Mathematics Class 12 — Part…
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Variables Separable and Homogeneous Differential Equations

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

This mathematics assessment will be based on: Variables Separable and Homogeneous Differential Equations
Targeting Class 12 level in General Mathematics, with Advanced difficulty.

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9.4 Methods of Solving First-Order, First-Degree Differential Equations

This section presents three classical solution methods, each for a particular structural form of \(\dfrac{dy}{dx}\):

  1. Variables-separable — when \(dy/dx\) factors as \(g(x)\cdot h(y)\).
  2. Homogeneous — when \(dy/dx=F(y/x)\).
  3. Linear — when \(dy/dx+P(x)\,y=Q(x)\) (next part).

9.4.1 Variables-Separable Method

Variables-separable form
A first-order DE \[\dfrac{dy}{dx}=g(x)\cdot h(y)\] can be rewritten by separation: \[\boxed{\;\dfrac{dy}{h(y)}=g(x)\,dx\;}\] Integrate both sides to get \(\int\dfrac{dy}{h(y)}=\int g(x)\,dx+C\). The result is the (possibly implicit) general solution.

The technique is the workhorse for all "rate-equals-function-of-position" laws. Many fundamental DEs are separable: \(dy/dx=ky\) (exponential growth), \(dT/dt=k(T-T_s)\) (Newton's cooling), \(dh/dt=-k\sqrt h\) (Torricelli's drain).

Worked Examples

Example 6. Solve \(\dfrac{dy}{dx}=\dfrac{1+y^2}{1+x^2}\).
Separate: \(\dfrac{dy}{1+y^2}=\dfrac{dx}{1+x^2}\). Integrate: \(\tan^{-1}y=\tan^{-1}x+C\), giving the general solution \(\tan^{-1}y-\tan^{-1}x=C\).
Example 7. Solve \(y\,\dfrac{dy}{dx}=x\) with initial condition \(y=1\) when \(x=0\).
Separate: \(y\,dy=x\,dx\). Integrate: \(y^2/2=x^2/2+C\), i.e. \(y^2-x^2=2C\). Apply IC: at \(x=0,y=1\): \(1-0=2C\Rightarrow C=1/2\). Particular solution: \(y^2-x^2=1\) (a hyperbola).
Example 8. Find the curve passing through (0, 1) whose differential equation is \(y'=e^{x+y}\).
\(dy/dx=e^x\cdot e^y\). Separate: \(e^{-y}\,dy=e^x\,dx\). Integrate: \(-e^{-y}=e^x+C\). At \((0,1)\): \(-e^{-1}=1+C\Rightarrow C=-1-1/e\). So \(-e^{-y}=e^x-1-1/e\), or \(e^{-y}+e^x=1+1/e\).
Example 9. Solve \(\dfrac{dy}{dx}=4x^3 e^{4y}\) with \(y(0)=0\).
Separate: \(e^{-4y}\,dy=4x^3\,dx\). Integrate: \(-e^{-4y}/4=x^4+C\). At \((0,0)\): \(-1/4=0+C\Rightarrow C=-1/4\). So \(-e^{-4y}/4=x^4-1/4\), giving \(4x^4+3e^{-4y}-3=0\) (after rearranging — verify by substitution).

9.4.2 Homogeneous Differential Equations

Homogeneous DE
A first-order DE \(\dfrac{dy}{dx}=F(x,y)\) is homogeneous if \(F\) is a homogeneous function of degree zero, i.e. \[F(\lambda x,\lambda y)=F(x,y)\text{ for all }\lambda.\] Equivalently, \(F\) depends only on the ratio \(y/x\) (or \(x/y\)). Examples:
• \(\dfrac{dy}{dx}=\dfrac{y}{x}\) — F = y/x. Yes.
• \(\dfrac{dy}{dx}=\dfrac{x^2+y^2}{xy}\) — degree-2 numerator and denominator both → ratio depends on y/x.
• \(\dfrac{dy}{dx}=x+y\) — NOT homogeneous (has unmatched degree).
Substitution: y = vx
For a homogeneous DE, set \(y=vx\) where \(v=v(x)\) is a new function. Then \(\dfrac{dy}{dx}=v+x\,\dfrac{dv}{dx}\). Substituting into \(dy/dx=F(y/x)=F(v)\): \[v+x\,\dfrac{dv}{dx}=F(v)\Rightarrow x\,\dfrac{dv}{dx}=F(v)-v\Rightarrow \dfrac{dv}{F(v)-v}=\dfrac{dx}{x}.\] This is now separable! Integrate, then back-substitute \(v=y/x\) to recover \(y\) as a function of \(x\).

