Differential Equations

Formation, classification, and solution of ODEs — variable separable, homogeneous, linear, exact, and real-world applications for JEE Mains & Advanced.

Order & Degree Formation Variable Separable Homogeneous Linear First Order Applications

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Order, Degree & Formation of ODE

Understanding order and degree is the first step. Formation of ODEs by eliminating arbitrary constants is a JEE favourite.

◆ Order & Degree

Definition — Order The order of a differential equation is the order of the highest derivative present. The degree is the power of the highest-order derivative, after the equation has been made free of radicals and fractions involving derivatives.

Important Notes

The degree is defined only when the equation is polynomial in its derivatives.
\(y'' + \sin(y') = 0\): order 2, degree not defined (sin of derivative).
\(\left(\frac{d^2y}{dx^2}\right)^3 + y = 0\): order 2, degree 3.

★ Example

Find the order and degree of \(\left[1 + \left(\frac{dy}{dx}\right)^2\right]^{3/2} = \frac{d^2y}{dx^2}\).

Square both sides: \(\left[1 + (y')^2\right]^3 = (y'')^2\).
Order = 2 (highest derivative is \(y''\)), Degree = 2 (power of \(y''\) after clearing radicals).

◆ Formation of ODE

Method Given a family of curves with \(n\) arbitrary constants, differentiate \(n\) times and eliminate all constants to obtain an ODE of order \(n\).

★ Example

Form the ODE representing the family of circles \(x^2 + y^2 + 2gx + 2fy + c = 0\) with \(g, f, c\) as arbitrary constants.

Three constants require three differentiations.
Differentiating once: \(2x + 2yy' + 2g + 2fy' = 0\) — (i)
Twice: \(2 + 2(y')^2 + 2yy'' + 2fy'' = 0\) — (ii)
From (ii): \(f = -\frac{1 + (y')^2 + yy''}{y''}\).
Thrice: Differentiate (ii) to get an equation involving only \(x, y, y', y'', y'''\).
The resulting ODE has order 3.

Variable Separable & Reducible Forms

The simplest type of ODE — separate variables to each side and integrate. JEE also tests substitutions that reduce equations to separable form.

◆ Variable Separable Method

Definition An ODE \(\frac{dy}{dx} = f(x,y)\) is separable if it can be written as \(g(y)\,dy = h(x)\,dx\). Integrate both sides to solve.

★ Example

Solve \(\frac{dy}{dx} = \frac{1+y^2}{1+x^2}\).

Separating: \(\frac{dy}{1+y^2} = \frac{dx}{1+x^2}\).
Integrating: \(\tan^{-1}y = \tan^{-1}x + C\).
Or equivalently: \(\tan^{-1}y - \tan^{-1}x = C\), i.e., \(\tan^{-1}\frac{y-x}{1+xy} = C\).
Solution: \(y - x = C(1 + xy)\).

◆ Reducible to Separable

Substitution Techniques

\(\frac{dy}{dx} = f(ax+by+c)\): Substitute \(v = ax+by+c\), then \(\frac{dv}{dx} = a + b\,f(v)\).
\(\frac{dy}{dx} = f\!\left(\frac{y}{x}\right)\): Homogeneous — substitute \(v = y/x\).

★ Example

Solve \(\frac{dy}{dx} = (x+y)^2\).

Let \(v = x + y\), so \(\frac{dv}{dx} = 1 + \frac{dy}{dx} = 1 + v^2\).
\(\frac{dv}{1+v^2} = dx \Rightarrow \tan^{-1}v = x + C\).
Solution: \(\tan^{-1}(x+y) = x + C\), or \(x + y = \tan(x + C)\).

Homogeneous Differential Equations

An equation is homogeneous if it can be expressed as \(dy/dx = \phi(y/x)\). The substitution \(y = vx\) converts it to a separable equation.

◆ Homogeneous Functions & Solution Method

Definition — Homogeneous ODE \(\frac{dy}{dx} = \frac{f(x,y)}{g(x,y)}\) is homogeneous if \(f\) and \(g\) are homogeneous functions of the same degree. Equivalently, \(\frac{dy}{dx} = \phi\!\left(\frac{y}{x}\right)\).

Solution Method Substitute \(y = vx\), so \(\frac{dy}{dx} = v + x\frac{dv}{dx}\). The equation becomes: \[v + x\frac{dv}{dx} = \phi(v) \quad\Rightarrow\quad \frac{dv}{\phi(v) - v} = \frac{dx}{x}\] which is separable.

★ Example

Solve \(x\,dy - y\,dx = \sqrt{x^2+y^2}\,dx\).

