UNIT 1Differential Equations (9 Hrs)
CO1, CO6. Higher order linear ODEs with constant coefficients → particular integrals → variation of parameters → Euler/Legendre → simultaneous equations.
1.1 Operator form and the Complementary Function
| Nature of roots of \(f(m)=0\) | Complementary Function |
|---|---|
| Real & distinct: \(m_1,m_2,m_3\) | \(c_1e^{m_1x}+c_2e^{m_2x}+c_3e^{m_3x}\) |
| Two equal roots \(m,m\) | \((c_1+c_2x)e^{mx}\) |
| Three equal roots \(m,m,m\) | \((c_1+c_2x+c_3x^2)e^{mx}\) |
| Complex pair \(\alpha\pm i\beta\) | \(e^{\alpha x}\left(c_1\cos\beta x+c_2\sin\beta x\right)\) |
| Repeated complex pair \(\alpha\pm i\beta\) twice | \(e^{\alpha x}\left[(c_1+c_2x)\cos\beta x+(c_3+c_4x)\sin\beta x\right]\) |
1.2 Particular Integral — all six standard forms (the heart of the unit)
For \(g(D^2)=1\), \(g(-a^2)=f'(-a^2)\) where \(f'\) means \(d/du\) with \(u=D^2\).
- Only \(e^{ax}\) → Rule 1.
- Only \(\sin/\cos\) → Rule 2 (watch resonance).
- Only polynomial \(x^n\) → Rule 3.
- \(e^{ax}\times\)(polynomial) → Rule 4.
- x × (sin/cos) or x × \(e^{ax}\) → Rule 4 or Rule 5.
- \(e^{ax}\sin bx\) → Rule 6.
- Product of any three / \(x^2e^{ax}\sin bx\) combinations → shift first (Rule 4), then Rule 5/6.
1.3 Method of Variation of Parameters
1.4 Homogeneous equations of Euler's and Legendre's type
| Roots in \(\theta\) | C.F. |
|---|---|
| \(m_1\neq m_2\) real | \(c_1x^{m_1}+c_2x^{m_2}\) |
| \(m,m\) equal | \((c_1+c_2\log x)x^{m}\) |
| \(\alpha\pm i\beta\) | \(x^{\alpha}\big[c_1\cos(\beta\log x)+c_2\sin(\beta\log x)\big]\) |
1.5 Simultaneous linear differential equations with constant coefficients
- Write both equations in \(D\) form (e.g. \(Dx-2y=e^t\)).
- Operate on the equations with suitable operators so that one dependent variable cancels; this yields a single higher-order equation in the other variable.
- Solve it; get the second variable from one of the original equations. Keep only the correct number of arbitrary constants.
Transform both equations using the given initial values, solve the pair of algebraic equations for \(\bar x,\bar y\), then invert.
1.6 Unit 1 — Worked Examples & Question Bank
Example 1 — \( (D^2-3D+2)y=e^{3x}\)
C.F.: \(m^2-3m+2=0\Rightarrow m=1,2\Rightarrow c_1e^{x}+c_2e^{2x}\).
P.I. (Rule 1): \(\dfrac{e^{3x}}{9-9+2}=\dfrac{e^{3x}}{2}\).
Answer: \(y=c_1e^{x}+c_2e^{2x}+\tfrac12 e^{3x}\).
Example 2 — \( (D^2+4)y=\sin 2x\) (resonance)
C.F.: \(c_1\cos 2x+c_2\sin 2x\).
Here \(f(-a^2)=f(-4)=0\), \(f(D^2)=(D^2+4)g\) with \(g=1\), \(a=2\).
P.I. \(=\dfrac{-x\cos 2x}{2(2)(1)}=-\dfrac{x}{4}\cos 2x\).
Answer: \(y=c_1\cos 2x+c_2\sin 2x-\tfrac{x}{4}\cos 2x\).
Example 3 — \( (D^2+3D+2)y=x^2\)
\(\text{P.I.}=\dfrac{1}{2}\Big[1+\dfrac{3D+D^2}{2}\Big]^{-1}x^2=\dfrac{1}{2}\Big[x^2-\dfrac{3(2x)+2}{2}+\dfrac{9(2)}{4}\Big]\)
\(=\dfrac12\big[x^2-3x-1+4.5\big]=\dfrac{x^2}{2}-\dfrac{3x}{2}+\dfrac{7}{4}\).
Answer: \(y=c_1e^{-x}+c_2e^{-2x}+\tfrac{x^2}{2}-\tfrac{3x}{2}+\tfrac74\). (Substitute back to verify: \(y''+3y'+2y=x^2\). ✓)
Example 4 — \( (D^2+1)y=e^{2x}\sin x\) (Rule 6)
\(f(a+ib)=f(2+i)=(2+i)^2+1=4+4i=4(1+i)\), so \(\dfrac{1}{f(2+i)}=\dfrac{1-i}{8}\).
