Lagrange and Charpit Methods MCQ Quiz - Objective Question with Answer for Lagrange and Charpit Methods - Download Free PDF

Concept:

The Lagrange's auxiliary equation of the PDE of the for Pp + qq = R is

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

Explanation:

Given PDE is

z(px - qy) = y2 - x2

⇒ xzp - yzq = y2 - x2

 Integrating we get

Also, $\frac{d(x+y)}{z(x-y)}=\frac{dz}{y^2-x^2}$

⇒ $\frac{d(x+y)}{z}=\frac{dz}{-(x+y)}$

⇒ (x + y) d(x + y) + zdz = 0 

Integrating both sides we get

$(x+y{)}^2+z^2=c_2$....(iii)

Taking (i) and (iii) we get teh solution

x2 + y2 + z2 = $f((x+y{)}^2+z^2)$

(3) is true.

Taking (ii) and (ii) we get the solution

$(x+y{)}^2+z^2=f(xy)$

(4) is true.

Concept:

The non-linear PDE of the form

f(x, y, z, p, q) = 0 satisfy Charpit equation

$\frac{dx}{f_p}=\frac{dy}{f_q}=\frac{dz}{p{f}_p+q{f}_q}$$=\frac{dp}{-(f_x+p{f}_z)}=\frac{dq}{-(f_y+q{f}_z)}$

Explanation:

Here f(x, y, z, p, q) = (p2 + q2)y - qz 

Using Charpit formula

$\frac{dx}{2py}=\frac{dy}{2qy-z}$$=\frac{dz}{2p^{2}y+2q^{2}y-qz}$$=\frac{dp}{0+pq}=\frac{dq}{-p^2-q^2+q^2}$

⇒ $\frac{dx}{2py}=\frac{dy}{2qy-z}$$=\frac{dz}{2p^{2}y+2q^{2}y-qz}$$=\frac{dp}{pq}=\frac{dq}{-p^2}$

Taking

$\frac{dp}{pq}=\frac{dq}{-p^2}$

⇒ pdp + qdq = 0

Integrating

p² + q² = a²....(i)

Putting in the given equation

a²y = qz

⇒ q = $\frac{a^{2}y}{z}$

Putting in (i)

p = $\sqrt{a^2-q^2}$ = $\frac{a}{z}\sqrt{z^2-a^2{y}^2}$

Putting p and q in

dz = pdx + qdy

⇒ dz = $\frac{a}{z}\sqrt{z^2-a^2{y}^2}$dx + $\frac{a^{2}y}{z}$dy

⇒ $\frac{zdz-a^{2}ydy}{\sqrt{z^2-a^2{y}^2}}$ = a dx

Integrating

$\sqrt{z^2-a^2{y}^2}$ = ax + b

Squaring both sides

z² = a²y² + (ax+b)²

Both (1) and (2) are correct.

Hence option (3) 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:

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

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 

ux - uuy = 0, x, y ∈ ℝ,

u(0, y) = 2y, y ∈ ℝ.

Using Lagrange's method

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

⇒ $\frac{dx}{1}$ = $\frac{dy}{-u}$ = $\frac{du}{0}$

Taking last two ratio

u = 0

⇒ u = c1...(i)

and putting u = c1 we get from first two terms

$\frac{dx}{1}$ = $\frac{dy}{-c_1}$  

⇒ dy + c1dx = 0

Integrating we get

⇒ y + c1x = c2

⇒ y + ux = c2...(ii)

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

u = ϕ(y + ux)

Using u(0, y) = 2y we get

2y = ϕ(y) ⇒ ϕ(y + ux) = 2(y + ux)

Hence solution is

u = 2(y + ux) 

⇒  u(1 - 2x) = 2y ⇒ u = $\frac{2y}{1-2x}$

Therefore u(1, 2) = $\frac{4}{1-2}$ = - 4

Option (2) is correct

Explanation:

Given z² = kxy, k ∈ ℝ ← system surface

⇒ $\mathrm{k}=\frac{{\mathrm{z}}^2}{\mathrm{x}\mathrm{y}}=\mathrm{f}(\mathrm{x},\mathrm{y},\mathrm{z})$      ........(i)

