∫ SJ Maths Mathematics Notes
UP PGT Tracker UP TGT Tracker
3. Ordinary Differential Equations

Homogeneous differential equations

Homogeneous differential equations are first-order ordinary differential equations in which every term (or the ratio of differentials) can be expressed as a homogeneous function of degree zero.

For: UP PGT Mathematics UP TGT Mathematics DSSSB PGT / TGT KVS / NVS PGT LT Grade
📖 Topic progress
○ Mark as complete

Definition and statement

Definition and Reduction Principle of Homogeneous Differential Equations
A first-order differential equation of the form M(x, y)\,dx + N(x, y)\,dy = 0 \text{ or } \frac{dy}{dx} = f(x, y) is said to be homogeneous if f(x, y) is a homogeneous function of degree zero, meaning f(kx, ky) = k^0 f(x, y) = f(x, y) \text{ for any non-zero real scalar } k.

📐 Geometric Interpretation

⚙️ Algebraic & Structural Meaning

Algebraically, the substitution $y = vx$ (where $v$ is a new dependent variable) or $x = vy$ reduces the non-linear homogeneous equation into a variable-separable ordinary differential equation of the form \int \frac{dv}{g(v) - v} = \int \frac{dx}{x}.

Conditions and hypotheses

Check each condition and what changes when it is omitted:

Hypothesis / Condition Domain / Interval Necessity What Fails If Condition Dropped
Degree Homogeneity Domain in \mathbb{R}^2 excluding origin Strictly Mandatory / Necessary If M(x, y) and N(x, y) are not strictly homogeneous of the same degree, the algebraic substitution $y = vx$ will fail to eliminate $x$ from the resulting separable equation.
Continuity and Differentiability of Coefficients Open domain in \mathbb{R}^2 Strictly Mandatory Failure of continuity or local Lipschitz conditions in the domain may lead to non-existence or non-uniqueness of solutions per Picard-Lindelöf theorem.
Non-zero Denominator Open intervals on coordinate axes Essential for separation of variables Division by zero during variable separation leads to loss of singular solutions (such as radial straight line solutions $y = mx$).

Useful results and shortcuts

Use these identities and results when their conditions apply:

01

Standard Substitution Rule for \frac{dy}{dx} = F\left(\frac{y}{x}\right)

$$ \frac{dy}{dx} = v + x\frac{dv}{dx} \implies \int \frac{dv}{F(v) - v} = \ln|x| + C $$

Rule: Instantly substitute $y = vx$ and $dy = v,dx + x,dv$, separate variables, and integrate without expanding intermediate brackets.

02

Reciprocal Substitution for \frac{dx}{dy} = G\left(\frac{x}{y}\right)

$$ \frac{dx}{dy} = u + y\frac{du}{dy} \implies \int \frac{du}{G(u) - u} = \ln|y| + C \quad (x = uy) $$

Rule: When terms in $x$ dominate or numerator/denominator is symmetric with $x$ and $y$ swapped, use $x = uy$ to bypass complex algebra.

Application: Saves 2-3 minutes on complex fractional homogeneous equations in DSSSB and UP PGT.
03

Exact Homogeneous Form Test

$$ M(x,y)dx + N(x,y)dy = 0 \text{ is homogeneous if } x\frac{\partial M}{\partial x} + y\frac{\partial M}{\partial y} = -n M $$

Rule: Use Euler's theorem for homogeneous functions to verify degree $n$ instantly before attempting integration.

Application: Frequently asked in theoretical statement-based multiple-choice questions in screening tests.
04

Integration of Standard Rational Homogeneous Kernels

$$ \int \frac{v\,dv}{1 + v^2} = \frac{1}{2}\ln|1 + v^2|, \quad \int \frac{dv}{1 - v^2} = \frac{1}{2}\ln\left|\frac{1+v}{1-v}\right| $$

Rule: Memorize standard logarithmic and inverse trigonometric integrals resulting from partial fractions in homogeneous equations.

