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 \({x^n}\frac{{{d^n}y}}{{d{x^n}}} + {x^{n - 1}}\frac{{{d^{n - 1}}y}}{{d{x^{n - 1}}}} + \ldots + y = 0\) 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

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

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, \(PI=\frac{1}{f\left( D \right)}\phi \left( x \right)\)

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

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

Explanation:

\(\frac{{{\rm{dy}}}}{{{\rm{dx}}}} + {\rm{y}} = {{\rm{e}}^{\rm{x}}},{\rm{y}}\left( 0 \right) = 1\)

Integrating factor, \({\rm{IF}} = {{\rm{e}}^{\smallint {\rm{dx}}}} = {{\rm{e}}^{\rm{x}}}\)

\(\begin{array}{l} {\rm{y}}\left( {{\rm{IF}}} \right) = \smallint \left( {{\rm{IF}}} \right){{\rm{e}}^{\rm{x}}}{\rm{dx}} + {\rm{C}}\\ {\rm{y}}.{{\rm{e}}^{\rm{x}}} = \frac{{{{\rm{e}}^{2{\rm{x}}}}}}{2} + {\rm{C}} \end{array}\)

At x=0, y=1

\(\therefore {\rm{C}} = \frac{1}{2}\)

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

Explanation:

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

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

\({\left. {\frac{{dy}}{{dx}}} \right|_{x = 1}} = 0\)

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

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

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

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

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

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

\(\therefore \smallint \frac{1}{x}dx - \smallint 4dt = c\)

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:

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

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

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

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

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

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

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

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

Concept:

Write the auxillary equation by replacing derivatives and dy / dx = m, \(\frac{{{d^2}y}}{{d{x^2}}} = {m^2}, \ldots \frac{{{d^n}y}}{{d{x^n}}} = {m^n}\) 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

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

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:

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

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