Euler's Equation MCQ Quiz - Objective Question with Answer for Euler's Equation - Download Free PDF

Concept:

Write $x^2\frac{d^{n}y}{d{x}^n}=D\left(D-1\right)\dots (D-\left(n-1\right),D=\frac{d}{dz},x=e^2$ 

Solve the auxiliary equation for values of m.

For separated roots: y = (C₁ + C₂ z)e^mz

Calculation:

m(m – 1) – 5m + 9 = 0

m² – 6m + 9 = 0 ⇒ (m – 3)² = 0

m = 3, 3

y = (C₁ + C₂ z) e^3z

y = (C₁ + C₂ In(x)) x³

Hence y(x) is unbounded on real number as ln(x) is unbounded.

So option(4) is false.

If y(1) = 1 then 1 = C1

then y = (1 + C2 In(x)) x3

Now y(x) → ∞ as x → ∞ 

So Option (1) is true and Option (2) & (3) are false.

Concept:

Explicit Forward Euler Method:

$\frac{dy}{dx}=f(x,y)$

yn+1 = yn + hf(xn, yn)

Calculation:

Given:

$\frac{\mathrm{d}\mathrm{y}}{\mathrm{d}\mathrm{t}}+2\mathrm{y}=0$ y(0) = 1, h = Δt = positive

t₀ = 1, y₀ = 1

$\frac{\mathrm{d}\mathrm{y}}{\mathrm{d}\mathrm{t}}=-2\mathrm{y}$

f(t, y) = -2y

y_n+1 = y_n + hf(t_n, y_n)

yn+1 = yn + Δt(-2y_n)

yn+1 = yn - Δt2y_n

yn+1 = yn (1 - 2Δt)

$\frac{{\mathrm{y}}_{\mathrm{n}+1}}{{\mathrm{y}}_{\mathrm{n}}}=1-2\Delta \mathrm{t}$

$\frac{|{\mathrm{Y}}_{\mathrm{n}+1}|}{|{\mathrm{y}}_{\mathrm{n}}|}\le 1$

⇒ |1 - 2Δt| ≤ 1

⇒ |1 - 2Δt| ≤ 1

-1 ≤ 1 - 2Δt ≤ 1

-2 ≤ -2Δt ≤ 0

0 ≤ 2Δt ≤ 2

0 ≤ Δt ≤ 1

Concept:

Using Euler's first order method

$y_{n+1}=y_n+hf(x_n,y_n)$

Analysis:

Step size, h = 0.1

${\mathrm{y}}_1^{\mathrm{p}}={\mathrm{y}}_0+\mathrm{h}\mathrm{f}({\mathrm{x}}_0,{\mathrm{y}}_0)$

$\frac{\mathrm{d}\mathrm{u}(\mathrm{x})}{\mathrm{d}\mathrm{x}}-\mathrm{y}(\mathrm{x})=\mathrm{x}$

y = 0 ⇒ x = 0, h = 0.1

y = 0.3

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

y₁ = y₀ + h(x₀ + y₀)

= 0 + 0.1 × 0      f(x₀ y₀) = 0 + 0

y₁ = 0

y₂ = y₁ + h(x₁ + y₁)

= 0 + 0.1 × [0.1 + 0]

= 0.01

y₃ = y₂ + 0.1 (x₂ + y₂)

= 0.01 + 0.1 (.2 + 0.01)

= 0.01 + 0.021

= 0.031

Given:

$u=\left(x-y\right)+\left(\frac{y^2}{x}\right)$

Consider:

$u\left(kx,ky\right)=kx-ky+\frac{k^2{y}^2}{kx}$

$u\left(kx,ky\right)=kx-ky+\frac{k{y}^2}{x}$

$u\left(kx,ky\right)=k\left[x-y+\frac{y^2}{x}\right]=k.u\left(x,y\right)$

u(x, y) is a homogeneous function with degree n = 1

By Euler’s theorem, we have:

$x^2\frac{{\partial }^{2}u}{\partial x^2}+2xy\frac{{\partial }^{2}u}{\partial x\partial y}+y^2\frac{{\partial }^{2}u}{\partial y^2}=n\left(n-1\right)u$

Concept:

Euler’s theorem for homogeneous function of degree (n)

$x\frac{\partial \upsilon }{\partial x}+y\frac{\partial \upsilon }{\partial y}=nu$

u(λx, λy) = λⁿ (x, y), n is degree.

