Engineering Mathematics Test 7

\(y\frac{{dy}}{{dx}} = 6{x^2} + 5\)

⇒ y dy = (6x² + 5) dx

By integrating both the sides, we get

\( \Rightarrow \frac{{{y^2}}}{2} = 2{x^3} + 5x + C\) 

⇒ y² = 4x³ + 10x + C

y(0) = 2

⇒ C = 4

Now, y² = 4x³ + 10x + 4

At x = 1, y² = 4 + 10 + 4 = 18

\( \Rightarrow y\left( 1 \right) = \sqrt {18} = \pm 3\sqrt 2 \)

(D² + 1) y = 0

A.E. is (D² + 1) = 0

⇒ D = ± i

Solution is, y = (C₁ cos t + C₂ sin t)

Given that, y(0) = 1

⇒ 1 = (C₁ + 0) ⇒ C₁ = 1

y’ = -C₁ sin t + C₂ cos t

y’(0) = 0

⇒ 0 = 0 + C₂ ⇒ C₂ = 0

⇒ y = cos t

\(\Rightarrow y\left( {\frac{\pi }{2}} \right) = \cos \left( {\frac{\pi }{2}} \right) = 0\)

Explanation:

\(\frac{{{\partial ^2}}}{{\partial {y^2}}} + z = 0\)

If z were function of y alone, the solution would have been z = A sin y + B cos y, where A and B are constants.

Since z is a function of x and y, A and B can be arbitrary functions of x.

Hence the solution of the given equations is,

z = f(x) sin y + ϕ(x) cos y

\(\frac{{\partial z}}{{\partial y}} = f\left( x \right)\cos y - \phi \left( x \right)\sin y\)

At, y = 0, z = e^x

⇒ e^x = ϕ(x)

At, \(y = 0,\frac{{\partial z}}{{\partial y}} = {e^{ - x}}\)

⇒ e^-x = f(x)

The solution is,

One dimensional wave equation: \(\frac{{{\partial ^2}u}}{{\partial {t^2}}} = {C^2}\frac{{{\partial ^2}u}}{{\partial {x^2}}}\)

Two-dimensional wave equation: \(\frac{{{\partial ^2}u}}{{\partial {t^2}}} = \frac{{{\partial ^2}u}}{{\partial {x^2}}} + \frac{{{\partial ^2}u}}{{\partial {y^2}}}\)

Important:

Heat equation: \(\frac{{\partial u}}{{\partial t}} = {C^2}\frac{{{\partial ^2}u}}{{\partial {x^2}}}\)

Concept:

The standard form of a first-order linear differential equation is,

Form 1: \(\frac{{dy}}{{dx}} + Py = Q\)

Where P and Q are the functions of x.

Integrating factor, \(IF = {e^{\smallint Pdx}}\)

Now, the solution for the above differential equation is,

\(y\left( {IF} \right) = \smallint IF.Qdx+C\)

Form 2: \(\frac{{dx}}{{dy}} + Px = Q\)

Where P and Q are the functions of y.

Integrating factor, \(IF = {e^{\smallint Pdy}}\)

Now, the solution for the above differential equation is,

\(x\left( {IF} \right) = \smallint IF.Qdy+C\)

Calculation:

y' + y tan x = cos x

Integrating factor \( = {e^{\smallint \tan x\;dx}}\) 

\( = {e^{ - \log \left| {\cos x} \right|}}\) 

\( = \frac{1}{{\cos x}}\) 

\(y\left( {\frac{1}{{\cos x}}} \right) = \smallint \frac{1}{{\cos x}} \cdot \cos x + C\) 

\( \Rightarrow \frac{y}{{\cos x}} = x + C\) 

⇒ y = x cos x + C cos x

At x = 0, y = 0

⇒ C = 0

⇒ y = x cos x

\(\frac{{dy}}{{dt}} - y = {e^{3t}}\) 

By applying the Laplace transform,

\(sy\left( s \right) - y\left( s \right) - y\left( 0 \right) = \frac{1}{{s - 3}}\) 

\( \Rightarrow sy\left( s \right) - y\left( s \right) - 2 = \frac{1}{{s - 3}}\) 

\( \Rightarrow \left( {s - 1} \right)y\left( s \right) = 2 + \frac{1}{{\left( {s - 3} \right)}}\) 

\( \Rightarrow y\left( s \right) = \frac{2}{{s - 1}} + \frac{1}{{\left( {s - 1} \right)\left( {s - 3} \right)}}\) 

\(= \frac{2}{{\left( {s - 1} \right)}} - \frac{1}{2}\left( {\frac{1}{{s - 1}} - \frac{1}{{s - 3}}} \right)\) 

\( = \frac{3}{2}\frac{1}{{\left( {s - 1} \right)}} + \frac{1}{2}\frac{1}{{\left( {s - 3} \right)}}\) 

By applying inverse Laplace transform,

\( \Rightarrow y\left( t \right) = \frac{3}{2}{e^t} + \frac{1}{2}{e^{3t}}\) 

= Ae^t + Be^3t

\(A + B = \frac{3}{2} + \frac{1}{2} = 2\)

The given partial differential equation can be written as

(4D² + 12DD’ + 9D’²) z = 0

Its auxiliary equation is,

4m² + 12m + 9 = 0

\( \Rightarrow m = \frac{{ - 3}}{2},\frac{{ - 3}}{2}\) 

The complete solution is,

z = f₁ (y – 1.5 x) + x f₂ (y – 1.5 x)

Concept:

For different roots of the auxiliary equation, the solution (complementary function) of the differential equation is as shown below.

