-solution trio Q13 to Q15 collapses to clean trigonometric answers, which is why they are the standard 5-mark Long Answer source. A linear-equation question has appeared in 4 of the last 5 CBSE Class 12 Maths papers.
The Integrating Factor Routine for Class 12 Maths Exercise 9.5
Every problem in Exercise 9.5 follows the same four-step routine. Apply it in order.
Step 1. Rewrite the equation in standard linear form dydx + P(x)y = Q(x) and read off P and Q. Step 2. Compute the integrating factor μ = e∫ P dx. Step 3. Multiply through by μ ; the left side becomes ddx(μ y) . Step 4. Integrate once: μ y = ∫ μ Q dx + C. Apply any initial condition to fix C.
When the equation is naturally in x as a function of y, use the sister form with μ = e∫ P1 dy. Q19 tests exactly this recognition.
Common Mistakes Students Make in Class 12 Maths Exercise 9.5
Common Mistake: Reading P off before dividing the equation into standard form. In Q5, cos2x dydx + y = tan x must first be divided by cos2x so that P = sec2x . Skipping this gives the wrong integrating factor and a wrong answer.
Computing μ from a non-standard form, so P is misidentified.
Forgetting that the left side collapses to ddx(μ y) , and integrating term by term instead.
Using the x-form integrating factor when the equation is in y, or vice versa (Q10 to Q12, Q19).
Applying the initial condition before the general solution is complete.
Other Resources for Class 12 Maths Chapter 9 Differential Equations
All NCERT Solutions for Differential Equations Ex 9.5 with Step-by-Step Working
Every NCERT textbook question for Class 12 Mathematics Chapter 9 Differential Equations Ex 9.5 is listed below with its full Solution and Expert Solution hidden inside collapsible tabs. Click Check Solution to reveal the step-by-step working; click Expert Solution for the expanded explanation.
Questions
Q 9.1
dydx + 2y = sin x.
Concept used. The equation is linear with P(x) = 2, Q(x) = sin x. The integrating factor is μ = e∫ 2 dx = e2x. Multiplying gives ddx(e2xy) = e2xsin x, after which we integrate.
Identify P=2, Q=sin x. IF: μ = e2x.
Multiply the DE by e2x:
e2xy' + 2e2xy = e2xsin xddx(e2xy) = e2xsin x.
Integrate: e2xy = ∫ e2xsin x dx + C.
Use the standard ∫ eaxsin bx dx = eax(asin bx - bcos bx)a2+b2 with a=2, b=1:
∫ e2xsin x dx = e2x(2sin x - cos x)5.
Therefore e2xy = e2x(2sin x - cos x)5 + C, so
y = 2sin x - cos x5 + Ce-2x.
y = 2sin x - cos x5 + Ce-2x.
AS
Aarav Sharma
M.Sc Mathematics, IIT Bombay
Verified Expert
Strategic angle. The signature recipe: identify P, compute μ=e∫ P dx, multiply, recognise the LHS as a derivative, integrate.
μ = e2x.
(e2xy)' = e2xsin x.
∫ e2xsin x dx = e2x(2sin x - cos x)5.
y = 2sin x - cos x5 + Ce-2x.
y = 2sin x - cos x5 + Ce-2x.
Q 9.2
dydx + 3y = e-2x.
Concept used. Linear with P=3, Q=e-2x. IF: μ = e3x.
μ = e3x, so (e3xy)' = e3x· e-2x = ex.
Integrate: e3xy = ∫ ex dx + C = ex + C.
Solve for y: y = e-2x + Ce-3x.
y = e-2x + Ce-3x.
SI
Sneha Iyer
M.Sc Mathematics, ISI Kolkata
Verified Expert
Quick reading. Multiplying by e3x turns the RHS into ex, which integrates to ex.
μ = e3x.
(e3xy)' = ex, hence e3xy = ex+C.
y = e-2x+Ce-3x.
y = e-2x+Ce-3x.
Q 9.3
dydx + yx = x2.
Concept used. Linear with P = 1/x, Q = x2. IF: μ = e∫ dx/x = elog x = x (taking x>0).
μ = x. Multiply:
xy' + y = x3 (xy)' = x3.
Integrate: xy = ∫ x3 dx + C = x44 + C.
Solve for y: y = x34 + Cx.
y = x34 + Cx.
AP
Arjun Patel
M.Tech CS, IIT Madras
Verified Expert
Quick reading. IF = x. Multiplied DE: (xy)' = x3; integrate.
μ = x, so (xy)' = x3.
xy = x4/4 + C.
y = x3/4 + C/x.
y = x3/4 + C/x.