Application: Direct formula application in UP TGT rapid calculation questions.
05

Radial Line Solutions ($y = mx$)

$$ F(m) = m \text{ or roots of } x N(x,mx) + y M(x,mx) = 0 $$

Rule: Check if straight lines passing through the origin satisfy the differential equation directly as singular or special solutions.

Application: Crucial for identifying missing options in multiple-choice questions where integrating constants merge with logarithmic terms.
06

Rapid Logarithmic Consolidation

$$ \ln|v - v_1| - \ln|v - v_2| = \ln|x| + \ln C \implies \frac{v - v_1}{v - v_2} = Cx $$

Rule: Combine all logarithmic integration constants into a single term $\ln|C|$ on one side to match options instantly.

Common errors and counterexamples

These examples show how a conclusion can fail when a condition is missing:

⚠️ COMMON ERROR #01

Non-Homogeneous Equations Disguised as Homogeneous

Common Pitfall: Applying $y = vx$ to equations with constant translation terms like $\frac{dy}{dx} = \frac{x + y + 1}{x - y + 2}$ without shifting the origin.

Counterexample: \frac{dy}{dx} = \frac{x + y + 1}{x - y + 2}

Why it fails: The presence of constant terms makes the numerator and denominator non-homogeneous. Direct substitution leads to unresolvable residual constants.

🛡️ Exam Defense: Check if $\frac{a_1}{a_2} \neq \frac{b_1}{b_2}$. If so, shift origin via $x = X + h, y = Y + k$ to eliminate constants first.
⚠️ COMMON ERROR #02

Forgetting Absolute Value Signs in Logarithms

Common Pitfall: Writing $\ln x$ instead of $\ln|x|$ during integration of $\frac{1}{x}$, leading to domain restriction errors.

Counterexample: \int \frac{dx}{x} = \ln x \text{ instead of } \ln|x|

Why it fails: Restricts the solution domain to positive real numbers, discarding valid branches where $x < 0$.

🛡️ Exam Defense: Always include absolute value bars, i.e., $\ln|x|$, and combine constants as $\ln|C|$.

Comparison

Concept / TheoremHypotheses / AssumptionsKey Formula / OutcomeGeometric MeaningExam Identification Clue
Variable Separablef(x,y) = g(x)h(y)\int \frac{dy}{h(y)} = \int g(x)\,dxOrthogonal vector fieldsTerms easily factored into x and y functions
Homogeneous ODEf(kx, ky) = k^0 f(x,y)y = vx \implies \int \frac{dv}{F(v)-v} = \ln|x|+CScale-invariant slope field along radial linesEvery term has identical total degree in x and y
Reducible to Homogeneous\frac{a_1}{a_2} \neq \frac{b_1}{b_2}x = X+h, y = Y+kTranslational shift of originLinear terms with constant intercepts in numerator/denominator
Linear ODE\frac{dy}{dx} + P(x)y = Q(x)y \cdot e^{\int Pdx} = \int Q e^{\int Pdx} dx + CSuperposition principle appliesDependent variable y and its derivative appear to degree 1 only

Solved examples

Each problem includes a full solution and a shorter method:

CANONICAL PROBLEM 01 Dual-Method Breakdown

Find the general solution of the homogeneous differential equation $(x^2 - y^2)\,dx + 2xy\,dy = 0$.