\(\text{P.I.}=\operatorname{Im}\Big[\dfrac{1-i}{8}e^{2x}(\cos x+i\sin x)\Big]=\dfrac{e^{2x}}{8}(\sin x-\cos x)\).
Answer: \(y=c_1\cos x+c_2\sin x+\dfrac{e^{2x}}{8}(\sin x-\cos x)\).
Example 5 — Variation of parameters: \(y''+a^2y=\sec ax\)
\(y_1=\cos ax,\ y_2=\sin ax,\ R=\sec ax\), and \(W=y_1y_2'-y_2y_1'=a\cos^2ax+a\sin^2ax=a\).
\(u_1=-\displaystyle\int\frac{\sin ax\cdot\sec ax}{a}dx=-\frac1a\int\tan ax\,dx=\frac{1}{a^2}\ln(\cos ax)\).
\(u_2=\displaystyle\int\frac{\cos ax\cdot\sec ax}{a}dx=\frac{x}{a}\).
Answer: \(y=c_1\cos ax+c_2\sin ax+\dfrac{\cos ax}{a^2}\ln(\cos ax)+\dfrac{x\sin ax}{a}\).
Example 6 — Euler: \(x^2y''-3xy'+4y=x^2\)
Put \(x=e^z,\ \theta=xD\): \((\theta^2-4\theta+4)y=x^2=e^{2z}\) since \(x^2=e^{2z}\).
C.F.: \((\theta-2)^2=0\Rightarrow y=(c_1+c_2z)e^{2z}=(c_1+c_2\log x)x^2\).
P.I.: \(f(D)=D^2-4D+4\), \(f(2)=0,\ f'(2)=0,\ f''=2\Rightarrow \text{P.I.}=\dfrac{z^2e^{2z}}{2}=\dfrac{x^2(\log x)^2}{2}\).
Answer: \(y=(c_1+c_2\log x)x^2+\tfrac12x^2(\log x)^2\).
Example 7 — Legendre: \((1+x)^2y''+(1+x)y'+y=4\cos\big(\log(1+x)\big)\)
Put \(1+x=e^z,\ \theta=(1+x)D\). Then \((1+x)^2y''=\theta(\theta-1)y\), giving \((\theta^2+1)y=4\cos z\).
C.F.: \(c_1\cos z+c_2\sin z\). P.I. resonance: \(\dfrac{1}{D^2+1}4\cos z=4\cdot\dfrac{z\sin z}{2\cdot 1}=2z\sin z\).
Answer: \(y=c_1\cos z+c_2\sin z+2z\sin z\), with \(z=\log(1+x)\).
Example 8 — Simultaneous: \(\dfrac{dx}{dt}=y,\quad \dfrac{dy}{dt}=x\)
In operator form \(Dx-y=0,\ Dy-x=0\). Eliminate \(y\): from the first, \(y=Dx\), so \(D(Dx)-x=0\Rightarrow(D^2-1)x=0\).
\(x=c_1e^{t}+c_2e^{-t}\), hence \(y=Dx=c_1e^{t}-c_2e^{-t}\).
Answer: \(x=c_1e^{t}+c_2e^{-t},\quad y=c_1e^{t}-c_2e^{-t}\).
- Solve \((D^2+4D+4)y=e^{-2x}+x^2\) and \((D^2+a^2)y=\sin ax\), explaining the resonance rule.
- Solve \(y''+y=\sec x\) (or \(x^2\)) by the method of variation of parameters.
- Solve \(x^2y''+xy'-y=x\log x\) (Euler type) and also \(x^2y''-2xy'+2y=x^3\).
- Solve \((1+x)^2y''+(1+x)y'+y=\sin\big(2\log(1+x)\big)\) (Legendre type).
- Solve the simultaneous system \(Dx-2y=e^{t},\ Dy+2x=e^{-t}\) with \(x(0)=y(0)=1\) (elimination or Laplace).
- Solve \((D^2+1)y=e^{2x}\cos x+x\sin x\), applying all relevant P.I. rules.
UNIT 2Vector Calculus (9 Hrs)
CO2, CO5. Gradient, divergence, curl → directional derivative → irrotational/solenoidal → vector integration → Green, Gauss, Stokes (no proofs) → cube/rectangular-box applications.
2.1 Gradient, divergence and curl
| Operator | Acts on | Formula | Result |
|---|---|---|---|
| Gradient \(\nabla\phi\) | scalar field \(\phi\) | \(\dfrac{\partial\phi}{\partial x}\mathbf i+\dfrac{\partial\phi}{\partial y}\mathbf j+\dfrac{\partial\phi}{\partial z}\mathbf k\) | vector |
| Divergence \(\nabla\cdot\mathbf F\) | vector \(\mathbf F=F_1\mathbf i+F_2\mathbf j+F_3\mathbf k\) | \(\dfrac{\partial F_1}{\partial x}+\dfrac{\partial F_2}{\partial y}+\dfrac{\partial F_3}{\partial z}\) | scalar |
| Curl \(\nabla\times\mathbf F\) | vector \(\mathbf F\) | \(\begin{vmatrix}\mathbf i&\mathbf j&\mathbf k\\ \partial_x&\partial_y&\partial_z\\ F_1&F_2&F_3\end{vmatrix}\) | vector |
2.2 Directional derivative
- Maximum value \(=|\nabla\phi|\), occurring along \(\nabla\phi\) itself.
