Engineering Mathematics Test 7

Concept:

Integrating factor of $\frac{dy}{dx}+Py=Q\left(x\right)$ is given by,

IF = e^{∫ Pdx}

Calculation:

P = 2 / sin x

IF = e^{∫ 2 / sin x dx} = e^{∫ 2 cosec xdx}

IF = e^{2 log (tan (x / 2))}

∴ IF = tan² (x / 2)

Concept:

The differential equation of the form $x^n\frac{d^{n}y}{d{x}^n}+x^{n-1}\frac{d^{n-1}y}{d{x}^{n-1}}+\dots +y=0$ is called the Euler-Cauchy differential equation.

To solve this, we substitute

x = e^z

and replace the derivatives by dy / dz and we write

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

Write auxillary equation and equate = 0

Calculation:

x²y’’ + 5y - 3xy’ = 0

⇒ D (D - 1) - 3D + 5 = 0

D² - 4D + 5 = 0

$D=\frac{4\pm \sqrt{16-20}}{2}\Rightarrow D=2\pm i$

∴ Solution is: y = e^2z [c₁ cos z + c₂ sin z]

Z = ln (x)

∴ y = x² [c₁ cos (ln x) + c₂ sin (ln x)]

Explanation:

Given:

Differential equation: (2D³ – 7D² + 7D – 2) y = e^-8x

Particular integral, $PI=\frac{1}{f\left(D\right)}\varphi \left(x\right)$

$=\frac{1}{2D^3-7D^2+7D-2}{e}^{-8x}$

$=\frac{1}{2{(-8)}^3-7{(-8)}^2+7(-8)-2}{e}^{-8x}$

Explanation:

$\frac{\mathrm{d}\mathrm{y}}{\mathrm{d}\mathrm{x}}+\mathrm{y}={\mathrm{e}}^{\mathrm{x}},\mathrm{y}\left(0\right)=1$

Integrating factor, $\mathrm{I}\mathrm{F}={\mathrm{e}}^{\int \mathrm{d}\mathrm{x}}={\mathrm{e}}^{\mathrm{x}}$

$\begin{array}{l}\mathrm{y}\left(\mathrm{I}\mathrm{F}\right)=\int \left(\mathrm{I}\mathrm{F}\right){\mathrm{e}}^{\mathrm{x}}\mathrm{d}\mathrm{x}+\mathrm{C}\\ \mathrm{y}.{\mathrm{e}}^{\mathrm{x}}=\frac{{\mathrm{e}}^{2\mathrm{x}}}{2}+\mathrm{C}\end{array}$

At x=0, y=1

$\therefore \mathrm{C}=\frac{1}{2}$

$\therefore \mathrm{y}\left(1\right)=\frac{\mathrm{e}}{2}+\frac{{\mathrm{e}}^{-1}}{2}=\frac{1}{2}\left[\mathrm{e}+{\mathrm{e}}^{-1}\right]$

Explanation:

$\frac{d}{dx}\left(x\frac{dy}{dx}\right)=x$

$\Rightarrow x\frac{dy}{dx}=\frac{x^2}{2}+c_1$

${\frac{dy}{dx}|}_{x=1}=0$

$\Rightarrow 0=\frac{1}{2}+c_1\Rightarrow c_1=-\frac{1}{2}$

$\Rightarrow \frac{dy}{dx}=\frac{x}{2}-\frac{1}{2x}$

$\Rightarrow y=\frac{x^2}{4}-\frac{1}{2}\ln x+c_2$

y(1) = 0

$\Rightarrow 0=\frac{1}{4}-0+c_2$

$c_2=-\frac{1}{4}$

$y=\frac{x^2}{4}-\frac{1}{2}\ln x-\frac{1}{4}$

Concept:

Variable separable differential equation.

f(x) dx + f(y) dy = 0

⇒ ∫ f(x) dx + ∫ f(y) dy = c

Calculation:

Given:

dx / dt = 4x

$\frac{1}{x}dx-4dt=0$

$\frac{1}{x}dx-4dt=0$

$\therefore \int \frac{1}{x}dx-\int 4dt=c$

ln (x) - 4t = c

At t = 0, x = 6

∴ ln (6) = c

Hence, the f(x) is, ln x = 4t + ln 6

⇒ ln (x / 6) = 4t

x = 6e^4t

At, t = 5

Explanation:

$x+y=t\Rightarrow \frac{dy}{dx}=\frac{dt}{dx}-1$

$\frac{dy}{dx}=\sin \left(x+y\right)+\cos \left(x+y\right)$

$\frac{dt}{dx}-1=\sin t+\cos t\Rightarrow \frac{dt}{dx}=1+\cos t+\sin t$

$\frac{dt}{dx}=2{\cos }^2\frac{t}{2}+2\sin \frac{t}{2}\cos \frac{t}{2}$

$\frac{dt}{dx}=2{\cos }^2\left(\frac{t}{2}\right)\left(1+\tan \frac{t}{2}\right)=\frac{2\left(1+\tan \frac{t}{2}\right)}{{\sec }^2\frac{t}{2}}$

$\int dx=\int \frac{{\sec }^2\frac{t}{2}}{2\left(1+\tan \left(\frac{t}{2}\right)\right)}dt$

$x=ln\left(1+\tan \frac{t}{2}\right)+c$

$x=\ln \left(1+\tan \frac{1}{2}\left(x+y\right)\right)+c$

Concept:

Write the auxillary equation by replacing derivatives and dy / dx = m, $\frac{d^{2}y}{d{x}^2}=m^2,\dots \frac{d^{n}y}{d{x}^n}=m^n$ making RHS = 0

→ Solve auxillary equation for ‘m’.

→ for complex values of m = α ± iB.

Complementary function = e^αx [c₁ cos B x + c₂ sin B x]

Calculation:

4y’’(x) + 64y(x) = 0

⇒ 4m² + 64 = 0

⇒ m² = -16 ⇒ m = ± 4i

∴ y = c₁ cos 4x + c₂ sin 4x

Now,

y(0) = 0 ⇒ 0 = c₁

∴ y = c­₂ sin 4x

y’(x) = 4c₂ cos 4x

y’(0) = 4c₂ = 1100

⇒ c₂ = 275

∴ y(x) = 275 sin 4x

∴ y(1) = - 208.12

Explanation:

y = 3e^2x + e^-2x – αx

$-\frac{dy}{dx}=6e^{2x}-2e^{-2x}-\alpha$

Given that, $\frac{dy}{dx}\left(0\right)=1$

⇒ 1 = 6 – 2 - α

∴ α = 3

$\frac{d^{2}y}{d{x}^2}+\beta y=4\alpha x$

Complementary solution

(D² + β) = 0      ----(1)

The given is, y = 3e^2x + e^-2x – αx

It indicates, 2 and -2 are the roots of auxiliary equations.

⇒ (D + 2) (D – 2) = 0

⇒ D² – 4 = 0

By comparing this equation with equation (1)

∴ β = - 4

Concept:

Particular integral in case of polynomial expression

$PI=\frac{f\left(x\right)}{f\left(D\right)}$

Factorize f(D) and use (1 + x)^-1 = 1 - x + x² - x³ + …

(1 - x)^-1 = 1 + x + x² + x³ + …

Calculation:

$PI=f\left(x\right)=\frac{1}{D^3-1}\left(x^5+3x^4-2x^3\right)$

PI = -1 (1 - D³)^-1 (x⁵ + 3x⁴ - 2x³)

∴ PI = -1 (1 + D³ + D⁶ + D⁹ + …) (x⁵ + 3x⁴ - 2x³)

Now,

f(x) = -(x⁵ + 3x⁴ - 2x³) - (60x² + 72x - 12)

∴ f(1) = -(1 + 3 - 2) - (60 + 72 - 12)

∴ f(1) = -2 - (60 + 60)

∴ f(1) = - 122