Chapter 9 — Differential Equations

A differential equation (DE) is an equation that relates an unknown function to its derivatives. In general form a DE is written as F(x, y, y', y'', ..., y^{(n)}) = 0 where y = y(x) and y^{(k)} denotes the k-th derivative. The order of a DE is the highest order derivative present. The degree is the power of the highest order derivative when the equation is a polynomial in derivatives.

Key solution concepts:

• General solution: a family of solutions containing n arbitrary constants for an n-th order DE. Example symbolically y = phi(x, C1, C2, ..., Cn).
• Particular solution: obtained from the general solution by giving specific values to the arbitrary constants, often using initial or boundary conditions.
• Singular solution: a solution that cannot be obtained from the general solution by choosing constant values; often an envelope of the family of general solutions.
• Initial value problem (IVP): a DE together with specified values of the function and possibly its derivatives at a point (e.g. y(x0)=y0, y'(x0)=y1). An IVP selects a unique particular solution when existence and uniqueness conditions hold.

Classification and common first-order types (useful for solving):

• Separable: dy/dx = g(x) h(y) — variables can be separated and integrated.
• Linear first-order: dy/dx + P(x)y = Q(x) — solved using an integrating factor.
• Homogeneous (in x,y): dy/dx = f(y/x) — use substitution y = vx.
• Exact: M(x,y) + N(x,y) y' = 0 where dM/dy = dN/dx; solved by finding a potential function.

Formation of a differential equation: if a family of curves depends on n arbitrary constants, differentiate enough times to eliminate the constants and obtain an n-th order DE satisfied by the family. Example: family y = ax + b leads to y'' = 0 after eliminating a and b.

Existence and uniqueness (first-order): for y' = f(x,y), if f and partial f/partial y are continuous near (x0,y0), there exists a unique solution through (x0,y0).

Geometric idea: solutions are integral curves in the xy-plane. Slope (direction) fields visualize the slope y' at many points and help sketch solution curves.

General solution: For an ordinary differential equation (ODE) of order n, a general solution is a family of functions containing n arbitrary constants. It represents all possible solutions of that order. Example format: y = f(x, C1, C2, ..., Cn).

Particular solution: A particular solution is obtained from the general solution by giving specific values to the arbitrary constants using initial or boundary conditions (ICs). Example: from y = Ce^{kx}, using y(0)=y0 gives C = y0 and the particular solution y = y0 e^{kx}.

Singular solution: A singular solution is a solution of the differential equation that cannot be obtained by choosing constants in the general solution. Often it is the envelope of the one-parameter family given by the general solution. To detect a singular solution when the general solution is given as y = f(x,C) (one-parameter family), differentiate with respect to C and eliminate C using the two equations

y = f(x,C),   ∂f/∂C = 0.

If elimination yields a new relation y = g(x) that satisfies the original DE but is not obtainable by any fixed C, g(x) is a singular solution (the envelope).

Important notes:

• An nth-order ODE has a general solution with n arbitrary constants; initial/boundary conditions fix those constants to give a unique particular solution (if conditions are sufficient).
• Clairaut's equation y = x p + f(p) (with p = dy/dx) gives a typical example: general solution y = C x + f(C) (family of straight lines) and the envelope (singular solution) found from x + f'(C) = 0.

What it means
Formation of a differential equation means finding a differential equation (DE) whose general solution is a given family of curves containing arbitrary constants. The order of the required DE equals the number of arbitrary constants in the family.

General procedure

1. Start with the given family of curves expressed as F(x,y,C1,C2,...,Cn)=0, where C1...Cn are arbitrary constants.
2. Differentiate the relation with respect to x successively n times to obtain n equations involving x, y and derivatives up to y^(n).
3. Eliminate the constants C1...Cn between the original equation and the n differentiated equations. The resulting relation between x, y and derivatives y', y'', ..., y^(n) is the required differential equation of order n.
4. Check: the general solution of the obtained DE should contain n arbitrary constants and reproduce the original family.

Important points

• Order of DE = number of arbitrary constants in the family.
• Elimination may require algebraic manipulation (solving a linear system or substitution).
• If an arbitrary constant appears multiplied by a function that can be zero identically, consider singular solutions/envelopes separately.

Short illustrated steps (example idea)

Example: y = ax^2 + bx + c  (three constants a,b,c)
1) y' = 2ax + b
2) y'' = 2a
Now eliminate a,b,c: from y'' = 2a => a = y''/2
from y' = 2ax + b => b = y' - 2ax = y' - x*y''
Substitute a,b into y = ax^2 + bx + c to eliminate a,b and solve for c (or just eliminate c by forming relation)
Since three constants, third order DE is y''' = 0, whose general solution is y = Ax^2 + Bx + C.