Now, we will solve it using lagrange A.E - 

(Recall: Pp + Qq = R)

$\frac{\partial \mathrm{f}}{\partial \mathrm{x}}=\frac{{\mathrm{z}}^2}{\mathrm{y}}(-\frac{1}{{\mathrm{x}}^2})=\frac{-{\mathrm{z}}^2}{{\mathrm{x}}^2\mathrm{y}}$

$\frac{\partial \mathrm{f}}{\partial \mathrm{y}}=\frac{{\mathrm{z}}^2}{\mathrm{x}}(-\frac{1}{{\mathrm{y}}^2})=\frac{-{\mathrm{z}}^2}{\mathrm{x}{\mathrm{y}}^2}$

$\frac{\partial \mathrm{f}}{\partial \mathrm{z}}=\frac{2\mathrm{z}}{\mathrm{x}\mathrm{y}}$

So, $\frac{\partial \mathrm{f}}{\partial \mathrm{x}}\mathrm{p}+\frac{\partial \mathrm{f}}{\partial \mathrm{y}}\mathrm{q}=\frac{\partial \mathrm{f}}{\partial \mathrm{z}}$ (p = zx, q = zy)

⇒ $\left(\frac{-{\mathrm{z}}^2}{{\mathrm{x}}^2\mathrm{y}}\right)\mathrm{p}+\left(\frac{-{\mathrm{z}}^2}{\mathrm{x}{\mathrm{y}}^2}\right)\mathrm{q}=\frac{\partial \mathrm{f}}{\partial \mathrm{z}}=\frac{2\mathrm{z}}{\mathrm{x}\mathrm{y}}$

⇒ $-\frac{\mathrm{z}}{\mathrm{x}\mathrm{y}}[\frac{\mathrm{z}}{\mathrm{x}}\mathrm{p}+\frac{\mathrm{z}}{\mathrm{y}}\mathrm{q}]=\frac{\mathrm{z}}{\mathrm{x}\mathrm{y}}[2]$

⇒ $\frac{\mathrm{z}}{\mathrm{x}}\mathrm{p}+\frac{\mathrm{z}}{\mathrm{y}}\mathrm{q}=-2$

⇒ (zy)p + (xz)q = -2xy (on taking LCM)     (ii)

(zy)p → P

(xz)q → Q

-2xy → R

So, by lagrange, A.E -

$\frac{\mathrm{d}\mathrm{x}}{\mathrm{P}}=\frac{\mathrm{d}\mathrm{y}}{\mathrm{Q}}=\frac{\mathrm{d}\mathrm{z}}{\mathrm{R}}$   

$\Rightarrow \frac{\mathrm{d}\mathrm{z}}{\mathrm{z}\mathrm{y}}=\frac{\mathrm{d}\mathrm{y}}{\mathrm{z}\mathrm{x}}=\frac{\mathrm{d}\mathrm{z}}{(-2\mathrm{x}\mathrm{y})}$

Now,

$\frac{\mathrm{d}\mathrm{x}}{\mathrm{z}\mathrm{y}}=\frac{\mathrm{d}\mathrm{y}}{\mathrm{z}\mathrm{x}}$

⇒ x dx = y dy

on integrating

$\frac{{\mathrm{x}}^2}{2}=\frac{{\mathrm{y}}^2}{2}+\mathrm{c}$

⇒ x² - y² = c₁

$\frac{\mathrm{d}\mathrm{x}}{\mathrm{z}\mathrm{y}}=\frac{-\mathrm{d}\mathrm{z}}{2\mathrm{x}\mathrm{y}}$

⇒ 2x dx = -z dz

on integrating

${\mathrm{x}}^2=\frac{-\mathrm{z}62}{2}+\mathrm{c}$

⇒ 2x² + z² = c₂

So, general solⁿ will be ϕ (c₁, c₂) = 0

⇒ ϕ (x² - y², 2x² + z²) = 0 ⇒ option (3) correct.

other possible form of solution and when c₁ & c₂ calculated.

ϕ(c1, c2), c₁ = ϕ(c2), c₂ = ϕ(c2), $\frac{{\mathrm{c}}_1}{2}=\varphi ({\mathrm{c}}_2)$...

