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Ordinary & Partial Differentiation

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ByBorhan<NSTU>
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References:

Integration Formulae and tricks

Credit: Udvash Engineering Concept Book

All Formulae: https://elearn.daffodilvarsity.edu.bd/pluginfile.php/578401/mod_resource/content/4/Formula of Differentiation and integration.pdf

Definition of Differential :

Dependent/Independent Variable:

dydx{\frac{dy}{dx}}

x = independent

y= dependent (y depends on x, y is a function of x)

Types :

Order:

Degree:

→ How ? → Squaring, cubing…^n-ing the whole equation

Problems:

NB: Don’t define degree if 1ydydx,dy2dx2dydx{\frac{1}{y}} {\frac{dy}{dx}}, {\frac{\frac{dy^2}{dx^2}}{\frac{dy}{dx}}} , first make it like multiple of yy, dydx\frac{dy}{dx} for the example.

Transcendental Function:

Definition: In mathematics, when a function is not expressible in terms of a finite combination of algebraic operation of addition, subtraction, division, or multiplication raising to a power and extracting a root…

Keywords: function which are return a infinite series

Example:

1) edydx=5ye^{\frac{dy}{dx}} = 5y

Order : 1

Degree : Undefined

2) sin(dydx)d2ydx2+6y=0sin(\frac{dy}{dx} ) {\frac{d^2y}{dx^2}} + 6y = 0

Order : 2

Degree : Not defined

3) siny+d3ydx3=22siny + \frac{d^3y}{dx^3} = 22

Order : 3

Degree : 1 (No derivative function in the sinesine)

Wronskian Theorem:

Let y1(x)y{_1}(x) and y2(x)y{_2}(x) are two solution of second order linear differential equation then Wronskian of y1,y2y{_1}, y{_2}  is

W(y1,y2)=y1y2y1y2=y1y2y1y2W(y{_1},y{_2}) = \begin{vmatrix} y{_1} & y{_2} \\ y'{_1} & y'{_2} \end{vmatrix} = y{_1}y'{_2} - y'{_1}y{_2}

W(y1,y2)==0W(y{_1},y{_2}) == 0 → Linearly Dependent Solution

W(y1,y2)0W(y{_1},y{_2}) \not= 0 → Linearly Independent Solution

If there’re y1,....,yny{_1},...., y{_n} , then the equation will be like below

W(y1,...,yn)=y1y2...yny1y2...yn.....yn1yn2...ynnW(y{_1},...,y{_n}) = \begin{vmatrix} y{_1} & y{_2} & ... & y{_n} \\ y'{_1} & y'{_2} & ... & y'{_n} \\ ..... \\ y^{n'}{_1} & y^{n'}{_2} & ... & y^{n'}{_n} \\ \end{vmatrix}

Linear Differential Equation:

Variable Separable Method:

Steps:

Problems:

Reducible to Variable Separable Method:

If the equation like that,

dydx=f(ax+by+c)...(i)\frac{dy}{dx} = f(ax+by+c) .. .(i)

C may be 0.

ax+by+c=vax + by + c = v or something else to solve the problem

dydx=dvdx+....\frac{dy}{dx} = \frac{dv}{dx} + .... .. (ii)

Put dydx\frac{dy}{dx}  from (ii) value in (i) and ax+by+c=vax + by + c = v

Trick: term which one is repeated more than one = v

Problems:

2019-20 → 2(a)

Mistake : xdx=uduxdx = udu

Linear Differentiation Equation:

Definition: If P and Q are only functions of x or constants then the differential equation of the form 𝑑𝑦/𝑑𝑥 +𝑃𝑦 = 𝑄 is called the first order linear differential equation.

dydx+yP=Q\frac{dy}{dx} + yP = Q

Where P and Q are constant or function of x. Then, linear differentiable equation is applicable. Otherwise Separable Method.