Calculation:

Given differential equation is 20D² + 4D + 1 = 0

\(\Rightarrow D=\frac{-4\pm \sqrt{16-80}}{40}=\frac{-4\pm 8i}{40}=-0.1\pm 0.2i\)

Solution of the given differential equation is:

\(y={{e}^{-0.1x}}\left( {{C}_{1}}\cos 0.2x+{{C}_{2}}\sin 0.2x \right)\)

\({y}'\left( x \right)={{e}^{-0.1x}}\left( -0.2{{C}_{1}}\sin 0.2x+0.2{{C}_{2}}\cos 0.2x \right)-0.1{{e}^{-0.1x}}\left( {{C}_{1}}\cos 0.2x+{{C}_{2}}\sin 0.2x \right)\)

y(0) = 3.2 ⇒ 3.2 = C₁

\({y}'\left( 0 \right)=0\Rightarrow 0=0.2{{C}_{2}}-0.1{{C}_{1}}\)

⇒ C₁ = 2 C₂

⇒ C₁ = 3.2 and C₂ = 1.6

Now the solution becomes,

\(y={{e}^{-0.1x}}\left( 3.2\cos 0.2x+1.6\sin 0.2x \right)\)

Concept:

Euler Cauchy Homogeneous linear equation:

\({x^n}\frac{{{d^n}y}}{{d{x^n}}} + {c_1}{x^{n - 1}}\frac{{{d^{n - 1}}y}}{{d{x^{n - 1}}}} + ... + {c_{n - 1}}x\frac{{dy}}{{dx}} + {c_n}y = f(x)\)

Take x = et ⇒ t = log x

\(\begin{array}{l} \frac{{dy}}{{dx}} = \frac{{dy}}{{dt}}.\frac{{dt}}{{dx}} = \frac{{dy}}{{dt}}.\frac{1}{x} \Rightarrow x\frac{{dy}}{{dx}} = \frac{{dy}}{{dt}} = Dy\\ {x^2}\frac{{{d^2}y}}{{d{x^2}}} = D(D - 1)y;\,{x^3}\frac{{{d^3}y}}{{d{x^3}}} = D(D - 1)(D - 2)y \end{array}\)

Calculation:

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

\( \Rightarrow \left( {D\left( {D - 1} \right) - 2} \right)y = {e^{2t}} + \frac{1}{{{e^t}}}\)

⇒ (D² – D - 2)y = e^2t + e^-t

\(PI = \frac{1}{{\left( {{D^2}D - 2} \right)}}\left( {{e^{2t}} + {e^{ - t}}} \right)\) 

\( = \frac{1}{{\left( {{D^2} - D - 2\;} \right)}}{e^{2t}} + \frac{1}{{\left( {{D^2} - D - 2} \right)}}{e^{ - t}}\) 

\( = \frac{t}{{2D - 1}}{e^{2t}} + \frac{t}{{2D - 1}}{e^{ - t}}\) 

\( = \frac{t}{3}{e^{2t}} + \frac{t}{{ - 3}}{e^{ - t}}\)  

\( = \frac{t}{3}\left( {{e^{2t}} - {e^{ - t}}} \right) = \frac{{ln\;x}}{3}\left( {{x^2} - \frac{1}{x}} \right)\) 

Concept:

Euler Cauchy Homogeneous linear equation:

\({x^n}\frac{{{d^n}y}}{{d{x^n}}} + {c_1}{x^{n - 1}}\frac{{{d^{n - 1}}y}}{{d{x^{n - 1}}}} + ... + {c_{n - 1}}x\frac{{dy}}{{dx}} + {c_n}y = f(x)\)

Take x = et ⇒ t = log x

\(\begin{array}{l} \frac{{dy}}{{dx}} = \frac{{dy}}{{dt}}.\frac{{dt}}{{dx}} = \frac{{dy}}{{dt}}.\frac{1}{x} \Rightarrow x\frac{{dy}}{{dx}} = \frac{{dy}}{{dt}} = Dy\\ {x^2}\frac{{{d^2}y}}{{d{x^2}}} = D(D - 1)y;\,{x^3}\frac{{{d^3}y}}{{d{x^3}}} = D(D - 1)(D - 2)y \end{array}\)

Calculation:

\({x^2}\frac{{{d^3}y}}{{d{x^3}}} - 4x\frac{{{d^2}y}}{{d{x^2}}} + 6\frac{{dy}}{{dx}} = 4\)

By multiplying with ‘x’ on both sides

\(\Rightarrow {x^3}\frac{{{d^3}y}}{{d{x^3}}} - 4{x^2}\frac{{{d^2}y}}{{d{x^2}}} + 6x\frac{{dy}}{{dx}} = 4x\)

The given equations become,

⇒ [D (D – 1) (D – 2) – 4 D (D – 1) + 6D] y = 4x

⇒ (D³ – 3D² + 2D – 4D² + 4D + 6D) y = 4x

⇒ (D³ – 7D² + 12D) y = 4x

Auxiliary equation: D³ – 7D² + 12D = 0

⇒ D (D² – 7D + 12) = 0

⇒ D = 0, 3, 4

Complementary solution (CF) = C₁ + C₂ e^3t + C₃ e^4t

CF = C1 + C2 x³ + C₃x⁴