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 

uux + uy = 0, x ∈ ℝ, y > 0,

u(x, 0) = x, x ∈ ℝ.

Using Lagrange's method

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

⇒ $\frac{dx}{u}$ = $\frac{dy}{1}$ = $\frac{du}{0}$

Solving this we get 

u = c₁...(i)

and putting u = c1 we get from first two terms

$\frac{dx}{c_1}$ = $\frac{dy}{1}$  

dx = c₁dy
⇒ x - c₁y = c₂

⇒ x - uy = c₂...(ii)

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

u = ϕ(x - uy)

Using u(x, 0) = x we get

x = ϕ(x)  so ϕ(x - uy) = x - uy

Hence solution is

u = x - uy ⇒  u(1 + y) = x ⇒ u = $\frac{x}{1+y}$

Therefore u(2, 3) = $\frac{2}{4}=\frac{1}{2}$

Option (3) is correct

Given -

Consider the Cauchy problem

$\mathrm{x}\frac{\partial \mathrm{u}}{\partial \mathrm{x}}+\mathrm{y}\frac{\partial \mathrm{u}}{\partial \mathrm{y}}=\mathrm{u};$

𝑢 = 𝑓(𝑡) on the initial curve Γ = (𝑡, 𝑡); 𝑡 > 0.

Concept -

If the PDE is the form of pP + qQ = R then 

(i) $\frac{p(t)}{x^{\prime }}=\frac{q(t)}{y^{\prime }}=\frac{R}{u^{\prime }(t)}$ ⇒ Infinite solutions 

(ii) $\frac{p(t)}{x^{\prime }}\ne \frac{q(t)}{y^{\prime }}\ne \frac{R}{u^{\prime }(t)}$ ⇒ Unique solution

(iii) $\frac{p(t)}{x^{\prime }}=\frac{q(t)}{y^{\prime }}\ne \frac{R}{u^{\prime }(t)}$ ⇒ No solution

where p = x, q = y and R = u  

Explanation -

we have the PDE is $\mathrm{x}\frac{\partial \mathrm{u}}{\partial \mathrm{x}}+\mathrm{y}\frac{\partial \mathrm{u}}{\partial \mathrm{y}}=\mathrm{u};$ and 𝑢 = 𝑓(𝑡) on the initial curve Γ = (𝑡, 𝑡); 𝑡 > 0.

So x = t, y = t and u = f(t) 

Now for Statement (P) -

$\frac{t}{1}=\frac{t}{1}\ne \frac{2t+1}{2}$

Hence there is no solution. So Statement P is false.

Now for Statement (Q) -

$\frac{t}{1}=\frac{t}{1}\ne \frac{2t-1}{2}$

Hence there is no solution. So Statement Q is false.

Hence option (4) is true.

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

Explanation:

Given z² = kxy, k ∈ ℝ ← system surface

⇒ $\mathrm{k}=\frac{{\mathrm{z}}^2}{\mathrm{x}\mathrm{y}}=\mathrm{f}(\mathrm{x},\mathrm{y},\mathrm{z})$      ........(i)

Now, we will solve it using lagrange A.E - 

(Recall: Pp + Qq = R)

$\frac{\partial \mathrm{f}}{\partial \mathrm{x}}=\frac{{\mathrm{z}}^2}{\mathrm{y}}(-\frac{1}{{\mathrm{x}}^2})=\frac{-{\mathrm{z}}^2}{{\mathrm{x}}^2\mathrm{y}}$

$\frac{\partial \mathrm{f}}{\partial \mathrm{y}}=\frac{{\mathrm{z}}^2}{\mathrm{x}}(-\frac{1}{{\mathrm{y}}^2})=\frac{-{\mathrm{z}}^2}{\mathrm{x}{\mathrm{y}}^2}$

$\frac{\partial \mathrm{f}}{\partial \mathrm{z}}=\frac{2\mathrm{z}}{\mathrm{x}\mathrm{y}}$