- Minimum value \(=-|\nabla\phi|\); zero along any direction perpendicular to \(\nabla\phi\).
- \(\nabla\phi\) is normal to the level surface \(\phi(x,y,z)=c\).
2.3 Irrotational and Solenoidal fields
Solenoidal: \(\nabla\cdot\mathbf F=0\). Then the flux through every closed surface is zero; \(\mathbf F=\nabla\times\mathbf A\) for some vector potential \(\mathbf A\).
2.4 Vector integration
| Type | Notation | Meaning |
|---|---|---|
| Line integral | \(\displaystyle\int_C \mathbf F\cdot d\mathbf r=\int_C(F_1dx+F_2dy+F_3dz)\) | work done along a curve \(C\); if \(\mathbf F=\nabla\phi\), value \(=\phi(B)-\phi(A)\), path-independent |
| Surface integral (flux) | \(\displaystyle\iint_S\mathbf F\cdot\hat{\mathbf n}\,dS\) | flux of \(\mathbf F\) across surface \(S\) |
| Volume integral | \(\displaystyle\iiint_V \phi\,dV\) or \(\iiint_V\mathbf F\,dV\) | integrate over a solid region \(V\) |
2.5 The three big theorems (statements only — no proofs needed)
- Boundary is a closed curve in a plane (triangle, rectangle, circle) → Green / Stokes.
- Boundary is a closed surface (cube, parallelepiped, sphere) → Gauss.
- For a box \(0\le x\le a,\ 0\le y\le b,\ 0\le z\le c\): the surface integral splits into 6 faces (\(x=0,a;\ y=0,b;\ z=0,c\)), and the volume integral becomes \(\int_0^a\!\int_0^b\!\int_0^c (\ )\,dx\,dy\,dz\).
2.6 Unit 2 — Worked Examples & Question Bank
Example 1 — Gradient and directional derivative
\(\phi=xy+yz+zx\). Then \(\nabla\phi=(y+z)\mathbf i+(x+z)\mathbf j+(y+x)\mathbf k\); at \(P(1,2,0)\): \(\nabla\phi=2\mathbf i+\mathbf j+3\mathbf k\).
Directional derivative at \(P\) along \(\mathbf i+2\mathbf j+2\mathbf k\): \(\dfrac{(2)(1)+(1)(2)+(3)(2)}{3}=\dfrac{10}{3}\).
Maximum value \(=\sqrt{4+1+9}=\sqrt{14}\), in the direction \(\dfrac{2\mathbf i+\mathbf j+3\mathbf k}{\sqrt{14}}\).
Example 2 — Irrotational field and its potential
\(\mathbf F=(y^2+2xz^2)\mathbf i+(2xy+z^2)\mathbf j+(2x^2z+2yz)\mathbf k\).
\(\nabla\times\mathbf F=\mathbf 0\) (verify each component), so \(\mathbf F\) is irrotational; find \(\phi\) with \(\mathbf F=\nabla\phi\):
\(\phi=\int(y^2+2xz^2)dx=xy^2+x^2z^2+f(y,z)\); comparing partials gives \(f=yz^2+\text{const}\).
\(\phi=xy^2+x^2z^2+yz^2+c\) (take \(c=0\)).
Example 3 — Solenoidal condition
\(\mathbf F=(x+3y)\mathbf i+(y-2z)\mathbf j+(x-\lambda z)\mathbf k\). \(\nabla\cdot\mathbf F=1+1-\lambda=0\Rightarrow\lambda=2\).
Example 4 — Green's theorem verification
Verify \(\oint_C(3x^2-8y^2)dx+(4y-6xy)\,dy\) where \(C\) bounds \(x=0,\ y=0,\ x+y=1\).
Here \(L=3x^2-8y^2,\ M=4y-6xy\), so \(\dfrac{\partial M}{\partial x}-\dfrac{\partial L}{\partial y}=-6y+16y=10y\).
\(\text{RHS}=\displaystyle\int_0^1\!\!\int_0^{1-x}10y\,dy\,dx=10\int_0^1\frac{(1-x)^2}{2}dx=\frac{5}{3}\cdot\ldots=\frac53\).
Line integral along the three edges also gives \(5/3\), so Green's theorem is verified.
Example 5 — Gauss divergence over a cube
\(\mathbf F=4xz\,\mathbf i-y^2\mathbf j+yz\,\mathbf k\) over the cube \(0\le x,y,z\le 1\).
\(\nabla\cdot\mathbf F=4z-2y+y=4z-y\). \(\displaystyle\iiint(4z-y)\,dV=4\cdot\tfrac12-\tfrac12=\tfrac32\).
