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GATE Mathematics Tutorial


GATE Differential Equation Tutorial

GATE Mathematics Tutorial

PART 1: Ordinary Differential Equations (ODE)

Example 1: First Order ODE (No CF)

Question:

\[ \frac{dy}{dx}=3x^2 \]

Solution:

Integrate both sides:

\[ y=\int 3x^2 dx \] \[ y=x^3+C \]
Final Answer: \[ \boxed{y=x^3+C} \]

This is a simple ODE. CF and PI concepts are not used.

Example 2: Second Order Homogeneous ODE (CF only)

Question:

\[ \frac{d^2y}{dx^2}-5\frac{dy}{dx}+6y=0 \]

Step 1: Auxiliary equation

\[ m^2-5m+6=0 \] Factor: \[ (m-2)(m-3)=0 \] Roots: \[ m=2,3 \]

Step 2: CF

\[ y_c=C_1e^{2x}+C_2e^{3x} \]
\[ \boxed{y=C_1e^{2x}+C_2e^{3x}} \]

Given PDE:

\(a\frac{\partial^2 f}{\partial x^2} + b\frac{\partial^2 f}{\partial y^2} = f(x,y)\)

Assume:

\(f=e^{\xi x+\eta y}\)

Step 1: Calculate derivatives

\[ \frac{\partial^2 f}{\partial x^2}=\xi^2 f \]

\[ \frac{\partial^2 f}{\partial y^2}=\eta^2 f \]

Step 2: Substitute into PDE

\[ a(\xi^2 f)+b(\eta^2 f)=f \]

Step 3: Take \(f\) common

\[ (a\xi^2+b\eta^2)f=f \]

Step 4: Cancel \(f\)

\[ \boxed{a\xi^2+b\eta^2=1} \]


Answer: Option A & B

Example 3: Non-Homogeneous ODE (CF + PI)

Question:

\[ y''-3y'+2y=e^x \]

Step 1: CF

Auxiliary equation: \[ m^2-3m+2=0 \] \[ (m-1)(m-2)=0 \] Therefore: \[ CF=C_1e^x+C_2e^{2x} \]

Step 2: PI

RHS is \(e^x\). Since \(e^x\) already exists in CF, failure case. Multiply trial by x: \[ PI=Axe^x \] Substitute: \[ A e^x=-e^x \] Therefore: \[ A=-1 \] \[ PI=-xe^x \]
Final solution: \[ \boxed{ y=C_1e^x+C_2e^{2x}-xe^x } \]

PART 2: Partial Differential Equations (PDE)

Example 4: PDE without CF (Exponential Test)

Question:

\[ a f_x+b f_y=f \] Assume: \[ f=e^{\xi x+\eta y} \]

Derivative:

\[ f_x=\xi f \] \[ f_y=\eta f \] Substitute: \[ a\xi f+b\eta f=f \] Cancel f:
\[ \boxed{a\xi+b\eta=1} \]

No CF and PI are required.

Example 5: PDE with CF only

Question:

\[ (D^2-4DD'+3D'^2)z=0 \]

Step 1: Auxiliary equation

\[ m^2-4m+3=0 \] \[ (m-1)(m-3)=0 \] Roots: \[ m=1,3 \]

Step 2: CF

\[ z=\phi_1(y+x)+\phi_2(y+3x) \]
\[ \boxed{ z=\phi_1(y+x)+\phi_2(y+3x) } \]

No PI because RHS = 0.

Example 6: PDE with CF + PI

Question:

\[ (D^2+3DD'+2D'^2)z=e^{x+2y} \]

Step 1: CF

Auxiliary equation: \[ m^2+3m+2=0 \] \[ (m+1)(m+2)=0 \] Therefore: \[ CF=\phi_1(y-x)+\phi_2(y-2x) \]

Step 2: PI

For exponential: \[ D=1,\quad D'=2 \] Substitute: \[ F(1,2)=1+6+8=15 \] Therefore: \[ PI=\frac{1}{15}e^{x+2y} \]
Final solution: \[ \boxed{ z=\phi_1(y-x)+\phi_2(y-2x) +\frac1{15}e^{x+2y} } \]

Example 7: PDE PI Failure Case

Question:

\[ (D-D')z=e^{x+y} \]

Normal PI:

\[ PI=\frac{e^{x+y}}{D-D'} \] Put: \[ D=1,D'=1 \] Denominator: \[ 1-1=0 \] Failure occurs.