So, $\frac{\partial \mathrm{f}}{\partial \mathrm{x}}\mathrm{p}+\frac{\partial \mathrm{f}}{\partial \mathrm{y}}\mathrm{q}=\frac{\partial \mathrm{f}}{\partial \mathrm{z}}$ (p = zx, q = zy)

⇒ $\left(\frac{-{\mathrm{z}}^2}{{\mathrm{x}}^2\mathrm{y}}\right)\mathrm{p}+\left(\frac{-{\mathrm{z}}^2}{\mathrm{x}{\mathrm{y}}^2}\right)\mathrm{q}=\frac{\partial \mathrm{f}}{\partial \mathrm{z}}=\frac{2\mathrm{z}}{\mathrm{x}\mathrm{y}}$

⇒ $-\frac{\mathrm{z}}{\mathrm{x}\mathrm{y}}[\frac{\mathrm{z}}{\mathrm{x}}\mathrm{p}+\frac{\mathrm{z}}{\mathrm{y}}\mathrm{q}]=\frac{\mathrm{z}}{\mathrm{x}\mathrm{y}}[2]$

⇒ $\frac{\mathrm{z}}{\mathrm{x}}\mathrm{p}+\frac{\mathrm{z}}{\mathrm{y}}\mathrm{q}=-2$

⇒ (zy)p + (xz)q = -2xy (on taking LCM)     (ii)

(zy)p → P

(xz)q → Q

-2xy → R

So, by lagrange, A.E -

$\frac{\mathrm{d}\mathrm{x}}{\mathrm{P}}=\frac{\mathrm{d}\mathrm{y}}{\mathrm{Q}}=\frac{\mathrm{d}\mathrm{z}}{\mathrm{R}}$   

$\Rightarrow \frac{\mathrm{d}\mathrm{z}}{\mathrm{z}\mathrm{y}}=\frac{\mathrm{d}\mathrm{y}}{\mathrm{z}\mathrm{x}}=\frac{\mathrm{d}\mathrm{z}}{(-2\mathrm{x}\mathrm{y})}$

Now,

$\frac{\mathrm{d}\mathrm{x}}{\mathrm{z}\mathrm{y}}=\frac{\mathrm{d}\mathrm{y}}{\mathrm{z}\mathrm{x}}$

⇒ x dx = y dy

on integrating

$\frac{{\mathrm{x}}^2}{2}=\frac{{\mathrm{y}}^2}{2}+\mathrm{c}$

⇒ x² - y² = c₁

$\frac{\mathrm{d}\mathrm{x}}{\mathrm{z}\mathrm{y}}=\frac{-\mathrm{d}\mathrm{z}}{2\mathrm{x}\mathrm{y}}$

⇒ 2x dx = -z dz

on integrating

${\mathrm{x}}^2=\frac{-\mathrm{z}62}{2}+\mathrm{c}$

⇒ 2x² + z² = c₂

So, general solⁿ will be ϕ (c₁, c₂) = 0

⇒ ϕ (x² - y², 2x² + z²) = 0 ⇒ option (3) correct.

other possible form of solution and when c₁ & c₂ calculated.

ϕ(c1, c2), c₁ = ϕ(c2), c₂ = ϕ(c2), $\frac{{\mathrm{c}}_1}{2}=\varphi ({\mathrm{c}}_2)$...

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 

uux + uy = 0, x ∈ ℝ, y > 0,

u(x, 0) = x, x ∈ ℝ.

Using Lagrange's method

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

⇒ $\frac{dx}{u}$ = $\frac{dy}{1}$ = $\frac{du}{0}$

Solving this we get 

u = c₁...(i)

and putting u = c1 we get from first two terms

$\frac{dx}{c_1}$ = $\frac{dy}{1}$  

dx = c₁dy
⇒ x - c₁y = c₂

⇒ x - uy = c₂...(ii)

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

u = ϕ(x - uy)

