Solving DEs

Differential Equations · free study notes · Updated

Solving Differential Equations – IIT‑JEE (Main + Advanced)

A concise, self‑contained guide to the most common DE techniques that appear in the JEE syllabus.


1. Simple Explanation

A differential equation (DE) is a relation that involves an unknown function (y(x)) and its derivatives.
The goal is to find all functions (y(x)) that satisfy the relation.

In the JEE context you will see:

Type Typical form What you need to know
First‑order (y' = f(x,y)) or (P(x) y' + Q(x) y = R(x)) Separable, Linear, Exact, Bernoulli, Integrating Factor
Second‑order (linear) (y'' + p(x)y' + q(x)y = r(x)) Homogeneous + Particular, Characteristic equation, Reduction of order
Higher‑order Rare Usually reducible to lower order
Non‑linear ((y')^2 = f(x,y)) Separable, Bernoulli, Homogeneous, or special substitutions

Key idea – Reduce the DE to a form you can integrate directly, or use a known formula (characteristic equation, integrating factor, etc.).


2. Key Points (Bullet‑style Cheat Sheet)

Category Method Conditions Resulting Integral
First‑order Separable Can be written ( \frac{dy}{dx} = g(x)h(y) ) (\displaystyle \int \frac{dy}{h(y)} = \int g(x),dx)
Linear (y' + p(x)y = q(x)) Integrating factor ( \mu(x)=e^{\int p(x)dx}) → (\displaystyle (\mu y)' = \mu q)
Exact (M(x,y)dx+N(x,y)dy=0) with (M_y=N_x) (\displaystyle \Phi(x,y)=C) where (\Phi_x=M, \Phi_y=N)
Bernoulli (y'+p(x)y = q(x)y^n) ( (n\ne0,1) ) Substitute (v=y^{1-n}) → linear in (v)
Homogeneous (F(\frac{y}{x})) or (M(x,y)dx+N(x,y)dy=0) with (M,N) homogeneous (v=\frac{y}{x}) → separable in (v,x)
Second‑order linear Homogeneous (y''+p(x)y'+q(x)y=0) Solve characteristic (m^2+ap+b=0) → (y = C_1e^{m_1x}+C_2e^{m_2x}) (or trig if complex)
Particular (und. coefficients) RHS a known function Assume form (polynomial, sine, exponential), plug‑in, solve for constants
Method of Und. Coefficients RHS is (P_n(x)e^{\lambda x}) or (P_n(x)\cos\beta x,, P_n(x)\sin\beta x) Try (y_p = x^s Q_n(x)e^{\lambda x}) etc.
Variation of Parameters General RHS; less common in JEE (y_p = u_1y_1+u_2y_2,\ u'_1y_1+u'_2y_2 = 0,\ u'_1y'_1+u'_2y'_2 = r(x))
Reduction of Order One solution (y_1) known Let (y = v(x)y_1) → first‑order for (v')

Tip: Always check if the DE is homogeneous (in the sense of order) or exact. These are the first checks before trying more complicated methods.


3. Worked Example (JEE‑style)

Problem
Solve the differential equation

[ x^2y'' - 3xy' + 4y = 0 ]

with the initial conditions (y(1)=2) and (y'(1)=5).


Step 1: Identify the equation

It is a second‑order linear homogeneous ODE with variable coefficients of the Cauchy–Euler type.


Step 2: Use the standard substitution

For a Cauchy–Euler equation (x^2y'' + axy' + by = 0), try (y = x^m).

Compute derivatives:

[ y' = mx^{m-1},\qquad y'' = m(m-1)x^{m-2} ]

Plug into the ODE:

[ x^2[m(m-1)x^{m-2}] - 3x[mx^{m-1}] + 4x^m = 0 ]

Simplify:

[ m(m-1)x^m - 3mx^m + 4x^m = 0 ] [ [m(m-1) - 3m + 4]x^m = 0 ]

Thus the indicial (characteristic) equation is

[ m^2 - m - 3m + 4 = m^2 - 4m + 4 = 0 ] [ (m-2)^2 = 0 \quad \Longrightarrow \quad m = 2 \text{ (double root)} ]


Step 3: Write the general solution

For a repeated root (m),

[ y(x) = C_1x^m + C_2x^m \ln x = C_1x^2 + C_2x^2\ln x ]


Step 4: Apply initial conditions

Your note

Page 1

Doubt hai is topic me?

Free AI teacher se poocho — 24/7, step by step.