Why this matters
Forming DEs connects geometric families of curves to the analytic language of derivatives. Many physical laws and models are obtained by recognizing that observed families of solutions satisfy a particular differential relation.

What is a separable differential equation?
A first-order differential equation is called separable if it can be written in the form dy/dx = f(x) g(y), i.e. all x-dependent factors and dx on one side and all y-dependent factors and dy on the other side.

General method (step-by-step)

1. Write the equation as dy/dx = G(x) H(y).
2. Separate variables: (1/H(y)) dy = G(x) dx. This requires H(y) ≠ 0 on the interval considered.
3. Integrate both sides: ∫ (1/H(y)) dy = ∫ G(x) dx + C.
4. Solve the resulting equation for y if possible. The result may be an explicit solution y = φ(x) or an implicit solution F(x,y) = C.
5. Use any initial condition y(x0) = y0 to determine the constant C and obtain a particular solution. Check for constant (singular) solutions that arise where H(y)=0.

Important remarks

• When integrating, absolute values appear from integrals like ∫ dy/y = ln|y| + C; keep track of domains and signs.
• If H(y)=0 for some y = y*, then y(x)=y* is a constant (singular) solution; include such solutions separately.
• Integration may require substitutions or partial fractions for rational expressions in y.
• Solutions can be implicit; sometimes an explicit form is not obtainable in elementary functions.

Existence/Uniqueness (brief)
If G and H are continuous and H(y0) ≠ 0, then by standard theorems a unique local solution through (x0,y0) exists. If H(y0)=0, constant solutions occur.

What is a homogeneous first-order differential equation?

A first-order differential equation is called homogeneous if the right-hand side can be written as a function of the ratio y/x (or if both differential terms are homogeneous functions of the same degree). Two common forms are:

• dy/dx = f(y/x), i.e. the slope at any point depends only on y/x.
• M(x,y) dx + N(x,y) dy = 0, where M and N are homogeneous functions of the same degree.

Key idea (substitution)

If dy/dx = f(y/x), put y = v x (so v = y/x). Then

dy/dx = v + x dv/dx

Substitute into the equation to get an equation in v and x. Typically this becomes separable in variables x and v and can be integrated. After integrating, replace v by y/x to obtain the solution in x and y.

Procedure (dy/dx = f(y/x))

1. Assume y = v x so that dy/dx = v + x dv/dx.
2. Substitute: v + x dv/dx = f(v).
3. Rearrange to separate variables: x dv/dx = f(v) - v.
4. Separate and integrate: ∫ 1/(f(v) - v) dv = ∫ dx/x + C.
5. Back-substitute v = y/x and simplify to get the general solution (often implicit).

Procedure (M dx + N dy = 0 with M,N homogeneous of same degree)

1. Check homogeneity: M(tx,ty) = t^n M(x,y), N(tx,ty) = t^n N(x,y).
2. Divide the equation by x^n (or y^n) to express M and N as functions of y/x: M(1,y/x) and N(1,y/x).
3. Write dy/dx = -M/N = F(y/x) and use substitution y = v x as above.

Important remarks

• Because substitution uses y/x, you must treat the line x = 0 separately (solutions through origin take special attention).
• Sometimes it is convenient to use x = u y (if y ≠ 0) leading to dx/dy = u + y du/dy.
• Solutions are often implicit; solve explicitly only when algebra permits.

Definition and standard form
A linear first-order ordinary differential equation (ODE) has the form dy/dx + P(x) y = Q(x), where P and Q are known functions of x and y(x) is the unknown function.

Integrating factor method (main method)
Multiply the equation by an integrating factor μ(x) chosen so the left-hand side becomes an exact derivative. Choose μ(x) = exp(∫P(x) dx). Then
μ(x) [dy/dx + P(x) y] = μ(x) Q(x) ⇒ d/dx[μ(x) y] = μ(x) Q(x).
Integrate both sides to get μ(x) y = ∫ μ(x) Q(x) dx + C, so
y(x) = (1/μ(x)) [∫ μ(x) Q(x) dx + C].

Step-by-step procedure

1. Write the equation in standard form dy/dx + P(x) y = Q(x).
2. Compute μ(x) = exp(∫P(x) dx) (choose any antiderivative inside the exponential).
3. Form d/dx[μ(x) y] and verify it equals μ(x) Q(x).
4. Integrate: μ(x) y = ∫ μ(x) Q(x) dx + C.
5. Solve for y(x): y = (1/μ)[∫ μ Q dx + C]. Apply initial condition to find C if given.

Homogeneous part and particular solution
The associated homogeneous equation dy/dx + P(x) y = 0 has solution y_h = C exp(-∫P dx). The general solution is y = y_p + y_h where y_p is any particular solution found by the integrating factor formula.