Step:

A given differential equation may not be integrable as such. But it may become integrable when it is multiplied by a function. Such a function is called the integrating factor (I.F).

dx = independent variable. If y is an independent variable then, it would be Pdy.

yIf=QIfdx+C yI{_f} = \int{Q I{_f}dx} + C

Problems:

Homogenous Differentiation Equation:

Definition: An equation of the form dydx=𝑓1(𝑥,𝑦)𝑓2(𝑥,𝑦)\frac{dy}{dx} = \frac{𝑓{_1}(𝑥, 𝑦) }{𝑓{_2}(𝑥, 𝑦) } in which 𝑓1(𝑥,𝑦)𝑓{_1}(𝑥, 𝑦)  and 𝑓2(𝑥,𝑦)𝑓{_2}(𝑥, 𝑦)  are homogeneous functions of x and y of the same degree can be reduced to an equation in which variables are separable by putting 𝑦=𝑣𝑥𝑦 = 𝑣𝑥 , dydx=v+xdvdx\frac{dy}{dx} = v + x\frac{dv}{dx}

Homogenous Function & It’s Degree: F(x,y)F(x,y) is homogenous function and nn is the degree of the function then if f(mx,my)=mnf(x,y)f(mx, my) = m^n f(x,y), where mm is a non-zero value.

Step to Solve a Homogenous Differentiation Equation:

Problems:

Exact Differentiation Equation:

Definition: M(x,y)dx+N(x,y)dy=0M(x,y) dx + N(x,y) dy = 0 the equation will be Exact equation if dM(x,y)dy=dN(x,y)dx\frac{dM(x,y)}{dy} = \frac{dN(x,y)}{dx} 

[NB: y is constant when diff.w.r.t x and x is constant when diff.w.r.t y]

Steps:

Problems:

Reducible to Exact Differential Equation:

Steps:

Rule 1:

Correct Ans: y2x1=cexy^2 - x -1 = ce^{-x}

Rule 2:

Rules 1: Homogenous Equation

Initial Value Problem:

First Order Equation:

1)dydx=y y(0)=31) \frac{dy}{dx} = y \\ \space \\ y(0) = 3

dyy=dxor,dyy=dxor,lny=x+Cor,y=ex+cor,y=ex.ecor,y=Aex   y(0)=3or,Ae0=3or,A=3 y=3ex\frac{dy}{y} = dx \\ or, \int\frac{dy}{y} = \int dx \\ or, lny = x + C \\ or, y = e^{x+c} \\ or, y = e^x . e^c \\ or, y = A e^x \\ \space \\ \space \\ \space \\ y(0) = 3 \\ or, Ae^0 = 3 \\ or ,A = 3 \\ \space \\ y = 3e^x

Linear Differential Equation with Constant Coefficient:

(dnydxn+P1dn1xdyn1+P2dn2xdyn2+....Pn)y=Q(\frac{d^ny}{dx^n} + P{_1}\frac{d^{n-1}x}{dy^{n-1}} + P{_2}\frac{d^{n-2}x}{dy^{n-2}} + .... P{_n})y = Q

is a Linear Differential Equation if P1,....,PnP1, ....,Pn are the function of x or or constant.

If P1,....,PnP{_1},....,P{_n}  are constant, then it is called Linear Differential Equation with Constant Coefficient.

If right hand side is zero or Q=0Q=0, then it is called Linear Homogenous Differential Equation.

D=ddxD = \frac{d}{dx}

D2=d2dx2D^2 = \frac{d^2}{dx^2}

..........

Dn=dndxnD^n = \frac{d^n}{dx^n}

(Dn+P1Dn1+...+Pn)y=Q(D^n + P{_1}D{_{n-1}} + ...+P{_n})y = Q

1) If Q=0Q=0,

y=CFy=CF

Where, CF = Complementary Function

Complementary Function:

Rule 1: Roots are real and distinct

y=C1er1x+...+Cnernxy = C{_1}e^{r{_1}x} + ...+C{_n}e^{r{_n}x} 

Where r1,....,rnr{_1},...., r{_n} are roots of the AE.