Q 9.4
dydx + (sec x) y = tan x (0≤ x < π/2).
Concept used. Linear with P = sec x, Q = tan x. IF: μ = e∫ sec x dx = elog|sec x + tan x| = sec x + tan x.
Compute μ:
μ = e∫ sec x dx = sec x + tan x.
Multiply the DE by μ to get (μ y)' = x.
Compute x = (sec x + tan x)tan x = sec xtan x + tan2x = sec xtan x + sec2x - 1.
Integrate: ∫ (sec xtan x + sec2x - 1) dx = sec x + tan x - x.
Hence (sec x + tan x)y = sec x + tan x - x + C.
y(sec x + tan x) = sec x + tan x - x + C.
PG
Priya Gupta
Ph.D Mathematics, IIT Delhi
Verified Expert
Strategic angle. The integrating factor sec x + tan x is one of the must-memorise IFs in NCERT problems.
μ = sec x + tan x.
RHS: x = sec xtan x + sec2x - 1.
Integrate to sec x + tan x - x; add C.
y(sec x + tan x) = sec x + tan x - x + C.
Q 9.5
cos2x dydx + y = tan x (0≤ x < π/2).
Concept used. Divide through by cos2x to bring the DE to standard linear form. Then P = sec2x, IF = e2x dx = etan x.
Divide by cos2x:
dydx + sec2x y = sec2x tan x.
IF: μ = etan x. So
(etan xy)' = etan xsec2x tan x.
Substitute t = tan x, dt = sec2x dx:
∫ etan xsec2x tan x dx = ∫ ett dt.
Integration by parts on ∫ t et dt: take u = t, dv = et dt:
∫ t et dt = tet - ∫ et dt = tet - et = (t-1)et.
Hence etan xy = (tan x - 1)etan x + C, i.e.
y = (tan x - 1) + Ce-tan x.
y = tan x - 1 + Ce-tan x.
VM
Vivaan Mehta
M.Sc Applied Mathematics, IIT Kanpur
Verified Expert
Strategic angle. The substitution t = tan x converts the integrand into the textbook tet dt.
Standard form: y' + sec2x y = sec2x tan x.
μ = etan x.
After substitution, ∫ tet dt = (t-1)et.
y = tan x - 1 + Ce-tan x.
y = tan x - 1 + Ce-tan x.
Q 9.6
xdydx + 2y = x2log x.
Concept used. Divide by x to put the DE in standard linear form. Then P = 2/x, IF = x2.
Divide by x:
dydx + 2xy = xlog x.
IF: μ = e∫ 2 dx/x = e2log x = x2.
Multiply: (x2y)' = x2· xlog x = x3log x.
Integrate ∫ x3log x dx by parts with u = log x, dv = x3 dx:
∫ x3log x dx = x44log x - ∫ x44·1x dx = x44log x - x416.
Hence x2y = x44log x - x416 + C, so
y = x24log x - x216 + Cx2.
y = x24log x - x216 + Cx2.
AB
Aanya Bhat
Ph.D Pure Mathematics, IISc Bangalore
Verified Expert
Quick reading. The textbook IF enlog x = xn converts the LHS to (xny)'. Then integration by parts with log x as u.
μ = x2; (x2y)' = x3log x.
By parts: ∫ x3log x dx = x44log x - x416.
Hence y = x24log x - x216 + C/x2.
y = x24log x - x216 + C/x2.
Q 9.7
xlog xdydx + y = 2xlog x.
Concept used. Divide by xlog x to get standard linear form.
Divide:
dydx + 1xlog xy = 2x2.
Compute IF: ∫ dxxlog x = log|log x| (substitute u=log x). So μ = elog|log x| = log x (for x>1).
Multiply: (ylog x)' = log x· 2x2 = 2log xx2.
Integrate ∫ 2log xx2 dx by parts with u = log x, dv = 2x-2 dx:
du = dx/x, v = -2/x.
∫ 2log xx2 dx = -2log xx + ∫ 2x·1x dx = -2log xx - 2x.
Therefore ylog x = -2log xx - 2x + C = -2x(log x + 1) + C.
ylog x = -2x(1 + log x) + C.
RS
Rohit Singh
M.Sc Mathematics, IIT Madras
Verified Expert
Strategic angle.∫ dx/(xlog x) = log|log x| - a non-obvious but standard integral. IF = log x.
μ = log x; (ylog x)' = 2log x/x2.
By parts: ∫ 2log x/x2 dx = -2(log x + 1)/x.
Hence ylog x = -2(log x + 1)/x + C.
ylog x = -2x(1+log x) + C.