Special Rule:

Multiply by x: \[ PI=xe^{x+y} \]
\[ \boxed{ z=\phi(x+y)+xe^{x+y} } \]

GATE Quick Decision Table

Equation Method
ODE: \(Ly=0\) Only CF
ODE: \(Ly=f(x)\) CF + PI
PDE: \(F(D,D')z=0\) Only CF
PDE: \(F(D,D')z=\phi(x,y)\) CF + PI
Check \(e^{\xi x+\eta y}\) Replace \(D\rightarrow\xi,D'\rightarrow\eta\)

PART 3: Laplace Transform (GATE Engineering Mathematics)

Example 8: Basic Laplace Transform

Question:

\[ \frac{dy}{dt}+2y=0 \] Given: \[ y(0)=3 \]

Step 1: Take Laplace Transform

Using: \[ L\{y'\}=sY(s)-y(0) \] we get: \[ sY(s)-3+2Y(s)=0 \]

Step 2: Solve for Y(s)

\[ (s+2)Y(s)=3 \] \[ Y(s)=\frac{3}{s+2} \]

Step 3: Inverse Laplace

\[ y=3e^{-2t} \]
\[ \boxed{y=3e^{-2t}} \]

Example 9: Second Order ODE using Laplace

Question:

\[ y''+y=0 \] Given: \[ y(0)=0,\qquad y'(0)=1 \]

Step 1

Take Laplace: \[ s^2Y(s)-1+Y(s)=0 \]

Step 2

\[ (s^2+1)Y(s)=1 \] Therefore: \[ Y(s)=\frac1{s^2+1} \]

Step 3

Inverse Laplace: \[ y=\sin t \]
\[ \boxed{y=\sin t} \]

PART 4: Fourier Series

General Fourier series: \[ f(x)=\frac{a_0}{2} +\sum_{n=1}^{\infty} (a_n\cos nx+b_n\sin nx) \] where \[ a_n=\frac{2}{L} \int_{-L}^{L}f(x)\cos(nx)dx \] \[ b_n=\frac{2}{L} \int_{-L}^{L}f(x)\sin(nx)dx \]

Example 10: Fourier Series of f(x)=x

For: \[ -\piCalculate \(b_n\) \[ b_n= \frac2\pi \int_0^\pi x\sin(nx)dx \] Integration gives: \[ b_n=\frac{2(-1)^{n+1}}{n} \] Therefore:
\[ \boxed{ x= 2\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}n \sin(nx) } \]

PART 5: Important PDE Equations

PDE Application Type
\[ u_t=\alpha^2u_{xx} \] Heat conduction Parabolic
\[ u_{tt}=c^2u_{xx} \] Wave motion Hyperbolic
\[ u_{xx}+u_{yy}=0 \] Electrostatics Elliptic

Example 11: Heat Equation using Separation of Variables

Question:

\[ u_t=\alpha^2u_{xx} \] Assume: \[ u=X(x)T(t) \]

Step 1

Substitute: \[ XT'=\alpha^2X''T \] Divide: \[ \frac{T'}{\alpha^2T} = \frac{X''}{X} \] Both sides equal constant: \[ =-\lambda^2 \]

Step 2: Solve Time Part

\[ T'+\alpha^2\lambda^2T=0 \] Therefore: \[ T=e^{-\alpha^2\lambda^2t} \]

Step 3: Solve Space Part

\[ X''+\lambda^2X=0 \] Solution: \[ X=A\sin\lambda x+B\cos\lambda x \]
General solution: \[ \boxed{ u= (A\sin\lambda x+B\cos\lambda x) e^{-\alpha^2\lambda^2t} } \]

Example 12: Wave Equation

Question:

\[ u_{tt}=c^2u_{xx} \] Assume: \[ u=X(x)T(t) \] Substitute: \[ XT''=c^2X''T \] Divide: \[ \frac{T''}{c^2T} = \frac{X''}{X} \] Set: \[ =-\lambda^2 \] Then: Space: \[ X''+\lambda^2X=0 \] Time: \[ T''+c^2\lambda^2T=0 \]
\[ \boxed{ u=(A\sin\lambda x+B\cos\lambda x) (C\sin c\lambda t+D\cos c\lambda t) } \]

Consider the following wave equation,

\[ \frac{\partial^2 f(x,t)}{\partial t^2} = 10000 \frac{\partial^2 f(x,t)}{\partial x^2} \]

Which of the given options is/are solutions to the above wave equation?