Calculation:

$u\left(x,y\right)=x^4\left[{\tan }^{-1}\left(\frac{y}{x}\right)+\left(\frac{1+{\left(\frac{y}{x}\right)}^5}{1+\left(\frac{y}{x}\right)}\right)\right]$

n = 4

Explanation:

If $u={\sin }^{-1}\left(\frac{x}{y}\right)+{\tan }^{-1}\left(\frac{y}{x}\right)$

$x\frac{\partial u}{\partial x}+y\frac{\partial u}{\partial y}=?$

$\frac{\partial u}{\partial x}=\frac{1}{\sqrt{1-{\left(\frac{x}{y}\right)}^2}}\times \frac{1}{y}+\frac{1}{1+{\left(\frac{y}{x}\right)}^2}\times y\left(\frac{-1}{x^2}\right)$

$\frac{\partial u}{\partial y}=\frac{-1}{\sqrt{1-{\left(\frac{x}{y}\right)}^2}}\times \left(\frac{x}{y^2}\right)+\frac{1}{1+{\left(\frac{1}{x}\right)}^2}\times \frac{1}{x}$

$x\frac{\partial u}{\partial x}+y\frac{\partial u}{\partial y}=\frac{x}{y}\times \frac{1}{\sqrt{1-{\left(\frac{x}{y}\right)}^2}}-x\times \frac{y}{x^2}\times \frac{1}{\left[1+{\left(\frac{y}{x}\right)}^2\right]}$

$-y\times \left(\frac{x}{y^2}\right)\times \frac{1}{\sqrt{1-{\left(\frac{x}{y}\right)}^2}}+y\times \frac{1}{x}\frac{1}{\left(1+{\left(\frac{y}{x}\right)}^2\right)}$

∴ $x\frac{\partial u}{\partial x}+y\frac{\partial u}{\partial y}=0$

Concept:

Since z is a homogeneous function of x, y of degree n, we can use Euler’s Theorem defined as:

$\mathrm{x}\frac{d\mathrm{z}}{\partial \mathrm{x}}+y\frac{\partial \mathrm{z}}{\partial \mathrm{y}}=\mathrm{n}\mathrm{z}$   ---(1)

Now, z = f(u)

$\frac{\partial \mathrm{z}}{\partial \mathrm{x}}={\mathrm{f}}^{\prime }\left(\mathrm{u}\right)\frac{\partial \mathrm{u}}{\partial \mathrm{x}}$, and

$\frac{\partial \mathrm{z}}{\partial \mathrm{y}}={\mathrm{f}}^{\prime }\left(\mathrm{u}\right)\frac{\partial \mathrm{u}}{\partial \mathrm{y}}$

Substituting this in Equation (1), we get:

$\mathrm{x}{\mathrm{f}}^{\prime }\left(\mathrm{u}\right)\frac{\partial \mathrm{u}}{\partial \mathrm{x}}+\mathrm{y}{\mathrm{f}}^{\prime }\left(\mathrm{u}\right)\frac{\partial \mathrm{u}}{\partial \mathrm{y}}=\mathrm{n}\mathrm{f}\left(\mathrm{u}\right)$

$\mathrm{x}\frac{\partial \mathrm{u}}{\partial \mathrm{x}}+\mathrm{y}\frac{\partial \mathrm{u}}{\partial \mathrm{y}}=\mathrm{n}\frac{\mathrm{f}\left(\mathrm{u}\right)}{{\mathrm{f}}^{\prime }\left(\mathrm{u}\right)}$

Calculation:

Given:

$\mathrm{u}={\sin }^{-1}(\frac{\mathrm{x}+\mathrm{y}}{\sqrt{\mathrm{x}}+\sqrt{\mathrm{y}}})$

u is not a homogenous function.

$\mathrm{z}=\mathrm{s}\mathrm{i}\mathrm{n}\mathrm{u}=\frac{\mathrm{x}+\mathrm{y}}{\sqrt{\mathrm{x}}+\sqrt{\mathrm{y}}}$

$\mathrm{z}=\frac{\mathrm{x}\left(1+\frac{\mathrm{y}}{\mathrm{x}}\right)}{\sqrt{\mathrm{x}}\left(1+\frac{\sqrt{\mathrm{y}}}{\sqrt{\mathrm{x}}}\right)}$

$\mathrm{z}={\mathrm{x}}^{\frac{1}{2}}\varphi \left(\frac{\mathrm{y}}{\mathrm{x}}\right)$

$\mathrm{f}(\mathrm{u})=\mathrm{t}\mathrm{a}\mathrm{n}\mathrm{u}$

Now, z is a homogeneous function of x and y of degree 1/2.