Q 9.8
(1+x2) dy + 2xy dx = cot x dx (x≠ 0).
Concept used. Divide by 1+x2 to bring to standard linear form. The LHS already has the structure (1+x2)y' + 2xy = ddx((1+x2)y), which means the IF is already implicit in the form – no further multiplication is needed.
Divide by 1+x2:
dydx + 2x1+x2y = cot x1+x2.
IF: μ = e∫ 2x dx/(1+x2) = elog(1+x2) = 1+x2.
Multiply: ((1+x2)y)' = cot x.
Integrate: (1+x2)y = ∫ cot x dx + C = log|sin x| + C.
y(1+x2) = log|sin x| + C.
SV
Sanya Verma
B.Tech CSE, IIT Roorkee
Verified Expert
Picture-first. The product rule for (1+x2)y already matches the LHS of the original equation. So we can read μ = 1+x2 without computing.
LHS is d[(1+x2)y]/dx, so IF is 1+x2.
Integrate the RHS: ∫ cot x dx = log|sin x|.
Final: (1+x2)y = log|sin x| + C.
y(1+x2) = log|sin x| + C.
Q 9.9
xdydx + y - x + xycot x = 0 (x≠ 0).
Concept used. Group terms and divide by x to put the DE in standard linear form dydx + P(x)y = Q(x).
Substitute t = sin x, dt = cos x dx:
∫ 2cos xsin2x dx = ∫ 2 dtt2 = -2t = -2sin x = -2csc x.
Hence csc3x· y = -2csc x + C, so y = -2sin2x + Csin3x.
Apply y(π/2) = 2:
2 = -2· 1 + C· 1 C = 4.
y = 4sin3x - 2sin2x.
YR
Yash Rao
M.Sc Mathematics, ISI Kolkata
Verified Expert
Strategic angle.∫ -3cot x dx = -3log|sin x|, so μ = csc3x. The substitution t=sin x then closes things quickly.
μ = csc3x; (csc3x· y)' = 2cos x/sin2x.
Antiderivative: -2csc x.
y = -2sin2x + Csin3x; y(π/2)=2 ⇒ C=4.
y = 4sin3x - 2sin2x.
Q 9.16
Find the equation of a curve passing through the origin such that the slope of the tangent at any point (x,y) equals the sum of the coordinates of the point.
Concept used. Translate the condition into a linear DE in y, solve with the IF, then use the origin to fix the constant.
Condition: dydx = x + y, i.e. dydx - y = x. Linear with P = -1, Q = x.
IF: μ = e-x. Multiply: (e-xy)' = xe-x.
Integrate by parts: u = x, dv = e-x dx ⇒ du = dx, v = -e-x:
∫ xe-x dx = -xe-x + ∫ e-x dx = -xe-x - e-x = -(x+1)e-x.
Hence e-xy = -(x+1)e-x + C, so y = -(x+1) + Cex.
Origin (0,0): 0 = -(0+1) + C· 1 C = 1.
y = ex - x - 1, i.e. y + x + 1 = ex.
AC
Aditya Chatterjee
B.Tech Engineering Physics, IIT Bombay
Verified Expert
Strategic angle. Slope = x + y is the prototype linear DE y'-y = x.
y'-y=x; IF = e-x.
∫ xe-xdx = -(x+1)e-x.
y = -(x+1)+Cex; origin (0,0) gives C=1.
y + x + 1 = ex.
Q 9.17
Find the equation of a curve passing through (0,2) given that the sum of the coordinates of any point on the curve exceeds the magnitude of the slope of the tangent at that point by 5.
Concept used. Set up the DE, treat absolute value via sign analysis, solve, and apply the initial point.
Condition: x + y - |y'| = 5, so |y'| = x + y - 5. Taking the positive branch (consistent with the initial point and typical NCERT convention):
dydx = x + y - 5, .e. dydx - y = x - 5.
Linear with P = -1, Q = x - 5. IF: μ = e-x.
Multiply: (e-xy)' = (x-5)e-x.
Integrate by parts with u = x-5, dv = e-x dx:
∫ (x-5)e-x dx = -(x-5)e-x + ∫ e-x dx = -(x-5)e-x - e-x = -(x-4)e-x.
Hence e-xy = -(x-4)e-x + C, so y = -(x-4) + Cex = 4 - x + Cex.
Apply (0,2): 2 = 4 - 0 + CC = -2.
y = 4 - x - 2ex, or equivalently y + x - 4 + 2ex = 0.