Worked Examples

Example 10. Solve \(\dfrac{dy}{dx}=\dfrac{y}{x}+\tan\!\dfrac{y}{x}\).
RHS depends only on \(y/x\) — homogeneous. Set \(y=vx\): \(v+x\,v'=v+\tan v\), so \(x\,v'=\tan v\), giving \(\dfrac{dv}{\tan v}=\dfrac{dx}{x}\), i.e. \(\cot v\,dv=\dfrac{dx}{x}\). Integrate: \(\ln|\sin v|=\ln|x|+\ln|C|\), so \(\sin v=Cx\). Back-substitute: \(\sin(y/x)=Cx\).
Example 11. Solve \((x^2+xy)\,dy=(x^2+y^2)\,dx\).
Rewrite: \(\dfrac{dy}{dx}=\dfrac{x^2+y^2}{x^2+xy}\). Numerator and denominator both degree 2 — homogeneous. Divide top and bottom by \(x^2\): \(\dfrac{dy}{dx}=\dfrac{1+(y/x)^2}{1+y/x}\). Set \(y=vx\): \(v+x\,v'=\dfrac{1+v^2}{1+v}\). So \(x\,v'=\dfrac{1+v^2}{1+v}-v=\dfrac{1+v^2-v-v^2}{1+v}=\dfrac{1-v}{1+v}\). Separate: \(\dfrac{1+v}{1-v}\,dv=\dfrac{dx}{x}\). The LHS = \(\dfrac{-(1-v)+2}{1-v}\,dv=\left(-1+\dfrac{2}{1-v}\right)\,dv\). Integrate: \(-v-2\ln|1-v|=\ln|x|+C\). Back-substitute \(v=y/x\): \(-y/x-2\ln|1-y/x|=\ln|x|+C\), or equivalently \(2\ln\left|\dfrac{x-y}{x}\right|+\dfrac{y}{x}+\ln|x|+C=0\).
Example 12. Solve \(\dfrac{dy}{dx}=\dfrac{x+y}{x-y}\) with \(y(1)=0\).
Homogeneous (degree 1 over degree 1). \(y=vx\): \(v+xv'=\dfrac{1+v}{1-v}\), so \(xv'=\dfrac{1+v-v(1-v)}{1-v}=\dfrac{1+v^2}{1-v}\). Separate: \(\dfrac{1-v}{1+v^2}\,dv=\dfrac{dx}{x}\). Split: \(\int\dfrac{1}{1+v^2}\,dv-\int\dfrac{v}{1+v^2}\,dv=\int\dfrac{dx}{x}\). LHS \(=\tan^{-1}v-\tfrac{1}{2}\ln(1+v^2)\). Hence \(\tan^{-1}(y/x)-\tfrac{1}{2}\ln(1+y^2/x^2)=\ln|x|+C\). At \((1,0)\): \(0-0=0+C\Rightarrow C=0\). Particular: \(2\tan^{-1}(y/x)=\ln(x^2+y^2)\) (after simplification).
Activity: Recognise and Solve
L4 Analyse
Materials: Pen, paper.
Predict: Look at \(\dfrac{dy}{dx}=\dfrac{xy}{x^2-y^2}\). Is this separable, homogeneous, or neither?
  1. Test homogeneity: replace \((x,y)\) with \((\lambda x, \lambda y)\). RHS becomes \(\dfrac{\lambda^2 xy}{\lambda^2(x^2-y^2)}=\dfrac{xy}{x^2-y^2}\) — unchanged. Homogeneous. ✓
  2. Substitute \(y=vx\): \(v+xv'=\dfrac{vx^2}{x^2(1-v^2)}=\dfrac{v}{1-v^2}\). Then \(xv'=\dfrac{v}{1-v^2}-v=\dfrac{v-v(1-v^2)}{1-v^2}=\dfrac{v^3}{1-v^2}\).
  3. Separate: \(\dfrac{1-v^2}{v^3}\,dv=\dfrac{dx}{x}\). Split: \(\int(v^{-3}-v^{-1})\,dv=\int\dfrac{dx}{x}\), giving \(-\dfrac{1}{2v^2}-\ln|v|=\ln|x|+C\). Back-substitute \(v=y/x\) and simplify.
  4. Lesson: the \(y=vx\) substitution always converts a homogeneous DE to a separable DE — that's the entire trick.
Spot-test for homogeneity in seconds: count the total degrees of every term in numerator and denominator. If they all match, it's homogeneous. Most JEE-style problems test recognition of this pattern more than the algebraic gymnastics that follow.