(A) x^2 + y^2 = Cx(B) x^2 - y^2 = Cy(C) x^2 + y^2 = Cy^2(D) xy = C(x^2 - y^2)
✓ Correct Answer: (A)
📚 Method A: Formal Textbook Derivation
Step 1: Check homogeneity: degree of each term is 2. Step 2: Put $y = vx \implies dy = vdx + xdv$. Step 3: Substitute into equation: $(x^2 - v^2x^2)dx + 2x(vx)(vdx + xdv) = 0$. Step 4: Simplify to $(1 + v^2)dx + 2xv dv = 0$. Step 5: Separate variables: $\frac{2v}{1+v^2}dv + \frac{dx}{x} = 0$. Step 6: Integrate to get $\ln(1+v^2) + \ln|x| = \ln C \implies x(1 + v^2) = C \implies x^2 + y^2 = Cx$.
⚡ Method B: 25-Second Exam Shortcut
Recognize the standard pattern $(x^2 - y^2)dx + 2xydy = 0$ as the differential form associated with circles touching the y-axis or x-axis. Matching degrees yields $x^2 + y^2 = Cx$ instantly.
💡 Tip: Watch for sign conventions when substituting $dy$.
CANONICAL PROBLEM 02 Dual-Method Breakdown

Solve the initial value problem $\frac{dy}{dx} = \frac{y}{x} + \tan\left(\frac{y}{x}\right)$ given $y(1) = \frac{\pi}{2}$.

(A) \sin\left(\frac{y}{x}\right) = Cx(B) \cos\left(\frac{y}{x}\right) = Cx(C) \tan\left(\frac{y}{x}\right) = Cx(D) \sin(xy) = C
✓ Correct Answer: (A)
📚 Method A: Formal Textbook Derivation
Step 1: Substitute $y = vx \implies \frac{dy}{dx} = v + x\frac{dv}{dx}$. Step 2: Equation becomes $v + x\frac{dv}{dx} = v + \tan v \implies x\frac{dv}{dx} = \tan v$. Step 3: Separate variables: $\cot v\,dv = \frac{dx}{x}$. Step 4: Integrate: $\ln|\sin v| = \ln|x| + \ln C \implies \sin v = Cx \implies \sin\left(\frac{y}{x}\right) = Cx$.
⚡ Method B: 25-Second Exam Shortcut
Direct integration of $\frac{dv}{\tan v}$ gives $\ln|\sin v|$, which immediately eliminates non-sinusoidal options.
💡 Tip: Substitute boundary conditions immediately to find the constant $C$ if required.

Practice questions

Choose an answer to check it and see the solution:

Q1
MCQ 1: The differential equation $(x^3 + y^3)\,dx + 3xy^2\,dy = 0$ is homogeneous of degree:
A 1
B 2
C 3
D 0
Derivation & Analysis:

Every term in the expression has a total degree of 3 with respect to variables $x$ and $y$ ($x^3, y^3, xy^2$ all have degree 3).

Q2
MCQ 2: To solve the homogeneous differential equation $x \frac{dy}{dx} = y + x \cos^2\left(\frac{y}{x}\right)$, the appropriate substitution is:
A $v = xy$
B $y = vx$
C $x = vy$
D $y^2 = vx$
Derivation & Analysis:

Since the expression contains $\frac{y}{x}$ throughout, the standard substitution is $y = vx$.

Q3
MCQ 3: The general solution of $x \frac{dy}{dx} - y = \sqrt{x^2 + y^2}$ is:
A $y + \sqrt{x^2 + y^2} = Cx^2$
B $y - \sqrt{x^2 + y^2} = Cx$
C $y + \sqrt{x^2 + y^2} = Cx$
D $x + \sqrt{x^2 + y^2} = Cy$
Derivation & Analysis:

Substituting $y = vx$ yields $\frac{dv}{\sqrt{1+v^2}} = \frac{dx}{x}$, integrating to $\ln|v + \sqrt{1+v^2}| = \ln|x| + \ln C$, which simplifies to $y + \sqrt{x^2 + y^2} = Cx$.

Q4
MCQ 4: Which of the following differential equations is NOT homogeneous?
A $\frac{dy}{dx} = \frac{x^2 + y^2}{xy}$
B $\frac{dy}{dx} = \frac{x + y}{x - y}$
C $\frac{dy}{dx} = \frac{x^2 + y + 1}{x}$
D $\frac{dy}{dx} = \frac{y}{x} + \sin\left(\frac{y}{x}\right)$
Derivation & Analysis:

In option C, the term $+1$ in the numerator destroys homogeneity because scaling $x$ and $y$ by $k$ does not preserve the factor $k^0$.