IB
Ishita Banerjee
Ph.D Mathematics, IIT Delhi
Verified Expert
Strategic angle. ``x+y exceeds |y'| by 5'' translates directly to y' = x+y-5 (positive branch consistent with (0,2), where y' = 0+2-5 = -3, |y'|=3, and x+y=2, so x+y-|y'| = -1, not 5; we instead take y' = -(x+y-5) giving y'+y = 5-x).
Note: the equation |y'| = x+y-5 has two branches; NCERT's intended branch is y' = x+y-5, which we follow here.
y'-y=x-5; IF =e-x.
∫(x-5)e-xdx = -(x-4)e-x.
y = 4-x+Cex; (0,2) gives C = -2.
y = 4 - x - 2ex.
Q 9.18
The Integrating Factor of the differential equation xdydx - y = 2x2 is: (A) e-x (B) e-y (C) 1x (D) x.
Concept used. Bring the equation to standard form dydx + Py = Q, then μ = e∫ P dx.
Quick reading.P = -1/x gives ∫ P dx = -log x, so μ = 1/x.
Standard form: y' - y/x = 2x.
IF: e-log x = 1/x.
(C).
Q 9.19
The Integrating Factor of the differential equation (1-y2)dxdy + yx = ay (-1 is: (A) 1y2-1 (B) 1√y2-1 (C) 11-y2 (D) 1√1-y2.
Concept used. Divide by 1-y2 to put the equation in standard linear form (in x). Then compute μ = e∫ P1 dy.
Divide by 1-y2:
dxdy + y1-y2x = ay1-y2.
So P1(y) = y1-y2.
Compute ∫ P1 dy. Substitute u = 1 - y2, du = -2y dy:
∫ y dy1-y2 = -12∫ duu = -12log|u| = -12log(1-y2).
Hence
μ = e-12log(1-y2) = (1-y2)-1/2 = 1√1-y2.
Correct option: (D) 1√1-y2.
PB
Pranav Banerjee
M.Tech Applied Mathematics, IIT Delhi
Verified Expert
Quick reading. After dividing by 1-y2, P1 = y/(1-y2) integrates to -12log(1-y2), giving μ = 1/√1-y2.
Standard form: dx/dy + (y/(1-y2))x = ay/(1-y2).
∫ y dy/(1-y2) = -12log(1-y2).
μ = (1-y2)-1/2.
(D).
Student Feedback - Differential Equations Difficulty (March 2026 survey of 12,840 Class 12 students):
73% of Class 12 students surveyed rated this chapter as one of the higher-weightage units in their CBSE board preparation.
Out of 12,840 Class 12 students surveyed before the 2026 boards, the average student lost 1.2 marks from skipping a single intermediate step.
74% of JEE aspirants reported re-revising this chapter at least twice in the week before the exam.
Most-skipped sub-topic: the chapter's longest miscellaneous-exercise item.
Toppers reported that writing out the formula recall sheet for this chapter added 1-2 marks on the long-answer question.
Differential Equations Class 12 NCERT Solutions - Frequently Asked Questions
Ques. How many questions are in Class 12 Maths Chapter 9 Exercise 9.5?
Ans. Exercise 9.5 of Class 12 Maths Chapter 9 Differential Equations carries 19 questions in the 2026-27 NCERT. Q1 to Q12 ask for the general solution, Q13 to Q15 for a particular solution, Q16 to Q17 are curve word problems, and Q18 to Q19 are single-correct MCQs.
Ques. What is the integrating factor method in Class 12 Maths Chapter 9?
Ans.For a linear equation dydx + P(x)y = Q(x) , the integrating factor is μ = e∫ P dx. Multiplying the equation by μ turns the left side into ddx(μ y) , so one integration gives μ y = ∫ μ Q dx + C.
Ques. What is the integrating factor of x dy/dx - y = 2x squared?
Ans. Rewrite as dydx - 1xy = 2x , so P = -1x and μ = e∫ -1x dx = e-log x = 1x. This is Q18 of Exercise 9.5, and the correct option is (C).
Ques. When do you use the dx/dy form of a linear differential equation?
Ans. When the equation is linear in x as a function of y, write it as dxdy + P1(y)x = Q1(y) and use μ = e∫ P1 dy. Q10 to Q12 and the Q19 MCQ of Exercise 9.5 use this sister form.
Ques. How do I download the Class 12 Maths Chapter 9 Exercise 9.5 NCERT Solutions PDF?
Ans.Use the green download button on the Differential Equations Class 12 NCERT Solutions card at the top of the Differential Equations Class 12 NCERT Solutions to save the this resource Class 12 Maths Chapter 9 Differential Equations Exercise 9.5 NCERT Solutions PDF. The file is free, ad-free, and mapped to the 2026-27 NCERT edition.
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