Competency-Based Questions

Scenario: Newton's Law of Cooling: a hot object cools at a rate proportional to the temperature difference with surroundings. \(dT/dt=-k(T-T_s)\) with \(T_s\) ambient temperature.
Q1. Solve Newton's cooling equation by separation of variables.
L3 Apply
Solution: \(\dfrac{dT}{T-T_s}=-k\,dt\). Integrate: \(\ln|T-T_s|=-kt+C\), so \(T-T_s=A\,e^{-kt}\). Hence \(T(t)=T_s+(T_0-T_s)e^{-kt}\) with \(T(0)=T_0\). Exponential decay to ambient.
Q2. (T/F) "\(\dfrac{dy}{dx}=x+y\) is a homogeneous DE." Justify.
L5 Evaluate
False. Replace \((x,y)\) with \((\lambda x,\lambda y)\): RHS becomes \(\lambda(x+y)\), which is \(\lambda\) times the original — degree 1, not 0. So NOT a homogeneous-of-degree-0 function. (However, it IS a linear DE — solvable in Part 3.)
Q3. Solve \(\dfrac{dy}{dx}=\dfrac{x+y}{x}\) (homogeneous).
L3 Apply
Solution: \(=1+y/x\). Set \(y=vx\): \(v+xv'=1+v\), so \(xv'=1\), \(dv=dx/x\), \(v=\ln|x|+C\). Back: \(y=x(\ln|x|+C)\).
Q4. Apply: a population grows at rate proportional to its size. Population doubles in 5 hours. Find time to triple.
L4 Analyse
Solution: \(dP/dt=kP\Rightarrow P(t)=P_0\,e^{kt}\). Doubles at t=5: \(2=e^{5k}\), so \(k=\ln 2/5\). Tripling: \(3=e^{kt}\Rightarrow t=\ln 3/k=5\ln 3/\ln 2\approx 7.92\) hours.
Q5. Design: a savings account compounds continuously at rate r per year. Find balance B(t) given B(0) = ₹10,000, and find r if B(10)=₹20,000.
L6 Create
Solution: \(dB/dt=rB\Rightarrow B(t)=10000\,e^{rt}\). \(B(10)=20000\Rightarrow e^{10r}=2\Rightarrow r=(\ln 2)/10\approx 0.0693\) i.e. about 6.93% per year. (This is the "rule-of-72": doubling time = 72/(rate%) ≈ 10.4 yrs at 7%.)

Assertion–Reason Questions

Assertion (A): \(\dfrac{dy}{dx}=\dfrac{y^2+x^2}{xy}\) is solved by the substitution \(y=vx\).
Reason (R): The RHS is a homogeneous function of degree zero.
(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 recognition test; A is the standard substitution.
Assertion (A): The DE \(\dfrac{dy}{dx}=g(x)h(y)\) is always solvable in closed form by separation.
Reason (R): Separation of variables converts the DE into integrals \(\int dy/h(y)\) and \(\int g(x)\,dx\) which always yield elementary functions.
(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: (d). A is technically false: the integrals \(\int g\) and \(\int 1/h\) may not be elementary (e.g. \(g=e^{-x^2}\) gives the non-elementary \(\sqrt\pi\,\text{erf}\)). R is true that separation is always the right first step. Subtle.
Assertion (A): The substitution \(y=vx\) on a homogeneous DE produces a separable DE in \(v\) and \(x\).
Reason (R): By chain rule, \(dy/dx=v+x(dv/dx)\), and substituting reduces the equation to \(x(dv/dx)=F(v)-v\), separable.
(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 computation justifying A.

Frequently Asked Questions — Variables Separable and Homogeneous Differential Equations

What is the variable-separable method?
If a first-order DE can be written dy/dx = g(x)·h(y), separate as dy/h(y) = g(x) dx and integrate.
What is a homogeneous differential equation?
A first-order DE dy/dx = F(x, y) is homogeneous if F(λx, λy) = F(x, y) for all λ — i.e. F depends only on y/x.
How do you solve a homogeneous DE?
Substitute y = vx; the DE becomes separable in v and x. Solve, then back-substitute v = y/x.
What is the difference between separable and homogeneous DEs?
In a separable DE the variables already separate cleanly. In a homogeneous DE they don't, but become separable after y = vx.
How do you check homogeneity?
Replace x with λx and y with λy. If the result equals the original (no λ remains), it's homogeneous of degree 0.
What is the general solution of dy/dx = ky?
y = A·e^(kx). The exponential growth/decay law.
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