Mathematics Class 12 — Part…
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Differential Equations Exercises and Summary

🎓 Class 12 Mathematics CBSE Theory Ch 9 — Differential Equations ⏱ ~15 min
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This MCQ module is based on: Differential Equations Exercises and Summary

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Targeting Class 12 level in General Mathematics, with Advanced difficulty.

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End-of-Chapter Exercises

Exercise 9.1 — Order and degree
1. \(\dfrac{d^4y}{dx^4}+\sin(y''')=0\). Order and degree.
Order 4. Degree undefined (sin of derivative).
2. \(y'+5y=0\). Order and degree.
Order 1, degree 1.
3. \((ds/dt)^4+3s\,d^2s/dt^2=0\).
Order 2, degree 1 (highest is d²s/dt², appears to the first power).
4. \((d²y/dx²)² + \cos(dy/dx) = 0\).
Order 2, degree undefined (cos of derivative).
7. \(y''' + 2y'' + y' = 0\).
Order 3, degree 1.
Exercise 9.2 — Verify solutions
2. Verify \(y=x^2+2x+C\) is a solution of \(y'-2x-2=0\).
\(y'=2x+2\); LHS = \(2x+2-2x-2=0\). ✓
5. Verify \(y=Ax\) is a solution of \(xy'=y\), and find the value of A so that \(y(1)=2\).
\(y'=A\); \(xy'=Ax=y\). ✓ Apply IC: \(2=A\cdot 1\Rightarrow A=2\). Particular: \(y=2x\).
Exercise 9.3 — Variables-separable
1. \(\dfrac{dy}{dx}=\dfrac{1-\cos x}{1+\cos x}\).
RHS = \(\tan^2(x/2)\) (half-angle). \(dy=\tan^2(x/2)\,dx=(\sec^2(x/2)-1)\,dx\). Integrate: \(y=2\tan(x/2)-x+C\).
2. \(\dfrac{dy}{dx}=\sqrt{4-y^2},\ -2
\(\dfrac{dy}{\sqrt{4-y^2}}=dx\). Integrate: \(\sin^{-1}(y/2)=x+C\), so \(y=2\sin(x+C)\).
5. \((e^x+e^{-x})\,dy-(e^x-e^{-x})\,dx=0\).
\(dy=\dfrac{e^x-e^{-x}}{e^x+e^{-x}}\,dx\). Integrate: \(y=\ln|e^x+e^{-x}|+C\) (since the numerator is the derivative of the denominator).
9. \(\dfrac{dy}{dx}=\sin^{-1}x\).
Integrate by parts: \(y=x\sin^{-1}x+\sqrt{1-x^2}+C\).
10. \(e^x\tan y\,dx+(1-e^x)\sec^2 y\,dy=0\).
Separate: \(\dfrac{\sec^2 y}{\tan y}\,dy=\dfrac{e^x}{e^x-1}\,dx\). Integrate: \(\ln|\tan y|=\ln|e^x-1|+\ln|C|\), so \(\tan y=C(e^x-1)\).
Exercise 9.4 — Homogeneous
1. \((x^2+xy)\,dy=(x^2+y^2)\,dx\). (Solved as Example 11.)
See worked Example 11 in Part 2: \(2\ln\left|\dfrac{x-y}{x}\right|+\dfrac{y}{x}+\ln|x|=C\).
3. \((x-y)\,dy-(x+y)\,dx=0\).
Rewrite \(\dfrac{dy}{dx}=\dfrac{x+y}{x-y}\). Set \(y=vx\): \(v+xv'=\dfrac{1+v}{1-v}\), so \(xv'=\dfrac{1+v^2}{1-v}\). Separate as in Example 12: \(\tan^{-1}(y/x)-\tfrac{1}{2}\ln(x^2+y^2)+\ln|x|=C\), or equivalently \(2\tan^{-1}(y/x)=\ln(x^2+y^2)+C'\).