Rewrite: \(\frac{dy}{dx} = \frac{y + \sqrt{x^2+y^2}}{x}\). Let \(y = vx\): \[v + x\frac{dv}{dx} = v + \sqrt{1+v^2} \Rightarrow x\frac{dv}{dx} = \sqrt{1+v^2}\] \[\frac{dv}{\sqrt{1+v^2}} = \frac{dx}{x} \Rightarrow \ln(v + \sqrt{1+v^2}) = \ln|x| + \ln C\] \[v + \sqrt{1+v^2} = Cx \Rightarrow \frac{y}{x} + \frac{\sqrt{x^2+y^2}}{x} = Cx\] Solution: \(y + \sqrt{x^2+y^2} = Cx^2\).

Linear First-Order ODE & Exact Equations

The integrating factor method for linear ODEs is the most important technique in JEE differential equations.

◆ Linear First-Order ODE

Definition — Linear ODE A first-order linear ODE has the form: \[\frac{dy}{dx} + P(x)\,y = Q(x)\]

Integrating Factor Method The integrating factor is \(\mu = e^{\int P(x)\,dx}\). The solution is: \[y \cdot \mu = \int Q(x) \cdot \mu\,dx + C\] JEE Shortcut: If the equation is \(\frac{dx}{dy} + P(y)\,x = Q(y)\), use IF \(= e^{\int P(y)\,dy}\) and solve for \(x\).

★ Example

Solve \(\frac{dy}{dx} + \frac{2y}{x} = x^2\), \(y(1) = 1\).

\(P = \frac{2}{x}\), IF \(= e^{\int 2/x\,dx} = e^{2\ln x} = x^2\).
Multiplying: \(\frac{d}{dx}(x^2 y) = x^4\).
\(x^2 y = \frac{x^5}{5} + C\). Using \(y(1) = 1\): \(1 = \frac{1}{5} + C \Rightarrow C = \frac{4}{5}\).
Solution: \(y = \frac{x^3}{5} + \frac{4}{5x^2}\).

◆ Exact Differential Equations

Definition — Exact Equation \(M(x,y)\,dx + N(x,y)\,dy = 0\) is exact if \(\frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}\). The solution satisfies \(F(x,y) = C\) where \(\frac{\partial F}{\partial x} = M\) and \(\frac{\partial F}{\partial y} = N\).

★ Example

Solve \((2xy + 3)\,dx + (x^2 + 4y)\,dy = 0\).

Check: \(\frac{\partial M}{\partial y} = 2x\), \(\frac{\partial N}{\partial x} = 2x\). Equal, so exact.
\(F = \int M\,dx = x^2y + 3x + g(y)\).
\(\frac{\partial F}{\partial y} = x^2 + g'(y) = x^2 + 4y \Rightarrow g'(y) = 4y \Rightarrow g(y) = 2y^2\).
Solution: \(x^2y + 3x + 2y^2 = C\).

Applications of Differential Equations

JEE tests formation and solving of ODEs in physical contexts: growth/decay, Newton's cooling, and orthogonal trajectories.

◆ Growth, Decay & Newton's Law of Cooling

Standard Models

Exponential growth/decay: \(\frac{dN}{dt} = kN \Rightarrow N = N_0 e^{kt}\). If \(k > 0\), growth; if \(k < 0\), decay.
Newton's cooling: \(\frac{dT}{dt} = -k(T - T_s)\) where \(T_s\) is surroundings temperature. Solution: \(T - T_s = (T_0 - T_s)e^{-kt}\).

★ Example

A radioactive substance has a half-life of 5 years. How long does it take for 90% of the substance to decay?

\(N = N_0 e^{-kt}\). At \(t = 5\): \(\frac{N_0}{2} = N_0 e^{-5k} \Rightarrow k = \frac{\ln 2}{5}\).
For 90% decay: \(0.1N_0 = N_0 e^{-kt} \Rightarrow -kt = \ln 0.1 = -\ln 10\).
\(t = \frac{\ln 10}{k} = \frac{5\ln 10}{\ln 2} = 5 \cdot \frac{\ln 10}{\ln 2} \approx 5 \times 3.32 = \mathbf{16.6 \text{ years}}\).

◆ Orthogonal Trajectories

Definition Two families of curves are orthogonal trajectories if every curve of one family intersects every curve of the other family at right angles.

Method

Find the ODE of the given family: \(\frac{dy}{dx} = f(x,y)\).
Replace \(\frac{dy}{dx}\) by \(-\frac{dx}{dy}\) (or \(-\frac{1}{f(x,y)}\)).
Solve the new ODE.

★ Example

Find the orthogonal trajectories of \(y = cx^2\).

Differentiating: \(\frac{dy}{dx} = 2cx\). From the curve, \(c = \frac{y}{x^2}\), so \(\frac{dy}{dx} = \frac{2y}{x}\).
For orthogonal trajectories: \(\frac{dy}{dx} = -\frac{x}{2y}\).
\(2y\,dy = -x\,dx \Rightarrow y^2 = -\frac{x^2}{2} + C\).
Orthogonal trajectories: \(x^2 + 2y^2 = C\) (family of ellipses).