Both the surface integral over the six faces and the volume integral give \(3/2\). ✓
Example 6 — Stokes' theorem over a rectangle
\(\mathbf F=(x^2-y^2)\mathbf i+2xy\,\mathbf j\) over \(0\le x\le a,\ 0\le y\le b\).
\(\nabla\times\mathbf F=(0,0,4y)\cdot\) so \(\iint_S(\nabla\times\mathbf F)\cdot\mathbf k\,dS=\int_0^a\!\!\int_0^b4y\,dy\,dx=2ab^2\).
The line integral round the rectangle \(=\;2ab^2\). ✓
- Find the constants \(a,b,c\) so that \(\mathbf F=(x+2y+az)\mathbf i+(bx-3y-z)\mathbf j+(4x+cy+2z)\mathbf k\) is irrotational, and find its scalar potential.
- Show \(\mathbf F=(y^2-z^2+3yz-2x)\mathbf i+(3xz+2xy)\mathbf j+(3xy-2xz+2z)\mathbf k\) is solenoidal and irrotational; find \(\phi\).
- Verify Gauss divergence theorem for \(\mathbf F=(x^2-yz)\mathbf i+(y^2-zx)\mathbf j+(z^2-xy)\mathbf k\) over the cube \(0\le x,y,z\le a\).
- Verify Stokes' theorem for \(\mathbf F=(2x-y)\mathbf i-yz^2\mathbf j-y^2z\mathbf k\) over the upper half surface of \(x^2+y^2+z^2=1\).
- Evaluate the flux of \(\mathbf F=x\mathbf i+y\mathbf j+z\mathbf k\) across the surface of the rectangular parallelepiped \(0\le x\le a,0\le y\le b,0\le z\le c\) (and confirm by Gauss).
- Verify Green's theorem for \(\oint_C(xy+y^2)dx+x^2dy\) where \(C\) is bounded by \(y=x\) and \(y=x^2\).
UNIT 3Laplace Transformation (9 Hrs)
CO1, CO3, CO4. Definition → standard transforms → properties → derivatives/integrals → \(e^{at}f(t),\,tf(t),\,f(t)/t\) → periodic, unit step, impulse → inverse → convolution → initial & final value theorems.
3.1 Definition and standard transforms
| \(f(t)\) | \(F(s)=L\{f(t)\}\) | \(f(t)\) | \(F(s)\) |
|---|---|---|---|
| \(1\) | \(\dfrac1s\) | \(\sinh at\) | \(\dfrac{a}{s^2-a^2}\) |
| \(t\) | \(\dfrac{1}{s^2}\) | \(\cosh at\) | \(\dfrac{s}{s^2-a^2}\) |
| \(t^n,\ n=0,1,2,\dots\) | \(\dfrac{n!}{s^{n+1}}\) | \(e^{at}\sin bt\) | \(\dfrac{b}{(s-a)^2+b^2}\) |
| \(t^{1/2}\) | \(\dfrac{\sqrt\pi}{2s^{3/2}}\) | \(e^{at}\cos bt\) | \(\dfrac{s-a}{(s-a)^2+b^2}\) |
| \(t^{-1/2}\) | \(\sqrt{\dfrac{\pi}{s}}\) | \(t\sin at\) | \(\dfrac{2as}{(s^2+a^2)^2}\) |
| \(e^{at}\) | \(\dfrac{1}{s-a}\) | \(t\cos at\) | \(\dfrac{s^2-a^2}{(s^2+a^2)^2}\) |
| \(\sin at\) | \(\dfrac{a}{s^2+a^2}\) | \(\sin at-at\cos at\) | \(\dfrac{2a^3}{(s^2+a^2)^2}\) |
| \(\cos at\) | \(\dfrac{s}{s^2+a^2}\) | \(\dfrac{\sin at}{t}\) | \(\cot^{-1}\dfrac{s}{a}=\tan^{-1}\dfrac{a}{s}\) |
3.2 Properties — shifting, scaling, \(e^{at}f(t),\,tf(t),\,f(t)/t\)
| Property | Result |
|---|---|
| Linearity | \(L\{af(t)+bg(t)\}=aF(s)+bG(s)\) |
| First shifting (damping) | \(L\{e^{at}f(t)\}=F(s-a)\) |
| Second shifting (unit step) | \(L\{f(t-a)u(t-a)\}=e^{-as}F(s)\) |
| Change of scale | \(L\{f(at)\}=\dfrac1aF\!\left(\dfrac{s}{a}\right),\ a>0\) |
| Multiplication by \(t\) | \(L\{t\,f(t)\}=-\dfrac{d}{ds}F(s)\) |
| Multiplication by \(t^n\) | \(L\{t^nf(t)\}=(-1)^n\dfrac{d^n}{ds^n}F(s)\) |
| Division by \(t\) | \(L\left\{\dfrac{f(t)}{t}\right\}=\displaystyle\int_s^\infty F(u)\,du\), if \(\lim_{t\to0^+}\dfrac{f(t)}{t}\) exists |
| General \(e^{at}\) with \(t^n\) | \(L\{t^ne^{at}\}=\dfrac{n!}{(s-a)^{n+1}}\) |
3.3 Transforms of derivatives and integrals
3.4 Periodic functions, unit step function and impulse function
3.5 Inverse Laplace transform and the convolution theorem
Methods, in order of speed:
- Partial fractions — for rational \(F(s)\), split into the table forms.