Rule 2: Roots are real and repeated

Suppose, roots are r=2,2,2r = -2,2,2

y=C1e2x+(C2+C3x)e2xy=C{_1}e^{-2x} +(C{_2}+C{_3}x)e^{2x}

Rule 3: Roots are imaginary

r=α±βi,....r = \alpha \pm \beta i, ....

y=eαx(C1cosβx+C2sinβx)+...y = e^{\alpha x}(C{_1}cos\beta x + C{_2}sin\beta x) + ...

1) If Q0Q\not=0

Q=Q(x)Q = Q(x), a function of x

y=CF+PIy=CF + PI

Where, CF = Complementary Function

PI = Particular Integral

Particular Integral: (NOT AS SIR DID)

f(D).y=Q(x)  or,y=Q(x)f(D)f(D) . y = Q(x) \\ \space \\ \\ \\ \space or, y = \frac{Q(x)}{f(D)}

Type 1: Q(x)=eaxQ(x) = e^{ax} , f(a)0f(a) \not= 0

  • CF as Constant Coefficient
  • PI=eaxf(a)PI = \frac{e^{ax}}{f(a)}
  • y=CF+PIy = CF + PI

Type 2: Q(x)=eaxQ(x) = e^{ax} , f(a)=0f(a) = 0

  • CF as Constant Coefficient
  • PI=xeaxf(a)PI = x\frac{e^{ax}}{f'(a)}
  • y=CF+PIy = CF + PI

Type 3: Q(x)=sinax,cosaxQ(x) = sinax, cosax 

  • CF
  • PI=sinaxf(D) PI = \frac{sinax}{f(D)}  Put D2=a2D^2 = -a^2
  • DDifferentiation,1DIntegrationD - Differentiation, \frac{1}{D} - Integration
  • y=CF+PIy = CF + PI

CT:02

Bernoulli Differential Equation:

Definition: If P and Q are only functions of x or constants, then the differential equation of the form 𝑑𝑦𝑑𝑥+𝑃𝑦=𝑄𝑦𝑛\frac{𝑑𝑦}{𝑑𝑥}+𝑃𝑦 = 𝑄𝑦^𝑛; 𝑛0𝑛 ≠ 0 is called Bernoulli’s equation

The Bernoulli Differentiation looks like,

dydx+Py=Qyn\frac{dy}{dx} + Py = Qy^n

How to solve ? → Convert it to LDE. We have to vanish yny^n  .

Steps:

Problems:

UC Method:

UC Function:

A function f(x)f(x) is UC function, if it is either

UC Set:

Definition: Given a UC function f(x)f(x). We call UC set of f(x)f(x), to the set of all UC functions consisting of f(x)f(x) itself and all linearly independent functions of which the successive derivatives of f(x)f(x) are either constant multiples or linear combinations.

Problems:

1)

(comparing/equating left-hand side to right –hand side… 2A = 4 ….)

2)

Trajectories:

Trajectories: A curve which cuts every member of a given family of curves is called trajectories.

Orthogonal Trajectories: If a curve cuts every member of given family at right angles (α=90\alpha=90), then it is called Orthogonal Trajectory.

Working Rule:

Oblique Trajectory: A curve that intersects the curve of family at a constant angle α90\alpha \not= 90 is called Oblique Trajectories

Working Rule:

Laplace Transform

Definition: Let f(t) be a function of t defined for 0t<0 \le t \lt \infin

Then, Laplace transform of f(t) denoted by L{f(t)}\mathscr{L}\{ f(t) \} or F(s)F(s), is defined by

L{f(t)}=F(s)=0estf(t)dt\mathscr{L}\{ f(t) \} = F(s) = \int_{0}^{\infin} e^{st} f(t) \,dt 

Reason why s > 0 || s < 0 || (a-s) < 0 →

Problems: