Mathematics Class 12 — Part…
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Linear Differential Equations

🎓 Class 12 Mathematics CBSE Theory Ch 9 — Differential Equations ⏱ ~15 min
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આ MCQ મોડ્યુલ આના પર આધારિત છે: Linear Differential Equations

આ મૂલ્યાંકન આના પર આધારિત હશે: Linear Differential Equations
Class 12 સ્તર, General Mathematics માં, Advanced કઠિનતા સાથે લક્ષ્ય.

મૂલ્યાંકન બનાવવામાં તેમની સામગ્રી સામેલ કરવા ચિત્રો, PDF અથવા Word દસ્તાવેજ અપલોડ કરો.

9.4.3 Linear Differential Equations

First-order linear DE — standard form
A first-order linear differential equation is one expressible in the form \[\boxed{\;\dfrac{dy}{dx}+P(x)\,y=Q(x)\;}\] where \(P\) and \(Q\) are functions of \(x\) alone (or constants). The unknown \(y\) appears to the first power and is not multiplied by its own derivative.

The companion form (with x as dependent and y as independent) is \(\dfrac{dx}{dy}+P_1(y)\,x=Q_1(y)\).

The integrating factor method

Derivation
We seek a function \(\mu(x)\) such that multiplying the equation by \(\mu\) makes the LHS a perfect derivative \(\dfrac{d}{dx}(\mu y)\). By the product rule, \[\dfrac{d}{dx}(\mu y)=\mu\,\dfrac{dy}{dx}+\mu'\,y.\] Comparing with \(\mu(y'+Py)=\mu y'+\mu P y\), we need \(\mu'=\mu P\), i.e. \(\dfrac{d\mu}{\mu}=P\,dx\), giving \(\mu=e^{\int P\,dx}\). After multiplying: \[\dfrac{d}{dx}\bigl(e^{\int P\,dx}\,y\bigr)=e^{\int P\,dx}\,Q.\] Integrating both sides yields the formula below. \(\square\)
Master formula
For \(\dfrac{dy}{dx}+P\,y=Q\):
Step 1. Compute the integrating factor \(\text{I.F.}=e^{\int P\,dx}\).
Step 2. Multiply the DE by I.F. The LHS becomes \(\dfrac{d}{dx}(\text{I.F.}\cdot y)\).
Step 3. Integrate both sides: \[\boxed{\;y\cdot\text{I.F.}=\int(\text{I.F.}\cdot Q)\,dx+C\;}\]