- First shifting — \(L^{-1}\{F(s-a)\}=e^{at}L^{-1}\{F(s)\}\).
- Second shifting — \(L^{-1}\{e^{-as}F(s)\}=f(t-a)u(t-a)\).
- Completing the square — for \((s-a)^2+b^2\) forms, gives \(e^{at}\sin/\cos bt\).
- Division by \(s\) — \(L^{-1}\{F(s)/s\}=\int_0^tf(u)du\).
- Convolution theorem — for products that resist the above.
3.6 Initial and final value theorems
3.7 Unit 3 — Worked Examples & Question Bank
Example 1 — \(L\{t\sin at\}\) and \(L\{t\cos at\}\)
\(L\{\sin at\}=\dfrac{a}{s^2+a^2}\). Using \(L\{tf(t)\}=-F'(s)\): \(L\{t\sin at\}=-\dfrac{d}{ds}\dfrac{a}{s^2+a^2}=\dfrac{2as}{(s^2+a^2)^2}\).
Similarly \(L\{t\cos at\}=-\dfrac{d}{ds}\dfrac{s}{s^2+a^2}=\dfrac{s^2-a^2}{(s^2+a^2)^2}\).
Example 2 — \(L\left\{\dfrac{\sin at}{t}\right\}\)
\(L\left\{\dfrac{\sin at}{t}\right\}=\displaystyle\int_s^\infty\frac{a}{u^2+a^2}du=\Big[\tan^{-1}\frac{u}{a}\Big]_s^\infty=\frac{\pi}{2}-\tan^{-1}\frac{s}{a}=\cot^{-1}\frac{s}{a}\).
Example 3 — First shifting
\(L\{e^{-2t}t^2\}=\dfrac{2!}{(s+2)^3}=\dfrac{2}{(s+2)^3};\qquad L\{e^{-3t}\cos 4t\}=\dfrac{s+3}{(s+3)^2+16}\).
Example 4 — Inverse by shifting: \(L^{-1}\left\{\dfrac{s+1}{s^2+2s+5}\right\}\)
\(s^2+2s+5=(s+1)^2+4\), so \(F(s-(-1))\) with \(F(s)=\dfrac{s}{s^2+4}\).
\(=e^{-t}\cos 2t\).
Example 5 — Convolution: \(L^{-1}\left\{\dfrac{1}{s(s^2+1)}\right\}\)
Take \(f(t)=1,\ g(t)=\sin t\). Then \(\displaystyle\int_0^t 1\cdot\sin(t-u)du=\big[\cos(t-u)\big]_0^t=1-\cos t\).
Answer \(=1-\cos t\). ✓ (Also obtainable by partial fractions.)
Example 6 — Periodic function
For the half-rectified sine \(f(t)=|\sin t|\) (period \(\pi\)): \(L\{f\}=\dfrac{1}{1-e^{-\pi s}}\displaystyle\int_0^{\pi}e^{-st}\sin t\,dt=\dfrac{1}{1-e^{-\pi s}}\cdot\dfrac{1+e^{-\pi s}}{s^2+1}=\dfrac{\coth\frac{\pi s}{2}}{s^2+1}.\)
Example 7 — Unit step
If \(f(t)=t\) for \(0
Example 8 — Initial and final value
For \(F(s)=\dfrac{1}{s(s+2)}\): \(f(0^+)=\lim_{s\to\infty}\dfrac{s}{s(s+2)}=0\) and \(\lim_{t\to\infty}f(t)=\lim_{s\to0}\dfrac{s}{s(s+2)}=\dfrac12\), matching \(f(t)=\tfrac12(1-e^{-2t})\).
- Find \(L\{t^2e^{-3t}\sin 4t\}\) and \(L\left\{\dfrac{e^{-2t}-e^{-5t}}{t}\right\}\).
- Find \(L\) of the periodic square wave of period \(2a\), and of the triangular wave of period \(2a\).
- Express \(f(t)=\begin{cases}t,&0
- Find \(L^{-1}\left\{\dfrac{s}{(s^2+a^2)(s^2+b^2)}\right\}\) using the convolution theorem.
- Find \(L^{-1}\left\{\dfrac{s^2}{(s^2+4)(s^2+9)}\right\}\) by partial fractions and verify by convolution.
- Prove \(L\left\{\dfrac{\sin t}{t}\right\}=\cot^{-1}s\) and hence evaluate \(\displaystyle\int_0^\infty\frac{\sin t}{t}dt=\frac{\pi}{2}\).