Using u(x, 0) = x we get

x = ϕ(x)  so ϕ(x - uy) = x - uy

Hence solution is

u = x - uy ⇒  u(1 + y) = x ⇒ u = $\frac{x}{1+y}$

Therefore u(2, 3) = $\frac{2}{4}=\frac{1}{2}$

Option (3) is correct

Explanation:

ux + x2uy = 0 with u(x, 0) = ex

Using Lagrange's auxiliary equation

$\frac{dx}{1}$ = $\frac{dy}{x^2}$ = $\frac{du}{0}$

Solving u = c₁

and taking first two

dy = x²dx

Integrating 

y = x³/3 - c₂

c2 = (x3 -3y)

hence solution is 

u = ϕ(x3 -3y)

using u(x, 0) = ex

ex =  ϕ(x3) so  ϕ(t) = $e^{t^{1/3}}$ 

hence u = $e^{(x^3-3y{)}^{1/3}}$

(3) is true

Concept:

A particular Quasi-linear partial differential equation of order one is of the form Pp + Qq = R, where P, Q and R are functions of x, y, z. Such a partial differential equation is known as Lagrange equation and it's Lagrange’s auxiliary equations is

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

Explanation:

Given (x - y)$\frac{\partial \mathrm{u}}{\partial \mathrm{x}}+(\mathrm{y}-\mathrm{x}-\mathrm{u})\frac{\partial \mathrm{u}}{\partial \mathrm{y}}=\mathrm{u}$

Then  Lagrange’s auxiliary equations is

$\frac{dx}{x-y}=\frac{dy}{y-x-u}=\frac{du}{u}$ 

Using

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

⇒ dx + dy + du = 0

⇒ x + y + u = c1 (integrating)

Also

$\frac{dx-dy+du}{x-y-y+x+u+u}=\frac{du}{u}$

⇒ $\frac{d(x-y+u)}{2(x-y+u)}=\frac{du}{u}$

⇒ ln|x - y + u|  - 2 ln|u| = lnc2 (Integrating)

⇒ $\frac{x-y+u}{u^2}$ = c2 

Hence general solution is

$\frac{x-y+u}{u^2}$ = ϕ(x + y + u)....(i)

Using u(x, 0) = 1 in (i)

x + 1 = ϕ (x + 1)

⇒ ϕ(x) = x

So ϕ(x + y + u) = x + y + u

The solution is

$\frac{x-y+u}{u^2}$ = x + y + u 

x - y + u = u²(x + y + u )

u2 (x + y + u) - (x – y + u) = 0

u2 (x + y + u) + (y – x – u) = 0

(2) is 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:

Given 

ux + uy = eu, u(x, 0) = 1 

Using Lagrange's method

$\frac{dx}{P}$ = $\frac{dy}{1}$ = $\frac{du}{R}$

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

Solving 1st two terms

$\frac{dx}{1}$ = $\frac{dy}{1}$

dy = dx

y - x = c1...(i)

Using 1st and 3rd term

$\frac{dx}{1}$ = $\frac{du}{e^u}$

dx = e-udu  

Integrating

x + e-u =  c2...(ii)

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

x + e-u = ϕ(y - x)

Using u(x, 0) = 1 we get

x + e-1= ϕ(- x)  so ϕ(x) = - x + e-1 so ϕ(y - x) = -y + x + e-1

Hence solution is

x + e-u = -y + x + e-1 ⇒ e-u = -y + e-1 ⇒ u = - ln(-y + e-1)

Hence ux = 0, uy = $\frac{1}{-y+e^{-1}}$

Therefore, $\mathrm{u}(\frac{1}{2\mathrm{e}},\frac{1}{2\mathrm{e}})$ = - ln($-\frac{1}{2e}+\frac{1}{e}$) = -ln($\frac{1}{2e}$) = ln(2e)

Option (1) is false

${\mathrm{u}}_{\mathrm{x}}(\frac{1}{2\mathrm{e}},\frac{1}{2\mathrm{e}})$ = 0

Option (2) is correct
${\mathrm{u}}_{\mathrm{y}}(\frac{1}{4\mathrm{e}},\frac{1}{4\mathrm{e}})$ = $\frac{1}{-\frac{1}{4e}+\frac{1}{e}}$ = $\frac{4e}{3}$

Option (3) is false

 ${\mathrm{u}}_{\mathrm{y}}(0,\frac{1}{4\mathrm{e}})$ = $\frac{1}{-\frac{1}{4e}+\frac{1}{e}}$ = $\frac{4e}{3}$

Option (4) is correct