Lagrange and Charpit Methods MCQ Quiz in తెలుగు - Objective Question with Answer for Lagrange and Charpit Methods - ముఫ్త్ [PDF] డౌన్‌లోడ్ కరెన్

Concept:

A PDE of the form

Pp + Qq = R satisfies Lagrange equation

$\frac{dx}{P}=\frac{dy}{Q}=\frac{dz}{R}$

Explanation:

PDE is

(x + y)z zx + (x − y)z zy = x2 + y2 

Using Lagrange equation we get

$\frac{dx}{(x+y)z}=\frac{dy}{(x-y)z}=\frac{dz}{(x^2+y^2)}$

$\frac{xdx-ydy-zdz}{(x^2+xy)z-(xy-y^2)z-z(x^2+y^2)}$ = $\frac{xdx-ydy-zdz}{0}$

So

xdx - ydy - zdz = 0

Integrating both sides

x2 - y2 - z2 = c₁ ...(i)

$\frac{ydx+xdy-zdz}{(xy+y^2)z+(x^2-xy)-z(x^2+y^2)}$ = $\frac{ydx+xdy-zdz}{0}$

So, ydx + xdy - zdz = 0

i.e., zdz - d(xy) = 0

Integrating both sides

z2 - 2xy = c₂...(ii)

Hence the general solution is 

F(x2 - y2 - z2, z2 - 2xy) = 0 for arbitrary C1 function F

(2) is correct

Explanation: 

 z = px + qy and f(x, y, z, p, q) = 0 both are compatible only when function is homogeneous in (x, y, z)

Therefore, the Correct Option is Option (3).

Explanation:

Given: $x\frac{\partial z}{\partial x}+y\frac{\partial z}{\partial y}=0$ i.e. xp + yq = 0

on comping with Pp + Qq = R, we have-

P = x, Q = y & R = 0

so, By Lagrange auxillary equation

 $\frac{dx}{p}=\frac{dy}{Q}=\frac{dz}{R}$

$\Rightarrow \frac{dx}{x}=\frac{dy}{y}=\frac{dz}{0}$

Now dz = 0

⇒ z = c₁

using first and 2^nd term

 $\frac{dx}{x}=\frac{dy}{y}$

Integrating, 

log |z| = log |y| + log c₂

$\Rightarrow \log \left(\frac{|x|}{|y|}\right)=\log c_2$

$\Rightarrow c_2=\frac{|x|}{|y|}$

Hence, general sol is -

c₁ ϕ(c₂) or c₂ = ϕ(c₁) or ϕ(c₁ c₂) = o

⇒ z = ϕ $\left(\frac{|x|}{|y|}\right)$ 

⇒ option (1) correct

Concept:

Let Pp + Qq = R be a PDE where P, Q, R are functions of x, y, z then by Lagrange's method

$\frac{dx}{P}$ = $\frac{dy}{Q}$ = $\frac{du}{R}$

Explanation:

yux + xuy = 0

Using Lagrange's auxiliary equation

$\frac{dx}{y}$ = $\frac{dy}{x}$ = $\frac{du}{0}$

Taking last two ratio

u = 0

⇒ u = c1

and taking first two ratio

$\frac{dx}{y}$ = $\frac{dy}{x}$

x dx = y dy

Integrating 

x² - y² = c2

Hence general solution is 

u = ϕ(x2 - y2)

It is passing through the curve x = t, y = 0, u = t

So, 

t =  ϕ(t2)

⇒  ϕ(t) = √t

⇒  ϕ(x2 - y2) = $\sqrt{x^2-y^2}$

Hence

u = $\sqrt{x^2-y^2}$ is the solution

(1) is true

Concept:

Let Pp + Qq = R be a PDE where P, Q, R are functions of x, y, z then by Lagrange's method

$\frac{dx}{P}$ = $\frac{dy}{Q}$ = $\frac{du}{R}$

Explanation:

Given 

exux + uy = u

Using Lagrange's method

$\frac{dx}{P}$ = $\frac{dy}{Q}$ = $\frac{du}{R}$

⇒ $\frac{dx}{e^x}$ = $\frac{dy}{1}$ = $\frac{du}{u}$

Taking first two ratio

e^-x dx = dy

Integrating

⇒ y + e^-x = c1...(i)

Taking last two ratio

$\frac{dy}{1}$ = $\frac{du}{u}$  

Integrating we get

⇒ log u - y = log c2

⇒ ue^-y = c2...(ii)

 From (i) and (ii) we get the general solution as

ue-y = ϕ(y + e-x)

or, u = eyϕ(y + e-x)

or, y + e-x = ϕ(ue-y) i.e., y = - e-x + ϕ(ue-y)

Option (4) is correct

Explanation:

Geometrical meaning of Lagrange's linear partial differential equation Pp + Qq = R is normal at point (x, y, z) of any surface f(x, y, z) = 0 is perpendicular to a line whose direction ratio's are P, Q, R.

(1) is true.