UNIT 4Applications of Laplace Transformation (9 Hrs)
CO6, CO4. Solve linear ODEs with constant coefficients, integral equations (convolution type) and simultaneous linear differential equations using Laplace transforms.
4.1 The general recipe
- Take Laplace transform of both sides of the equation (using \(L\{y'\}=sY-y(0)\), \(L\{y''\}=s^2Y-sy(0)-y'(0)\)).
- Substitute the given initial/boundary conditions.
- Solve the resulting algebraic equation (or system) for \(Y(s)\).
- Take the inverse Laplace transform using partial fractions, shifting, or convolution.
4.2 Linear ODE with constant coefficients
Solved — \(y''+3y'+2y=e^{-t},\quad y(0)=0,\ y'(0)=0\)
Transforming: \((s^2Y-s\cdot0-0)+3(sY-0)+2Y=\dfrac{1}{s+1}\)
\((s^2+3s+2)Y=\dfrac{1}{s+1}\Rightarrow Y=\dfrac{1}{(s+1)^2(s+2)}\).
Partial fractions: \(\dfrac{1}{(s+1)^2(s+2)}=\dfrac{1}{s+1}-\dfrac{1}{(s+1)^2}-\dfrac{1}{s+2}\).
\(y=e^{-t}-te^{-t}-e^{-2t}=e^{-t}(1-t)-e^{-2t}\).
Solved — \(y''+4y=\sin 2t,\ y(0)=0,\ y'(0)=0\)
\((s^2+4)Y=\dfrac{2}{s^2+4}\Rightarrow Y=\dfrac{2}{(s^2+4)^2}\).
Using \(L^{-1}\left\{\dfrac{2a^3}{(s^2+a^2)^2}\right\}=\sin at-at\cos at\) with \(a=2\): \(\dfrac{2}{(s^2+4)^2}=\dfrac{1}{16}\cdot\dfrac{2\cdot 8}{(s^2+4)^2}\).
\(y=\dfrac{1}{16}\big(\sin 2t-2t\cos 2t\big)\).
4.3 Integral equations — convolution type
Solved — \(y(t)=1+\displaystyle\int_0^t y(u)\,du\)
\(Y=\dfrac1s+\dfrac{Y}{s}\Rightarrow Y\Big(1-\dfrac1s\Big)=\dfrac1s\Rightarrow Y=\dfrac{1}{s-1}\).
\(y(t)=e^{t}\). ✓
Solved — \(y(t)=t+\displaystyle\int_0^t \sin(t-u)\,y(u)\,du\)
\(Y=\dfrac{1}{s^2}+\dfrac{1}{s^2+1}Y\Rightarrow Y\dfrac{s^2}{s^2+1}=\dfrac1{s^2}\Rightarrow Y=\dfrac{s^2+1}{s^4}=\dfrac{1}{s^2}+\dfrac{1}{s^4}\).
\(y(t)=t+\dfrac{t^3}{6}\).
Solved — \(t=\displaystyle\int_0^t e^{t-u}y(u)\,du\)
\(\dfrac{1}{s^2}=\dfrac{1}{s-1}Y\Rightarrow Y=\dfrac{s-1}{s^2}=\dfrac1s-\dfrac{1}{s^2}\).
\(y(t)=1-t\).
4.4 Simultaneous linear differential equations
Solved — \(\dfrac{dx}{dt}+y=e^{t},\quad \dfrac{dy}{dt}+x=e^{t}\), with \(x(0)=y(0)=1\)
Transforming: \(sX-1+Y=\dfrac{1}{s-1}\) and \(sY-1+X=\dfrac{1}{s-1}\).
Add: \((s+1)(X+Y)-2=\dfrac{2}{s-1}\Rightarrow X+Y=\dfrac{2}{(s+1)(s-1)}+\dfrac{2}{s+1}=\dfrac{2s}{(s+1)(s-1)}\cdot\ldots\)
Subtract: \((s-1)(X-Y)=0\Rightarrow X=Y\).
Hence \(X=Y=\dfrac{1}{s+1}+\dfrac{1}{s-1}\) (after simplifying \(X\) from the sum).
\(x(t)=y(t)=e^{-t}+e^{t}=2\cosh t\).
- Add/subtract symmetric systems instead of Cramer's rule — it is much shorter.
- Always keep \(Y(s)\) in a form matching the standard table before inverting.
- For a repeated pole \((s-a)^2\), expect a \(te^{at}\) term.
- For \((s^2+a^2)^2\), expect \(\sin at-at\cos at\) type behaviour (resonance).
4.5 Unit 4 — Question Bank
- Solve \(y''-4y'+3y=e^{2t}\) with \(y(0)=0,y'(0)=1\) using Laplace transforms.
- Solve \(y''+y'+y=1\), \(y(0)=0,\ y'(0)=0\).