Since $\mathrm{x}\frac{\partial \mathrm{u}}{\partial \mathrm{x}}+\mathrm{y}\frac{\partial \mathrm{u}}{\partial \mathrm{y}}=\mathrm{n}\frac{\mathrm{f}\left(\mathrm{u}\right)}{{\mathrm{f}}^{\prime }\left(\mathrm{u}\right)}$, we can write:

$\mathrm{x}\frac{\partial \mathrm{u}}{\partial \mathrm{x}}+\mathrm{y}\frac{\partial \mathrm{u}}{\partial \mathrm{y}}=\frac{1}{2}\times \frac{\mathrm{s}\mathrm{i}\mathrm{n}\mathrm{u}}{\mathrm{c}\mathrm{o}\mathrm{s}\mathrm{u}}$

$\mathrm{x}\frac{\partial \mathrm{u}}{\partial \mathrm{x}}+\mathrm{y}\frac{\partial \mathrm{u}}{\partial \mathrm{y}}=\frac{1}{2}\tan \mathrm{u}$

Explanation:

Given DE

$x^2\frac{d^{2}y}{d{x}^2}-2x\frac{dy}{dx}+2y=0$

$\mathrm{\&}y\left(1\right)=0,y\left(2\right)=2$

⇒ [x²D² – 2xD + 2] y = 0

$Let\theta =\frac{d}{dz}\mathrm{\&}x=e^z$

Let xD = θ; x²D² = θ (θ - 1);

[θ × (θ - 1) - 2θ + 2]y = 0

⇒ [θ² - 3θ + 2] y = 0

The Auxiliary Equation is θ² - 3θ + 2 = 0

⇒ (θ - 2) × (θ - 1) = 0

⇒ θ = 1, 2

→ Obtained roots of Auxiliary Equation are real and distinct

∴ The solution of eq (1) is

y = C₁e^z + C₂e^2z

⇒ y = C₁x + C₂x² [∵ x = e^z]      ………..…(2)

Given ∴ y = 0 at x = 1

⇒ 0 = C₁ + C₂ ⇒ C₁ = -C₂

& y = 2 at x = 2

⇒ 2 = 2C₁ + 4C₂ ⇒ 2 = -2C₂ + 4C₂

C₂ = 1 and C₁ = -1

The solution is

y = -x + x² ⇒ y(3) = -3 + 3²

⇒ y(3) = 6.

$\frac{du}{dt}=3t^2+1$

Forward Euler Method

y₁ = y₀ + hf(t)

at t = 0, u = 0

y₁ = 0 + 2(3 × 0² + 1)

u₁ = 2

$\begin{array}{l}\frac{du}{dt}=3t^2+1\\ du=\underset{0}{\overset{2}{\int }}\left(3t^2+1\right)dt\\ u_1={t^3+t]}_0^2\end{array}$

u₁ = 10

Explanation:

Euler’s Equation of motion in the differential form given by the following equation:

$\frac{dp}{\rho }+gdz+vdv=0$

This equation is based on the assumptions that the flow is ideal and viscous forces are zero.

The integration of the Euler’s equation of motion with respect to displacement along a streamline gives the Bernoulli equation.

Additional Information

Bernoulli's Equation:

• Bernoulli's principle states that the sum of pressure energy, kinetic energy, and potential energy per unit volume of an incompressible, non- viscous fluid in a streamlined irrotational flow remains constant along a streamline.
• This means that in steady flow the sum of all forms of mechanical energy in a fluid along a streamline is the same at all points on that streamline.

$P+\frac{1}{2}\rho V^2+\rho gh=Aconstant$

Concept:

Using Euler's first order method

$y_{n+1}=y_n+hf(x_n,y_n)$

Analysis:

Step size, h = 0.1

${\mathrm{y}}_1^{\mathrm{p}}={\mathrm{y}}_0+\mathrm{h}\mathrm{f}({\mathrm{x}}_0,{\mathrm{y}}_0)$

$\frac{\mathrm{d}\mathrm{u}(\mathrm{x})}{\mathrm{d}\mathrm{x}}-\mathrm{y}(\mathrm{x})=\mathrm{x}$

y = 0 ⇒ x = 0, h = 0.1

y = 0.3

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

y₁ = y₀ + h(x₀ + y₀)