Related transforms
Bernoulli equation y' + P(x) y = Q(x) y^n (n ≠ 0,1) can be converted to linear by substitution z = y^{1-n}. Many other nonlinear first-order ODEs can be reduced to linear form by suitable substitutions.

Remarks
- For constant coefficient P(x)=a (a constant) the integrating factor is μ = e^{ax} and the homogeneous solution is Ce^{-ax}.
- Physically, linear first-order equations model transients approaching a steady-state (exponential decay/growth toward a particular solution).

What it is
Reduction techniques and substitutions are standard ways to transform a differential equation into a simpler (often separable or linear) form so it can be solved. Typical substitutions reduce complexity by (a) removing explicit dependence on one variable, (b) exploiting homogeneity, or (c) converting a nonlinear equation into linear form.

Common types of substitutions

• Homogeneous substitution: If dy/dx = F(x,y) and F(tx,ty) = F(x,y) for all t (i.e. F is homogeneous of degree 0), use y = v x (or x = u y). Then dy/dx = v + x dv/dx and the equation becomes separable in v and x.
• Translation to make homogeneous: If F(x,y) = G(x+a, y+b) where G is homogeneous, shift variables X = x + a, Y = y + b and then use Y = v X.
• Bernoulli equation: dy/dx + P(x) y = Q(x) y^n (n ≠ 0,1). Use substitution z = y^{1-n} to convert it into a linear first-order equation in z.
• Linear first-order (integrating factor): dy/dx + P(x)y = Q(x). Use integrating factor μ(x)=exp(∫P(x)dx) to obtain d/dx[μ y] = μ Q and integrate.
• Substitution for expressions in linear combinations: If dy/dx = f((ax+by+c)/(a'x+b'y+c')), set u = (ax+by+c)/(a'x+b'y+c') to reduce variables.
• Reduction of order when independent variable is missing: For second-order equations that do not contain x explicitly, set p = dy/dx and treat d^2y/dx^2 = dp/dx = p dp/dy. This reduces the order by one.
• Integrating factor depending only on x or y: For M(x,y) dx + N(x,y) dy = 0 not exact, if ( (∂M/∂y - ∂N/∂x)/N ) is a function of x alone there exists μ(x); similarly if ( (∂N/∂x - ∂M/∂y)/M ) depends on y alone there exists μ(y).

Why it works
Substitutions use structure (scaling, symmetry, or special nonlinear form) to convert the differential equation into one of the standard solvable types — separable, exact, or linear.

When to try which

• Check homogeneity: if F(tx,ty)=F(x,y) try y=vx.
• If equation matches Bernoulli form, use z=y^{1-n}.
• If you see linear combinations ax+by+c in arguments, try a ratio substitution u=(ax+by+c)/(a'x+b'y+c').
• If independent variable x is absent in a higher-order equation, set p=dy/dx to reduce order.

Overview

Applications and modelling with differential equations means translating a real physical, biological or engineering situation into a differential equation that describes how a quantity changes, solving that equation (often with an initial condition) and interpreting the solution.

Modelling steps

• Identify the dependent variable (e.g., population P(t), temperature T(t), amount of salt A(t), velocity v(t)).
• Express the rate of change of that variable (d/dt of the variable) in terms of the variable and other known functions/parameters.
• Form the differential equation and specify initial/boundary conditions.
• Solve the differential equation using an appropriate method (separation of variables, integrating factor for linear first-order equations, known solutions for special forms) and interpret the result.

Common modelling types and their intuition

• Exponential growth/decay: rate proportional to the current amount (dN/dt = kN). Produces exponential increase if k>0 or exponential decay if k<0.
• Newton's law of cooling: rate of change of temperature proportional to difference from ambient (dT/dt = -k(T - T_a)). Solution: temperature approaches ambient exponentially.
• Mixing problems (tank problems): change in amount = inflow concentration × inflow rate − outflow concentration × outflow rate. These are linear first-order ODEs.
• Motion with linear resistance: mass × acceleration = driving force − resistance proportional to velocity (m dv/dt = mg − kv). Velocity approaches terminal value exponentially.
• Logistic growth: limited growth with carrying capacity K (dP/dt = rP(1 − P/K)) producing an S-shaped curve (sigmoid) instead of unbounded exponential.

Solution methods (brief)

• Separable equations: if dy/dx = f(x)g(y), rearrange to integrate: ∫ dy/g(y) = ∫ f(x) dx.
• Linear first-order: dy/dx + P(x)y = Q(x). Use integrating factor μ(x) = exp(∫P(x) dx) and write (μ y)' = μ Q → y = (1/μ) (∫ μ Q dx + C).
• Standard second-order (e.g., simple harmonic motion): d²x/dt² + ω² x = 0 has solution x = A cos(ωt) + B sin(ωt).