Exercise 9.5 — Linear DEs
1. \(\dfrac{dy}{dx}+2y=\sin x\).
P=2, I.F.=\(e^{2x}\). \(\dfrac{d}{dx}(e^{2x}y)=e^{2x}\sin x\). Integrate by parts: \(\int e^{2x}\sin x\,dx=\dfrac{e^{2x}(2\sin x-\cos x)}{5}\). So \(y=\dfrac{2\sin x-\cos x}{5}+Ce^{-2x}\).
3. \(\dfrac{dy}{dx}+\dfrac{y}{x}=x^2\).
P=1/x, I.F.=\(e^{\ln x}=x\). \(\dfrac{d}{dx}(xy)=x^3\). Integrate: \(xy=x^4/4+C\), so \(y=x^3/4+C/x\).
4. \(\dfrac{dy}{dx}+y\sec x=\tan x\).
I.F.=\(e^{\int\sec x\,dx}=\sec x+\tan x\). Multiply: \(\dfrac{d}{dx}((\sec x+\tan x)y)=\tan x(\sec x+\tan x)=\sec x\tan x+\tan^2 x=\sec x\tan x+\sec^2 x-1\). Integrate: \((\sec x+\tan x)y=\sec x+\tan x-x+C\). So \(y=1-\dfrac{x-C}{\sec x+\tan x}\).
7. \(x\,dy/dx+y-x+xy\cot x=0\).
Rewrite: \(dy/dx+(1/x+\cot x)y=1\). I.F.=\(e^{\int(1/x+\cot x)\,dx}=e^{\ln|x|+\ln|\sin x|}=x\sin x\). Multiply: \(\dfrac{d}{dx}(xy\sin x)=x\sin x\). Integrate by parts: \(\int x\sin x\,dx=-x\cos x+\sin x+C\). So \(xy\sin x=-x\cos x+\sin x+C\), giving \(y=\dfrac{-\cos x}{\sin x}+\dfrac{1}{x}+\dfrac{C}{x\sin x}=-\cot x+\dfrac{1}{x}+\dfrac{C}{x\sin x}\).
Miscellaneous Exercise — selected
3. \((1-y^2)(1+\ln x)\,dx+2xy\,dy=0\).
Separable: \(\dfrac{2y\,dy}{y^2-1}=\dfrac{(1+\ln x)\,dx}{x}\). Integrate: \(\ln|y^2-1|=\dfrac{(1+\ln x)^2}{2}+C\), giving \(y^2-1=A\,e^{(1+\ln x)^2/2}\) (after exponentiating).
5. Solve \(\dfrac{dy}{dx}+y\cot x=4x\,\text{cosec}\,x\), \(y(\pi/2)=0\).
I.F.=\(\sin x\). \(\dfrac{d}{dx}(y\sin x)=4x\,\text{cosec}\,x\cdot\sin x=4x\). Integrate: \(y\sin x=2x^2+C\). At \(x=\pi/2,\ y=0\): \(0=\pi^2/2+C\Rightarrow C=-\pi^2/2\). So \(y=\dfrac{2x^2-\pi^2/2}{\sin x}\).
Activity: Identify Method First, Solve Second
L4 Analyse
Materials: Pen, paper.
Predict: For each DE below, decide which technique applies before computing.
  1. \(\dfrac{dy}{dx}=\dfrac{x}{y}\) — separable (already split).
  2. \(\dfrac{dy}{dx}=\dfrac{x+y}{x}\) — homogeneous (degrees match).
  3. \(\dfrac{dy}{dx}+y=e^x\) — linear.
  4. \((y\sin x)\,dx+(\cos x)\,dy=0\) — rewrite \(\dfrac{dy}{dx}=-y\tan x\); separable: \(dy/y=-\tan x\,dx\); integrate \(\ln|y|=\ln|\cos x|+C\), so \(y=A\cos x\).
  5. \(\dfrac{dy}{dx}+y\,\dfrac{1}{x}=\dfrac{\sin x}{x}\) — linear.
Decision tree: (1) variables already split? → separable. (2) homogeneous-degree-zero? → set y = vx. (3) linear in y or x? → integrating factor. The first two minutes spent classifying save twenty minutes of misdirected algebra.

Consolidation Competency-Based Questions

Scenario: A culture of bacteria grows at a rate proportional to its current population. Initial count 100; after 2 hours, count is 200.
Q1. Write and solve the DE for population P(t).
L3 Apply
Solution: \(dP/dt=kP\Rightarrow P=P_0 e^{kt}=100 e^{kt}\). At t=2: 200=100·e^{2k}, so k=ln2/2. Hence \(P(t)=100\cdot 2^{t/2}\).
Q2. (T/F) "Every linear DE is separable." Justify.
L5 Evaluate
False. Linear DEs need not be separable. Example: \(dy/dx + y = x\) is linear, but the RHS doesn't factor as g(x)·h(y); separation fails. Linear DEs require the integrating-factor method.
Q3. Solve \(\dfrac{dy}{dx}=2y\) with y(0)=3.
L3 Apply
Solution: Separable: \(dy/y=2dx\), \(\ln y=2x+C\), \(y=Ae^{2x}\). At x=0, y=3: A=3. So y=3e^{2x}.
Q4. Apply: a body cools from 100°C to 60°C in 20 minutes in a 20°C room. Find the temperature after 40 minutes (Newton's cooling).
L4 Analyse
Solution: \(T(t)=20+(T_0-20)e^{-kt}=20+80e^{-kt}\). At t=20: 60=20+80e^{-20k}, so e^{-20k}=1/2, k=ln2/20. At t=40: \(T=20+80\cdot(1/2)^2=20+20=40°C\).
Q5. Design: solve \(\dfrac{dy}{dx}+y\tan x=\sin 2x\), \(y(0)=1\).
L6 Create
Solution: I.F.=\(e^{\int\tan x\,dx}=\sec x\). Multiply: \(\dfrac{d}{dx}(y\sec x)=\sin 2x\sec x=2\sin x\). Integrate: \(y\sec x=-2\cos x+C\), so \(y=-2\cos^2 x+C\cos x\). At x=0, y=1: \(1=-2+C\Rightarrow C=3\). Hence \(y=3\cos x-2\cos^2 x\).

Consolidation Assertion–Reason

Assertion (A): \(\dfrac{dy}{dx}=\dfrac{x+y}{x-y}\) is a homogeneous DE.
Reason (R): Both numerator and denominator have the same degree (1) in x and y.
(a) Both true, R explains A.
(b) Both true, R doesn't explain A.
(c) A true, R false.
(d) A false, R true.
Answer: (a). Equal-degree numerator and denominator make the ratio degree-zero homogeneous.
Assertion (A): The DE \(\dfrac{dy}{dx}+P(x)y=Q(x)\) has integrating factor \(e^{\int P\,dx}\).
Reason (R): Multiplying by I.F. makes the LHS \(\dfrac{d}{dx}(I\!.\!F\!.\,y)\), allowing direct integration.
(a) Both true, R explains A.
(b) Both true, R doesn't explain A.
(c) A true, R false.
(d) A false, R true.
Answer: (a). R is the engineering reason for A's choice of I.F.

Chapter Summary

Key concepts at a glance
  • Differential equation: equation involving derivatives. ODE: one independent variable.
  • Order: highest derivative present.
  • Degree: highest power of the highest-order derivative (when polynomial in derivatives); else undefined.
  • General solution: contains as many arbitrary constants as the order. Particular solution: obtained by giving specific values to those constants.
  • Family of curves → DE: differentiate \(n\) times and eliminate \(n\) constants.
  • Variables-separable: \(dy/dx=g(x)h(y)\Rightarrow\int dy/h(y)=\int g(x)\,dx+C\).
  • Homogeneous: \(dy/dx=F(y/x)\); substitute \(y=vx\), get separable in \(v,x\).
  • Linear: \(dy/dx+P(x)y=Q(x)\); I.F. = \(e^{\int P\,dx}\); solution \(y\cdot\text{I.F.}=\int(\text{I.F.}\cdot Q)\,dx+C\).
  • Dual linear: \(dx/dy+P_1(y)x=Q_1(y)\) — same method with x,y swapped.

Historical Note

The study of differential equations began with the calculus of Newton (1665) and Leibniz (1684). Newton's Principia (1687) implicitly contained many DEs as the equations of motion of celestial bodies. The Bernoulli family — Jacob, Johann, Daniel — developed solution techniques in the early 18th century, including the substitution methods we still use.

Leonhard Euler (1707–1783) systematised the integrating-factor method for first-order linear DEs and developed the theory of higher-order linear equations with constant coefficients. Lagrange and Laplace applied DEs to celestial mechanics. Cauchy (1820s) gave the first rigorous existence-uniqueness proofs.

By the late 19th century, Henri Poincaré initiated the qualitative theory: rather than seeking explicit formulas (which often don't exist), study the geometry of solution curves. This led to dynamical systems, chaos theory, and the modern computational study of DEs that powers weather forecasting, planetary trajectories and AI.

Frequently Asked Questions — Differential Equations Exercises and Summary

What is the chapter summary of Class 12 Maths Differential Equations?
A differential equation involves a function and its derivatives. Order = highest derivative; degree = power of highest-order derivative (when polynomial). Three solution methods: variables-separable, homogeneous (substitute y = vx), linear (I.F. = e^(∫P dx)).
Who developed differential equations?
Newton, Leibniz, Bernoullis, Euler, Lagrange, Laplace, Cauchy, and Poincaré are the major figures spanning 1665–1900.
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