- Solve \(y(t)=1+t-\displaystyle\int_0^t(t-u)y(u)du\). [Hint: \(Y=\frac{1}{s}+\frac{1}{s^2}-\frac{Y}{s^2}\)]
- Solve \(y(t)=e^{-t}+\displaystyle\int_0^te^{u-t}y(u)du\). [Ans. \(y=e^{-t}+t\)]
- Solve the system \(\dfrac{dx}{dt}=2x-3y,\ \dfrac{dy}{dt}=x-2y\) with \(x(0)=8,\ y(0)=3\).
- Solve \(\dfrac{dx}{dt}+2x+3y=0,\ \dfrac{dy}{dt}+3x+2y=2e^{2t},\ x(0)=y(0)=0\).
UNIT 5Fourier Transformation (9 Hrs)
CO3, CO4. Infinite Fourier transform → sine/cosine transforms → properties → inversion → convolution → Parseval → finite sine/cosine transforms.
5.1 The infinite Fourier transform
5.2 Fourier sine and cosine transforms
| \(f(x)\) | \(F_s(s)\) | \(F_c(s)\) |
|---|---|---|
| \(e^{-ax},\ a>0\) | \(\sqrt{\dfrac{2}{\pi}}\dfrac{s}{s^2+a^2}\) | \(\sqrt{\dfrac{2}{\pi}}\dfrac{a}{s^2+a^2}\) |
| \(x\,e^{-ax}\) | \(\sqrt{\dfrac{2}{\pi}}\dfrac{2as}{(s^2+a^2)^2}\) | \(\sqrt{\dfrac{2}{\pi}}\dfrac{s^2-a^2}{(s^2+a^2)^2}\) |
| \(\dfrac{1}{x}\) | \(\sqrt{\dfrac{\pi}{2}}\) | — |
| \(f(x)=\begin{cases}1,&|x| | \(F(s)=\sqrt{\dfrac{2}{\pi}}\dfrac{\sin as}{s}\) | |
| \(f(x)=\begin{cases}1,&0 | \(\sqrt{\dfrac{2}{\pi}}\dfrac{1-\cos as}{s}\) | \(\sqrt{\dfrac{2}{\pi}}\dfrac{\sin as}{s}\) |
5.3 Properties (with the \(e^{isx}\) convention)
| Property | Result |
|---|---|
| Linearity | \(\mathcal F\{af+bg\}=aF(s)+bG(s)\) |
| Change of scale | \(\mathcal F\{f(ax)\}=\dfrac1{|a|}F\!\left(\dfrac{s}{a}\right)\) |
| Shifting in \(x\) | \(\mathcal F\{f(x-a)\}=e^{ias}F(s)\) |
| Modulation | \(\mathcal F\{e^{iax}f(x)\}=F(s+a)\) |
| Derivative | \(\mathcal F\{f'(x)\}=-isF(s)\), \(\mathcal F\{f''(x)\}=-s^2F(s)\) (for \(f,f'\to0\) as \(|x|\to\infty\)) |
| Multiplication by \(x\) | \(\mathcal F\{x\,f(x)\}=i\dfrac{d}{ds}F(s)\) |
| Sine transform of \(f'\) | \(\mathcal F_s\{f'\}=-\dfrac{s}{\sqrt{2\pi}}\,2?\ \) → use the direct form: \(\mathcal F_s\{f''\}=-s^2F_s(s)+\sqrt{\dfrac{2}{\pi}}\,s\,f(0)\) |
| Cosine transform of \(f''\) | \(\mathcal F_c\{f''\}=-s^2F_c(s)-\sqrt{\dfrac{2}{\pi}}\,f'(0)\) |
5.4 Convolution theorem and Parseval's identity
Use: evaluating definite integrals of the form \(\int_0^\infty\frac{\sin^2ax}{x^2}dx\), \(\int_0^\infty\frac{dx}{(a^2+x^2)^2}\), etc.
5.5 Finite Fourier sine and cosine transforms
These are exactly the tools used to solve the heat equation \(u_t=c^2u_{xx}\) and the wave equation on a finite rod with boundary values at \(x=0\) and \(x=l\).
5.6 Unit 5 — Worked Examples & Question Bank
Example 1 — Fourier transform of the rectangular pulse
\(f(x)=1\) for \(|x|
Deduce (by inversion at \(x=0\)): \(\displaystyle\int_0^\infty\frac{\sin as}{s}ds=\frac{\pi}{2}\).
Example 2 — Sine and cosine transforms of \(e^{-ax}\)
\(\mathcal F_s\{e^{-ax}\}=\sqrt{\dfrac{2}{\pi}}\displaystyle\int_0^\infty e^{-ax}\sin sx\,dx=\sqrt{\dfrac{2}{\pi}}\cdot\dfrac{s}{a^2+s^2}\);
\(\mathcal F_c\{e^{-ax}\}=\sqrt{\dfrac{2}{\pi}}\cdot\dfrac{a}{a^2+s^2}\).