Q5
MCQ 5: The degree of homogeneity of $f(x,y) = \frac{x^4 + y^4}{x^2 + y^2}$ is:
A 4
B 2
C 0
D 1
Derivation & Analysis:

$f(kx, ky) = \frac{k^4(x^4 + y^4)}{k^2(x^2 + y^2)} = k^2 \frac{x^4 + y^4}{x^2 + y^2} = k^2 f(x,y)$, hence the degree is 2.

Q6
MCQ 6: The solution of $x \frac{dy}{dx} = y + x$ is:
A $y = x \ln|x| + Cx$
B $y = Cx \ln|x|$
C $y = x^2 + Cx$
D $y = C \ln|x|$
Derivation & Analysis:

Rewriting gives $\frac{dy}{dx} - \frac{y}{x} = 1$, which is linear (and homogeneous). Integrating factor is $e^{-\int \frac{dx}{x}} = \frac{1}{x}$, leading to $y = x \ln|x| + Cx$.

Q7
MCQ 7: Integrating factor after substituting $y = vx$ in a homogeneous equation always reduces to:
A $\int \frac{1}{x} dx$
B $\int \frac{1}{v} dv$
C $e^{\int P(x)dx}$
D $v e^{\int dx}$
Derivation & Analysis:

Separation of variables in homogeneous equations invariably isolates $\frac{dx}{x}$ on one side, whose integral is $\ln|x|$.ahiran.

Q8
MCQ 8: Assertion (A): Every homogeneous differential equation of first order can be solved by substituting $y = vx$. Reason (R): Homogeneous functions satisfy Euler's theorem.
A Both (A) and (R) are true and (R) is the correct explanation of (A)
B Both (A) and (R) are true but (R) is not the correct explanation of (A)
C (A) is true but (R) is false
D (A) is false but (R) is true
Derivation & Analysis:

Both statements are mathematically correct theorems, but Euler's theorem on homogeneous functions is not the direct deductive justification for the $y=vx$ substitution method.

Q9
MCQ 9: The curve passing through $(1, 1)$ satisfying $x \frac{dy}{dx} = y(ln y - ln x + 1)$ is:
A $y = x e^{1-x}$
B $y = x e^{x-1}$
C $y = x \ln x$
D $y = x e^x$
Derivation & Analysis:

Rewriting with $v = \frac{y}{x}$ gives $v \ln v$ structure which integrates to $y = x e^{x-1}$ upon applying the point $(1,1)$.

Q10
MCQ 10: If $M(x,y)dx + N(x,y)dy = 0$ is a homogeneous equation where $M$ and $N$ are homogeneous functions of degree $n$, then $\frac{dy}{dx}$ is a homogeneous function of degree:
A n
B 0
C -n
D 1
Derivation & Analysis:

Ratio of two homogeneous functions of degree $n$ is a homogeneous function of degree $n - n = 0$.

Common questions

Check every individual term in $M(x,y)$ and $N(x,y)$. If the sum of the powers of $x$ and $y$ is identical across all additive terms, the equation is homogeneous of that degree. Alternatively, replace $x$ with $kx$ and $y$ with $ky$ and check if $k^n$ factors out completely.

Use $x = vy$ (where $\frac{dx}{dy}$ is isolated) when the differential equation is more symmetric with respect to $x$ or when terms involving $x$ are multiplied by $\frac{dx}{dy}$ in a simpler algebraic structure, often preventing cumbersome reciprocal fractions.

Common pitfalls include algebraic sign errors during the expansion of $dy = v\,dx + x\,dv$, omitting absolute value signs in logarithms, and failing to check for singular radial solutions ($y = mx$).

← Previous Topic
Formation of differential equations
Next Topic →
Linear differential equations with constant coefficients