= 0 + 0.1 × 0      f(x₀ y₀) = 0 + 0

y₁ = 0

y₂ = y₁ + h(x₁ + y₁)

= 0 + 0.1 × [0.1 + 0]

= 0.01

y₃ = y₂ + 0.1 (x₂ + y₂)

= 0.01 + 0.1 (.2 + 0.01)

= 0.01 + 0.021

= 0.031

Concept:

Euler's theorem for the homogeneous equation of degree 'n' with 'x' and 'y' as variables states that:

$x\frac{\partial u}{\partial x}+y\frac{\partial u}{\partial y}=nu$

Analysis:

We have:

$u=x^{n-1}\cdot yf\left(\frac{y}{x}\right)$

$u=x^n\cdot \left(\frac{y}{x}\right)f\left(\frac{y}{x}\right)$

u is a homogenous function of degree n.

$x\frac{\partial u}{\partial x}+y\frac{\partial u}{\partial y}=nu$

Now differentiate partially w.r.t. x again, we get:

$x\frac{{\partial }^{2}u}{\partial x}+\frac{\partial u}{\partial x}+y\frac{{\partial }^{2}u}{\partial x\partial y}=n\left(\frac{\partial u}{\partial x}\right)$

$\therefore x\frac{{\partial }^{2}u}{\partial x}+y\frac{{\partial }^{2}u}{\partial x\partial y}=\left(n-1\right)\frac{\partial u}{\partial x}$

Explanation:

$u={\tan }^{-1}\left(\frac{y^3-x^3}{x^2+y^2+xy}\right)$

$\tan u=\frac{y^3-x^3}{x^2+y^2+xy}=\frac{\left(y-x\right)\left(x^2+y^2+xy\right)}{\left(x^2+y^2+xy\right)}$

Now,

$\tan u=y\left[1-\left(\frac{x}{y}\right)\right]$ is a homogeneous function of degree ‘1’

$\therefore x\frac{\partial u}{\partial x}+y\frac{\partial u}{\partial y}=\frac{nf\left(u\right)}{f^{\prime }\left(u\right)}=1\times \frac{\tan u}{{\sec }^{2}u}$

$=\frac{\tan u}{1+{\tan }^{2}u}=\frac{\left(y-x\right)}{1+{\left(y-x\right)}^2}$

$=\frac{\left(3-1\right)}{1+{\left(3-1\right)}^2}=\frac{2}{5}=0.4$

Given that, $y^{\prime }=\frac{dy}{dx}=\sqrt{x^2+y^2}$

y(1) = 2 ⇒ y₀ = 2, x₀ = 1

h = 0.1

According to Euler’s method, $y_{n+1}=y_n+h{y}^{\prime }\left(x_n,y_n\right)$

First iteration:

y₀ = 2, x₀ = 1, h = 0.1

$y^{\prime }\left(x_0,y_0\right)=\sqrt{1^2+2^2}=\sqrt{5}$

$y_1=y_0+h{y}^{\prime }\left(x_0,y_0\right)=2+0.1\left(\sqrt{5}\right)=2.22$

Second iteration:

y₁ = 2.22, x₁ = x₀ + h = 1 + 0.1 = 1.1, h = 0.1

$y^{\prime }\left(x_1,y_1\right)=\sqrt{{1.1}^2+{2.2}^2}=\sqrt{6}$

Concept:

Explicit Forward Euler Method:

$\frac{dy}{dx}=f(x,y)$

y_n+1 = y_n + hf(x_n, y_n)

Calculation:

Given:

$\frac{dy}{dx}=y^2=f\left(x,y\right)$

y(0) = 1 which means x = 0, y = 1. Also, h = 0.1

x₀ = 0, x₁ = 0.1, x₂ = 0.2, x₃ = 0.3, x₄ = 0.5, x₅ = 0.5

Using forward Euler method:

yn+1 = yn + hf(xn, yn)

y₁ = y₀ + h f(x₀, y₀) = y₀ + hy₀ = 1 + 0.1 × (1)² = 1.1

y₂ = y₁ + h f(x₁, y₁) = 1.1 + 0.1 × (1.1)² = 1.221

y₃ = y₂ + h f(x₂, y₂) = 1.221 + 0.1 × (1.221)² = 1.37008

y₄ = y₃ + h f(x₃, y₃) = 1.37008 + 0.1 × (1.37008)² = 1.5578

y₅ = y₄ + h f(x₄, y₄) = 1.5578 + 0.1 × (1.5578)² = 1.8005

y5 = y(0.5) = 1.8005