  • (A) \[ f(x,t)=e^{-(x-100t)^2}+e^{-(x+100t)^2} \]
  • (B) \[ f(x,t)=e^{-(x-100t)} +\frac12e^{-(x+1000t)} \]
  • (C) \[ f(x,t)=e^{-(x-100t)} +\sin(x+100t) \]
  • (D) \[ f(x,t)=e^{j100\pi(-100x+t)} + e^{j100\pi(100x+t)} \]

Step 1: Identify the Wave Speed

The given wave equation is \[ \frac{\partial^2 f}{\partial t^2} = 10000 \frac{\partial^2 f}{\partial x^2}. \] Since \[ 10000=100^2, \] the wave speed is \[ c=100. \] Hence, the general solution of the one-dimensional wave equation is \[ f(x,t)=F(x-100t)+G(x+100t), \] where \(F\) and \(G\) are any twice differentiable functions. Therefore, every valid solution must be expressible as a function of
  • \(x-100t\)
  • \(x+100t\)

Step 2: Verify Each Option

Option (A)

\[ f(x,t) = e^{-(x-100t)^2} + e^{-(x+100t)^2} \] Observe that \[ F(u)=e^{-u^2}, \qquad G(v)=e^{-v^2}. \] Therefore, \[ f(x,t)=F(x-100t)+G(x+100t), \] which exactly matches the general solution.

✔ Option (A) is a solution.

Option (B)

\[ f(x,t) = e^{-(x-100t)} + \frac12e^{-(x+1000t)} \] The second term contains \[ x+1000t \] instead of \[ x+100t. \] Let \[ g=e^{-(x+1000t)}. \] Then \[ g_{tt} = 1000^2g = 10^6g. \] Also, \[ g_{xx}=g. \] Hence \[ 10000g_{xx} = 10^4g. \] Since \[ 10^6g \neq 10^4g, \] the wave equation is not satisfied.

✘ Option (B) is NOT a solution.

Option (C)

\[ f(x,t) = e^{-(x-100t)} + \sin(x+100t) \] The first term depends on \[ x-100t. \] The second term depends on \[ x+100t. \] Thus, \[ f(x,t) = F(x-100t) + G(x+100t), \] which matches the general solution.

✔ Option (C) is a solution.

Option (D)

\[ f(x,t) = e^{j100\pi(-100x+t)} + e^{j100\pi(100x+t)} \] Consider the first exponential. Its argument is \[ t-100x, \] which is **not** of the form \[ x-100t \quad\text{or}\quad x+100t. \] Differentiate: \[ f_{tt}=k^2e^{ku}, \] where \[ u=t-100x, \qquad k=j100\pi. \] Also, \[ f_{xx} = 10000k^2e^{ku}. \] Hence, \[ 10000f_{xx} = 10^8k^2e^{ku} \neq f_{tt}. \] Therefore the wave equation is not satisfied.

✘ Option (D) is NOT a solution.

Final Answer

✔ (A) is correct.
✔ (C) is correct.

Correct Options: \[ \boxed{A \text{ and } C} \]

GATE Differential Equation Decision Guide

Question Pattern Method
\[ F(D)y=0 \] CF only
\[ F(D)y=f(x) \] CF + PI
\[ F(D,D')z=0 \] PDE CF
\[ F(D,D')z=\phi(x,y) \] PDE CF + PI
Check: \[ e^{\xi x+\eta y} \] Replace: \[ D\rightarrow\xi,\quad D'\rightarrow\eta \]
Initial conditions given Usually Laplace transform
Boundary conditions + PDE Separation of variables/Fourier series
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