Example 3 — Parseval's identity in action
Using Example 1 with \(a=1\): \(|F(s)|^2=\dfrac{2}{\pi}\dfrac{\sin^2s}{s^2}\) and \(\int_{-\infty}^{\infty}|f(x)|^2dx=2\). Parseval gives
\(2=\dfrac{2}{\pi}\displaystyle\int_{-\infty}^{\infty}\frac{\sin^2s}{s^2}ds\Rightarrow\int_0^\infty\frac{\sin^2s}{s^2}ds=\frac{\pi}{2}.\)
Example 4 — Finite sine transform of a constant
\(\mathcal F_s\{1\}=\displaystyle\int_0^l\sin\frac{n\pi x}{l}dx=\dfrac{l}{n\pi}\big(1-(-1)^n\big)=\begin{cases}\dfrac{2l}{n\pi},&n\text{ odd}\\[4pt]0,&n\text{ even}\end{cases}\)
Example 5 — Finite sine transform of \(x\)
\(\mathcal F_s\{x\}=\displaystyle\int_0^l x\sin\frac{n\pi x}{l}dx=\dfrac{l^2}{n\pi}(-1)^{n+1},\quad n=1,2,3,\dots\)
- Find the Fourier transform of \(f(x)=\begin{cases}a^2-x^2,&|x|
- Find \(\mathcal F_s\{e^{-ax}\}\) and \(\mathcal F_c\{e^{-ax}\}\); using Parseval, evaluate \(\displaystyle\int_0^\infty\frac{dx}{(a^2+x^2)^2}\) with \(a>0\).
- Find the Fourier sine and cosine transforms of \(f(x)=\begin{cases}\sin x,&0
- Apply the convolution theorem to find the Fourier inverse of \(F(s)G(s)\) and illustrate with \(F=G=\sqrt{2/\pi}\,\dfrac{\sin s}{s}\).
- Find the finite Fourier sine and cosine transforms of \(f(x)=x\) on \((0,\pi)\); use them to solve \(u_t=u_{xx}\) with \(u(0,t)=u(\pi,t)=0,\ u(x,0)=x\).
- Using Parseval's identity, evaluate \(\displaystyle\int_0^\infty\frac{dx}{(1+x^2)(4+x^2)}\) (convolution) and \(\displaystyle\int_0^\infty\frac{dx}{(a^2+x^2)^2}\).
KITExam Strategy & One-Page Revision Sheet
| Part | Structure | Marks | Time budget (of 180 min) |
|---|---|---|---|
| Part A | 10 questions × 2 marks, no choice (≈2 per unit) | 20 | ≈35 min |
| Part B | 2 questions from each unit, internal choice, 16 marks each → 5 to attempt | 80 | ≈25–28 min each |
- CO1 (list properties, recall ODEs): Unit 1 C.F. table + Unit 3 property table = guaranteed Part A marks.
- CO2 (directional derivative, irrotational/solenoidal): Unit 2 — practise \(\nabla\phi\), \(\nabla\times\mathbf F\), potential-finding.
- CO3 (explain & solve LT/FT): Unit 3 + Unit 5 — always write the definition and the convention first.
- CO4 (inverse LT/FT problems): partial fractions, shifting, convolution, Parseval.
- CO5 (Green/Stokes/Gauss): Unit 2 verification problems with cube/parallelepiped — write "LHS = …, RHS = …, hence verified".
- CO6 (integral & differential equations by LT): Unit 4 — one full-length problem almost always appears.
- Unit 1: six P.I. rules (especially the resonance form \(-\frac{x\cos ax}{2a g(-a^2)}\)), \(W\) in variation of parameters, \(\theta=xD\) substitution.
- Unit 2: all three theorem statements verbatim, unit normal, and the six faces of a box.
- Unit 3: standard table, \(L\{tf\}=-F'\), \(L\{f/t\}=\int_s^\infty F\), \(L\{f'\}=sF-f(0)\), periodic formula, \(L\{\delta(t-a)\}=e^{-as}\).
- Unit 4: transform → solve algebraically → invert; convolution for integral equations; add/subtract for systems.
- Unit 5: state your kernel convention, sine vs cosine choice, Parseval, and the two finite-transform derivative formulas.
- Erwin Kreyszig, Advanced Engineering Mathematics, 10th ed., John Wiley & Sons, 2012.
- B. S. Grewal, Higher Engineering Mathematics, 41st ed., Khanna Publications, 2011.
- N. P. Bali & Manish Goyal, A Textbook of Engineering Mathematics, 8th ed., Laxmi Publications, 2011.
- T. Veerarajan, Engineering Mathematics for First Year, 2nd ed., Tata McGraw Hill, 2008.
- W. E. Boyce & R. C. DiPrima, Elementary Differential Equations and Boundary Value Problems, 9th ed., Wiley India, 2009.
- M. K. Venkataraman, Engineering Mathematics — First Year, 2nd ed., National Publishing Co., 2000.
- S. L. Ross, Differential Equations, 3rd ed., Wiley India, 2009.
Prepared as a study aid for the syllabus above. All formulas are standard; work every worked example once by hand — the marks in this course come from correct operator algebra and clean partial fractions, not from memorising statements.