math unit 3 forumlas

Question 1

If arbitary function is z, we need go diff it using u and v and normal arbituary functuon

Answer 1

P=dz/dx, q=dz/dy
VERY IMPPPP
Ux/Uy=Vx/Vy

Z, p,q should be order 1, should not contain zdz/dx=zdz/dy,dz/dx dz/dy
d represents partial symbols

Arbituary function
Manipulate it in a way such that relation is in btw z and p+q
Ex:
ex z=(x-a)^2+(y-b)^2
dz/dx=p=2(x-a) => x-a = p/2
dz/dy=q=2(y-b) => y-b = q/2

equate the equation in terms of p and q
(p/2)^2+(q/2)^2

used to remove constant

Else should be solved in terms of a relation of p and q

Question 2

auxiliary eqn and legrandes LDE:

Answer 2

Constants can be removed while solving ex
-x^2/2=y^2/2=C1/2
Thus =x^2+y^2=C1

dx/p=dy/q+dz/r
after this step put it in the form:
Pp+Qq+R

ex y^p-xyq=x(3-2y)
dx/y^2-dy/xy=dz/x(3-2y)
dx(xy)=dy(y^2)
-x^2/2=y^2/2=C1/2

NOTE: when integration ex its xdy or sin(x-2y) etc, try to relate that with C1 if u can and then integrate

Question 3

seperation method

Answer 3

Ex: question x^2∂x/∂y+y^2∂u/∂y=0

u=X(x)*Y(y)

∂u/∂x=Y(y)X’(x)
∂u/∂y=Y’(y)X(x)

x^2(X’Y)+y^2(Y’X)=0
and divide by XY
x^2X’/X+y^2Y’/Y=0
equate to constant k
x^2X’/X=K
-y^2Y’/Y = K

X’/X=kx^-2
Y’/Y=ky^-2
∫X’/Xdx = ∫kx^-2dx
logx=K(-1/x)+(C1)log
X/C1=e^-k/x
X=C1e^-k/x

if the eqn is X’/X+Y’/Y
its should be X’/X = K; -Y’Y=K
u(x,t)=X(x).T(t)

∫Y’/Ydx = ∫ky^-2dx
doing the same we get
Y=C2e^k/y

thus u=XY
C1e^-k/x*C2e^k/y
u=Ce^k(1/y-1/x)

even if its like
X’Y=2XY’+XY u still DIVIDE BY XY

Question 4

Problems on homogeneous and non homogenous Linear PDE with constant
coefficients

Answer 4

roots mo,m1,m2 distinct

ϕ(y+m1X)+ϕ2(y+m2x) +….

m=m2, m3=m4=m5
m6

ϕ(y+m1x)+xϕ2(y+m2x)
+ϕ3(y+m3x)+xϕ4(y+m4x)+x^2ϕ5(y+m5x)+ϕ6(y+m6x)

Dx=>m, Dy=1

Question 5

Type 1,2,3

Answer 5

e^ax+by
Dx->a,Dy->b
when e^ax+by/0, differentiate in terms of x

TYPE 2
sin(a+b) => D^2x=-a^2, D^2x=-b^2 DxDy=> -a*b

TYPE 3
division using x+y
its should be in the order for ex Dx^2+3DxDy+2Dy^2
ex: Dx^3-2Dx^2Dy
NOTE: Dx always come before Dy and DxDy btw Dx^2, Dy^2

order inside :x,y,c

TYPE 4
X*e^a+b Dx-> Dx+1,
Dy-> Dy+0

if e^a+b*sin(a+b)
replacement is
D^2x=-b^2 DxDy=> -6

Question 6

Linear PDE with constant coeffcients when the order is not same

Answer 6

F(Dx, Dy) is reducible if it can be written as the product of linear factors in the form (aDx+bDy+c)

f(Dx,Dy) is irreducable if we cant factorise as aDx+bDy+C

(Dx^2-Dy^2)=> Dx+Dy(Dx-Dy)

which is reduced

Ex:(Dx^3-Dy^2) => which is not reducable

z=e^-c/by[ϕ,(ay-bx)]+xϕ2,(ay-bx)+…. b≠0

z=e^-c/ax[ϕ,(ay-bx)]+xϕ2,(ay-bx)+….a≠0

Question 7

Legrand’s Multipliers Types

Answer 7

x,y,z
x^2,y^2,z^2
1/x,1/y,/1/z
1/x^2,1/y^2,1/z^2

d(xy) =xdy+ydx VERY IMPP
used very commonly