Worked Examples

Example 13. Solve \(\dfrac{dy}{dx}+y=\sin x\).
\(P=1,\ Q=\sin x\). \(\text{I.F.}=e^{\int 1\,dx}=e^x\). Multiply: \(\dfrac{d}{dx}(e^x y)=e^x\sin x\). Integrate: \(e^x y=\int e^x\sin x\,dx=\dfrac{e^x(\sin x-\cos x)}{2}+C\) (standard integration by parts). So \(y=\dfrac{\sin x-\cos x}{2}+C\,e^{-x}\).
Example 14. Solve \(\dfrac{dy}{dx}-y=\cos x\).
\(P=-1,\ Q=\cos x\). \(\text{I.F.}=e^{\int -1\,dx}=e^{-x}\). \(\dfrac{d}{dx}(e^{-x}y)=e^{-x}\cos x\). Integrate (parts): \(e^{-x}y=\dfrac{e^{-x}(\sin x-\cos x)}{2}+C\). So \(y=\dfrac{\sin x-\cos x}{2}+C\,e^x\).
Example 15. Solve \(x\,\dfrac{dy}{dx}+2y=x^2\) for \(x\ne 0\).
Divide by \(x\): \(\dfrac{dy}{dx}+\dfrac{2}{x}y=x\). \(P=2/x,\ Q=x\). \(\text{I.F.}=e^{\int 2/x\,dx}=e^{2\ln|x|}=x^2\). Multiply: \(\dfrac{d}{dx}(x^2 y)=x^3\). Integrate: \(x^2 y=x^4/4+C\). So \(y=\dfrac{x^2}{4}+\dfrac{C}{x^2}\).
Example 16. Solve \((x-3y^2)\,dy=y\,dx\) (treat x as a function of y).
Rewrite: \(\dfrac{dx}{dy}=\dfrac{x-3y^2}{y}=\dfrac{x}{y}-3y\), i.e. \(\dfrac{dx}{dy}-\dfrac{1}{y}\,x=-3y\). \(P_1=-1/y,\ Q_1=-3y\). \(\text{I.F.}=e^{\int -1/y\,dy}=e^{-\ln|y|}=1/y\). Multiply: \(\dfrac{d}{dy}(x/y)=-3\). Integrate: \(x/y=-3y+C\), so \(x=-3y^2+Cy\).
Example 17. Find the particular solution of \(\dfrac{dy}{dx}+y\cot x=2x+x^2\cot x\), \(x\ne 0\), with \(y(\pi/2)=0\).
\(P=\cot x,\ Q=2x+x^2\cot x\). \(\text{I.F.}=e^{\int\cot x\,dx}=e^{\ln|\sin x|}=\sin x\). Multiply: \(\dfrac{d}{dx}(y\sin x)=(2x+x^2\cot x)\sin x=2x\sin x+x^2\cos x\). Note \((x^2\sin x)'=2x\sin x+x^2\cos x\). Hence \(y\sin x=x^2\sin x+C\). Apply IC at \(x=\pi/2\): \(0=(\pi/2)^2+C\), so \(C=-\pi^2/4\). Particular solution: \(y\sin x=x^2\sin x-\pi^2/4\), or \(y=x^2-\dfrac{\pi^2}{4\sin x}\).
Example 18. Find the equation of a curve passing through (0, 2) such that the sum of the y-coordinate and the slope at any point equals \(x\).
"Slope" = \(dy/dx\). Equation: \(y+dy/dx=x\), i.e. \(\dfrac{dy}{dx}+y=x\). Linear with \(P=1,\ Q=x\). I.F. = \(e^x\). \(\dfrac{d}{dx}(e^x y)=xe^x\). Integrate: \(e^x y=xe^x-e^x+C=(x-1)e^x+C\). So \(y=x-1+Ce^{-x}\). Apply (0,2): \(2=0-1+C\Rightarrow C=3\). Curve: \(y=x-1+3e^{-x}\).
Activity: The "linear DE" Recipe Card
L3 Apply
Materials: Pen, paper.
Predict: Memorise the 4-step recipe and try it on \(dy/dx+y\tan x=\sec x\).
  1. Standard form ✓ (\(P=\tan x,\ Q=\sec x\)).
  2. I.F. = \(e^{\int\tan x\,dx}=e^{-\ln|\cos x|}=\sec x\).
  3. Multiply: \(\dfrac{d}{dx}(\sec x\cdot y)=\sec^2 x\). [check: LHS = \((\sec x)y'+(\sec x\tan x)y=\sec x(y'+y\tan x)=\sec x\cdot\sec x=\sec^2 x\) ✓]
  4. Integrate: \(y\sec x=\tan x+C\), so \(y=\sin x+C\cos x\).
  5. Now try \(\dfrac{dy}{dx}+y/x=x^2\) yourself. (I.F. = \(x\); answer \(y=x^3/4+C/x\).)
The 4-step recipe is mechanical once you spot "linear in y". Watch out for: (a) needing to divide first to get the standard form (when there's a coefficient on \(dy/dx\)); (b) the dual form \(dx/dy+P_1 x=Q_1\) when y is the independent variable; (c) integrating factors involving \(\ln\) — they often simplify to powers like \(x^n\) or \(\sin x\), \(\sec x\) etc.

Competency-Based Questions

Scenario: An RC electrical circuit has voltage equation \(R\,dQ/dt+Q/C=V_0\) (Q = charge, R = resistance, C = capacitance, V₀ = applied voltage).
Q1. Solve the RC circuit equation with Q(0) = 0.
L3 Apply
Solution: \(\dfrac{dQ}{dt}+\dfrac{1}{RC}Q=\dfrac{V_0}{R}\). Linear with \(P=1/(RC),\ Q=V_0/R\). I.F. = \(e^{t/(RC)}\). Solve to get \(Q(t)=CV_0(1-e^{-t/(RC)})\). Capacitor charges exponentially to \(CV_0\) with time-constant \(\tau=RC\).
Q2. Identify which of the following are linear DEs:
L3 Apply
  • (a) \(dy/dx+y=x^2\)
  • (b) \(dy/dx+y^2=x\)
  • (c) \(y'+y\sin x=\cos x\)
  • (d) \(yy'=1\)
Linear: (a) and (c). (b) has y² (not linear); (d) has y·y' (not linear).
Q3. (T/F) "If \(P(x)=k\) (a constant) in dy/dx + Py = Q, then I.F. = \(e^{kx}\)." Justify.
L5 Evaluate
True. \(\int k\,dx=kx\), so I.F. \(=e^{kx}\). This is the simplest case — relevant for any constant-coefficient first-order linear DE.
Q4. Solve \(dy/dx + 2xy = 4x\), \(y(0)=0\).
L4 Analyse
Solution: \(P=2x,\ Q=4x\). I.F. = \(e^{x^2}\). \(\dfrac{d}{dx}(e^{x^2}y)=4xe^{x^2}\). Note \(\dfrac{d}{dx}(2e^{x^2})=4xe^{x^2}\). So \(e^{x^2}y=2e^{x^2}+C\), giving \(y=2+Ce^{-x^2}\). At \(x=0,y=0\): \(0=2+C\Rightarrow C=-2\). Hence \(y=2-2e^{-x^2}\).
Q5. Design: a tank initially holds 100 L of pure water. Salt solution (concentration 1 kg/L) flows in at 5 L/min; mixed solution flows out at 5 L/min. Set up and solve the DE for amount A(t) of salt at time t.
L6 Create
Solution: Volume stays at 100 L. Concentration in tank = A/100 kg/L. Rate in = 5·1 = 5 kg/min; rate out = 5·(A/100) = A/20. So \(dA/dt = 5 - A/20\), i.e. \(dA/dt + A/20 = 5\). Linear, P=1/20, Q=5. I.F. = \(e^{t/20}\). Solve: \(A(t)=100+Ce^{-t/20}\). \(A(0)=0\Rightarrow C=-100\). So \(A(t)=100(1-e^{-t/20})\) kg. Approaches 100 kg as \(t\to\infty\).

Assertion–Reason Questions

Assertion (A): The integrating factor for \(dy/dx + y\cot x = 2x\) is \(\sin x\).
Reason (R): I.F. = \(e^{\int\cot x\,dx}=e^{\ln|\sin x|}=\sin x\).
(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). Direct computation.
Assertion (A): Multiplying \(dy/dx + Py = Q\) by I.F. = \(e^{\int P\,dx}\) makes the LHS a perfect derivative.
Reason (R): By product rule, \((I\!.\!F\!.\,y)'=I\!.\!F\!.(y'+Py)\) provided \(I\!.\!F\!.'=P\cdot I\!.\!F\!.\), which is satisfied by the chosen I.F.
(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 precise reason; A is the consequence.
Assertion (A): The DE \((y-x)dy/dx=1\) is linear in x as a function of y.
Reason (R): Rewriting: \(dx/dy+x=y\), which has the standard linear form \(dx/dy+P_1 x=Q_1\) with \(P_1=1, Q_1=y\).
(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). The trick: when an equation isn't linear in y but is linear in x, swap roles. R is exactly that move.

Frequently Asked Questions — Linear Differential Equations

What is a linear differential equation?
A first-order linear DE has the form dy/dx + P(x)·y = Q(x), with y appearing to the first power and not multiplied by its derivative.
What is the integrating factor (I.F.)?
I.F. = e^(∫P dx). Multiplying converts the LHS into d/dx(I.F. · y).
What are the steps to solve a linear DE?
(1) Standard form; (2) Compute I.F.; (3) Multiply through; (4) Integrate to get I.F.·y = ∫(I.F.·Q) dx + C.
What if the DE is dx/dy + P₁·x = Q₁?
Same method with x and y swapped. I.F. = e^(∫P₁ dy).
Why does multiplying by e^(∫P dx) help?
Because then LHS becomes the derivative of (I.F. · y) — integrable in one line.
Are linear DEs always solvable in closed form?
Yes, in principle — but the integrals of P and Q may not be elementary.
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Mathematics Class 12 — Part II
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