Interpretation

After solving, check units, limiting behaviour (t → ∞), special cases and whether the solution makes physical sense (e.g., nonnegative populations, bounded concentrations).

What is an Initial Value Problem (IVP)?
An IVP is a differential equation together with values of the unknown function (and possibly its derivatives) at a specific point. Example: find y(x) satisfying y'' - 3y' + 2y = e^x with y(0)=1 and y'(0)=0. The differential equation determines a family of solutions; the initial conditions pick the unique solution from that family.

General approach for linear ODEs with constant coefficients

1. Write the differential equation in standard form: L[y] = f(x), where L is a linear differential operator with constant coefficients (for example: y'' + ay' + by = f(x)).
2. Find the complementary function (CF), y_c, by solving the homogeneous equation L[y]=0 (solve characteristic polynomial to get exponentials, sines/cosines or polynomial solutions).
3. Find a particular integral (PI or y_p) — one specific solution of the non‑homogeneous equation L[y]=f(x). Methods: method of undetermined coefficients (for RHS like polynomials, exponentials, sines/cosines or their products) or variation of parameters (more general).
4. The general solution is y = y_c + y_p. Apply the initial conditions to determine the arbitrary constants in y_c and obtain the unique solution of the IVP.

Method of Undetermined Coefficients (summary)
If f(x) is of the form P(x)e^{ax}cos(bx) or P(x)e^{ax}sin(bx) (P polynomial), choose a trial y_p of the same form multiplied by x^s where s is the multiplicity of (a+ib) as a root of the characteristic equation. Plug the trial into the ODE and solve for the unknown coefficients.

Existence and uniqueness (brief)
For first order IVP y' = F(x,y) with F continuous and satisfying a Lipschitz condition in y near (x0,y0), there exists a unique local solution through (x0,y0). For linear ODEs with continuous coefficients, existence and uniqueness hold globally.

Why particular integrals matter
In physical systems (forced oscillators, circuits, thermal systems) solutions split into transient part (y_c, usually decays with time) and steady/state forced response (y_p). The particular integral often represents the steady-state effect of the forcing term.

In Class 12 Differential Equations, "Standard Examples and Problem Types" groups typical differential-equation forms and the standard methods to solve them. Recognising the form of an equation and choosing the right method is the key skill. Below are the common problem types, how to identify them, and the standard solution idea for each.

• Separable equations: equations of the form dy/dx = f(x)g(y). Solve by separating variables: ∫ dy/g(y) = ∫ f(x) dx + C.
• Homogeneous first-order equations: dy/dx = F(y/x) or where both numerator and denominator are homogeneous functions of same degree. Use substitution y = ux (so dy/dx = u + x du/dx) to reduce to separable or linear form.
• Linear first-order equations: dy/dx + P(x)y = Q(x). Use the integrating factor μ(x) = e^{∫P(x) dx}. Then (μ y)' = μ Q(x) and integrate.
• Bernoulli equation: dy/dx + P(x) y = Q(x) y^n (n ≠ 0,1). Use substitution v = y^{1-n} to transform it into a linear equation in v.
• Exact equations: M(x,y) dx + N(x,y) dy = 0. Check exactness: ∂M/∂y = ∂N/∂x. If exact, find potential function Φ(x,y) with Φ_x = M and Φ_y = N and set Φ(x,y) = C. If not exact, sometimes an integrating factor μ(x) or μ(y) (or rarely μ(x,y)) makes it exact.
• Higher-order linear equations with constant coefficients: e.g., ay'' + by' + cy = 0. Solve the characteristic equation ar^2 + br + c = 0. Use homogeneous solution forms for distinct/ repeated/ complex roots. For nonhomogeneous RHS, use method of undetermined coefficients or variation of parameters to get a particular solution.
• Orthogonal trajectories: Given a one-parameter family of curves F(x,y,c)=0, differentiate to eliminate c, get differential equation of the family, then replace dy/dx by -dx/dy to get the orthogonal family and solve.
• Modelling / application problems: Standard models include exponential growth/decay (dP/dt = kP), Newton's law of cooling (dT/dt = -k(T - T_s)), mixing problems (d(concentration)/dt = inflow - outflow), RC charging/discharging (first-order linear), and simple harmonic motion (y'' + ω^2 y = 0).

Strategy checklist for solving problems: (1) Identify type (separable/linear/homogeneous/exact/Bernoulli/constant-coefficient). (2) Apply the standard substitution or integrating factor. (3) Integrate and apply initial/boundary conditions if given. (4) Check the solution by differentiation.