★ Key Takeaways

A family of curves with \(n\) arbitrary constants gives an ODE of order \(n\).
For homogeneous equations, the substitution \(y = vx\) always reduces to a separable form.
The integrating factor \(e^{\int P\,dx}\) transforms a linear ODE into an exact derivative.
For exact equations, verify \(\partial M/\partial y = \partial N/\partial x\) first, then integrate systematically.
Orthogonal trajectories: replace \(dy/dx\) with \(-dx/dy\) in the ODE of the given family.

📝 Practice Problems

Problem 1

Find the order and degree of the ODE \(\frac{d^2y}{dx^2} = \left[1 + \left(\frac{dy}{dx}\right)^2\right]^{3/2}\).

Show Solution ▼

Square both sides: \((y'')^2 = [1 + (y')^2]^3\). Order = 2 (highest derivative is \(y''\)). Degree = 2 (power of \(y''\) after clearing radicals).

Problem 2

Solve \(\frac{dy}{dx} = e^{x+y}\) with \(y(0) = 0\).

Show Solution ▼

\(\frac{dy}{dx} = e^x \cdot e^y\). Separating: \(e^{-y}\,dy = e^x\,dx\). Integrating: \(-e^{-y} = e^x + C\). Using \(y(0) = 0\): \(-1 = 1 + C \Rightarrow C = -2\). Solution: \(-e^{-y} = e^x - 2\), or \(e^{-y} = 2 - e^x\). Thus \(y = -\ln(2 - e^x)\), valid for \(x < \ln 2\).

Problem 3

Solve \((x^2 - y^2)\,dx + 2xy\,dy = 0\).

Show Solution ▼

This is homogeneous (both terms are degree 2). Rewriting: \(\frac{dy}{dx} = \frac{y^2 - x^2}{2xy}\). Let \(y = vx\): \(v + xv' = \frac{v^2 - 1}{2v}\). So \(xv' = \frac{v^2 - 1 - 2v^2}{2v} = \frac{-(1+v^2)}{2v}\). Separating: \(\frac{2v\,dv}{1+v^2} = -\frac{dx}{x}\). Integrating: \(\ln(1+v^2) = -\ln|x| + C\). So \(x(1+v^2) = A\), i.e., \(x + y^2/x = A\). Solution: \(x^2 + y^2 = Ax\).

Problem 4

Solve the Bernoulli equation \(\frac{dy}{dx} + y = y^2 e^x\).

Show Solution ▼

Divide by \(y^2\): \(y^{-2}\frac{dy}{dx} + y^{-1} = e^x\). Let \(v = y^{-1}\), \(\frac{dv}{dx} = -y^{-2}\frac{dy}{dx}\). So \(-\frac{dv}{dx} + v = e^x\), i.e., \(\frac{dv}{dx} - v = -e^x\). IF \(= e^{-x}\). Solution: \(ve^{-x} = \int -e^x \cdot e^{-x}\,dx = -x + C\). So \(\frac{e^{-x}}{y} = -x + C\), i.e., \(\frac{1}{y} = (C-x)e^x\). Solution: \(y = \frac{e^{-x}}{C - x}\).

Problem 5

A body cools from 80°C to 60°C in 10 minutes in a room at 20°C. How long will it take to cool from 60°C to 40°C?

Show Solution ▼

Newton's cooling: \(T - 20 = (T_0 - 20)e^{-kt}\). First case: \(60 - 20 = (80 - 20)e^{-10k}\), so \(e^{-10k} = \frac{2}{3}\). Second case: \(40 - 20 = (60 - 20)e^{-kt}\), so \(e^{-kt} = \frac{1}{2}\). From the first: \(e^{-k} = (2/3)^{1/10}\). So \((2/3)^{t/10} = 1/2\), giving \(t = 10 \cdot \frac{\ln 2}{\ln(3/2)} = 10 \cdot \frac{0.693}{0.405} \approx \mathbf{17.1 \text{ minutes}}\).

🎯 Interactive Quiz

1. The degree of the differential equation \(\left(\frac{d^3y}{dx^3}\right)^2 + \left(\frac{d^2y}{dx^2}\right)^3 + \frac{dy}{dx} + y = 0\) is:

A 3

B 2

C 1

D Not defined

2. The family of curves \(y = ae^{bx}\) (where \(a, b\) are arbitrary constants) satisfies a differential equation of order:

A 1

B 2

C 3

D 4

3. The integrating factor of \(\frac{dy}{dx} + y\tan x = \sec x\) is:

A \(\cos x\)

B \(\sec x\)

C \(e^{\sin x}\)

D \(\tan x\)

4. The orthogonal trajectories of \(xy = c\) are:

A \(x^2 + y^2 = c\)

B \(x^2 - y^2 = c\)

C \(y = cx^2\)

D \(y = ce^x\)

5. The solution of \(\frac{dy}{dx} + 2y = e^{-x}\) with \(y(0) = 1\) at \(x = 1\) is:

A \(e^{-1}\)

B \(e^{-2}\)

C \(\frac{1}{2}\)

D \(2e^{-1}\)