Engineering Mathematics Test 6

The order of differential equation is the order of the highest derivative appearing in it.

The degree of a differential equation is the degree of the highest derivative accruing in it, after the equation has been expressed in a form free from radicals as far as the derivatives are concerned.

The degree of equation is the power of the highest differentiation. Hence the degree is 1.

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:

\(\left( {{x^3} - x} \right)\frac{{dy}}{{dx}} - \left( {3{x^2} - 1} \right)y = {x^5} - 2{x^3} + x\)

\(\frac{{dy}}{{dx}} - \frac{{\left( {3{x^2} - 1} \right)}}{{\left( {{x^3} - x} \right)}}y = \frac{{\left( {{x^5} - 2{x^3} + x} \right)}}{{\left( {{x^3} - x} \right)}}\)

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

\( = {e^{\int - \frac{{\left( {3{x^2} - 1} \right)}}{{\left( {{x^3} - x} \right)}}dx}}\)

\( = {e^{ - In\;\left( {{x^3} - x} \right)}}\)

\( = \frac{1}{{\left( {{x^3} - x} \right)}}\)

The solution is,

\(y\left( {\frac{1}{{{x^3} - x}}} \right) = \smallint \frac{{{x^5} - 2{x^3} + x}}{{{{\left( {{x^3} - x} \right)}^2}}}dx + C\)

\( = \smallint \frac{{{x^5} - 2{x^3} + x}}{{\left( {{x^6} - 2{x^4} + {x^2}} \right)}}dx + C\)

\( = \smallint \frac{{{x^5} - 2{x^3} + x}}{{x\left( {{x^5} - 2{x^3} + x} \right)}}dx + C\)

\( = \smallint \frac{1}{x}dx + C\)

\( \Rightarrow y\left( {\frac{1}{{{x^3} - x}}} \right) = \log x + C\)

The given solution is,

\(\frac{y}{{A\left( x \right)}} = \log x + C\)

By comparing both the equations,

A(x) = x³ - x

Concept:

M dx + N dy = 0, will be exact differential equation if:

\(\frac{{\partial {\rm{M}}}}{{\partial {\rm{y}}}} = \frac{{\partial {\rm{N}}}}{{\partial {\rm{x}}}}\)

Where M and N are functions of x and y.

Calculation:

M = α xy³ + y cos x

N = x²y² + β sin x

\(\frac{{\partial M}}{{\partial y}} = 3\;\alpha \;x{y^2} + \cos x\)

\(\frac{{\partial N}}{{\partial x}} = 2x{y^2} + \beta \cos x\)

The differential equation to be exact,

3 α xy² + cos x = 2xy² + β cos x

\( \Rightarrow 3\alpha = 2 \Rightarrow \alpha = \frac{2}{3}\)

Concept:

General equation for DE,

\(\frac{{{d^n}y}}{{d{x^n}}} + {k_1}\frac{{{d^{n - 1}}y}}{{d{x^{n - 1}}}} + {k_2}\frac{{{d^{n - 2}}y}}{{d{x^{n - 2}}}} + \ldots {k_n}y = 0\)

Then its corresponding Auxiliary equation will be

AE: Dⁿ + k₁ D^n-1 + … k_n = 0

Then the solution of above DE will be as follows:

Calculation:

Given DE,

\(\frac{{{d^4}y}}{{d{x^4}}} - \frac{{2{d^3}y}}{{d{x^3}}} + \frac{{2{d^2}y}}{{d{x^2}}} - \frac{{2dy}}{{dx}} + y = 0\)

Its AE: (D⁴ – 2D³ + 2D² – 2D + D) = 0

It roots = 1, 1, i, -i

Then it has two equal roots and one pair of imaginary roots.

Concept:

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

Calculation:

(D² + a²)y = sin ax

Auxiliary equation: (D² + a²) = 0

⇒ D = ± ja

Complementary function is,

y = C₁ cos ax + C₂ sin ax

Particular integral is,

\(PI = \frac{1}{{\left( {{D^2} + {a^2}} \right)}}\sin ax\) 

Put D² = -a², f(D) = 0

\( = \frac{x}{{2D}}\sin ax\) 

\( = \frac{D}{{2{D^2}}}\sin ax\) 

Put D² = -a²

\( = \frac{{ - x}}{{2{a^2}}}D\left( {\sin ax} \right) = \frac{{ - x}}{{2{a^2}}}\left( {\cos ax} \right)\left( a \right)\) 

\( = \frac{{ - x}}{{2a}}\cos ax\) 

Now, the total solution is,

\(y = {C_1}\cos ax + {C_2}\sin ax + \frac{{ - x}}{{2a}}\cos ax\) 

The given solution is,

y = C₁ cos ax + C₂ sin ax + A(x)

By comparing the above two equations.

Given differential equation is,

dx – (x + y + 1) dy = 0

\(\frac{{dy}}{{dx}} = \frac{1}{{\left( {x + y + 1} \right)}}\) 

Put (x + y + 1) = t

\( \Rightarrow 1 + \frac{{dy}}{{dx}} = \frac{{dt}}{{dx}}\) 

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

Now, the differential equation becomes

\(\frac{{dt}}{{dx}} - 1 = \frac{1}{t}\) 

\(\frac{{dt}}{{dx}} = \frac{1}{t} + 1\) 

\(\frac{{dt}}{{dx}} = \frac{{t + 1}}{t}\) 

\( \Rightarrow dt\left( {\frac{t}{{t + 1}}} \right) = dx\) 

\( \Rightarrow \left( {1 - \frac{1}{{t + 1}}} \right)dt = dx\) 

\( \Rightarrow \smallint \left( {1 - \frac{1}{{t + 1}}} \right)dt = dx\) 

⇒ t = In (t + 1) = x + C₁

⇒ (x + y + 1) – In (x + y + 2) = x + C₁

⇒ y + 1 – C₁ = In (x + y + 2)

⇒ In (x + y + 2) = y + k

⇒ (x + y + 2) = e^y - e^k

(D² – 3D + 2)y = x e^3x

 A.E. is (D² + 3D + 2) = 0

⇒ (D - 1) (D - 2) = 0

⇒ D = 1, 2

C.F. = C₁e^x + C₂e^2x

\(\begin{array}{l} P.I. = \frac{1}{{\left( {{D^2} - 3D + 2} \right)}}\left( {x\;{e^{3x}}} \right)\\ = {e^{3x}}\frac{1}{{{{\left( {D + 3} \right)}^2} - 3\left( {D + 3} \right) + 2}}x\;\\ = {e^{3x}}\frac{1}{{{D^2} + 3D + 2}}\left( x \right) \end{array}\)

\(\begin{array}{l} = \frac{{{e^{3x}}}}{2}{\left[ {1 + \left( {\frac{{3D + {D^2}}}{2}} \right)} \right]^{ - 1}}x\\ = \frac{{{e^{3x}}}}{2}\left( {1 - \frac{{3D}}{2} \ldots } \right)x\\ = \frac{{{e^{3x}}}}{2}\left( {x - \frac{3}{2}} \right) \end{array}\)

Concept:

In the equation M dx + N dy = 0

If \(\frac{{\left( {\frac{{\partial M}}{{\partial y}} - \frac{{\partial N}}{{\partial x}}} \right)}}{N}\) be a function of x only = f(x), then \({e^{\smallint f\left( x \right)dx}}\) is an integrating factor.

Calculation:

Given differential equation is

(x + y³)dx + 6xy² dy = 0

M = x + y³, N = 6xy²

\(\frac{{\partial M}}{{\partial y}} = 3{y^2}\) 

\(\frac{{\partial N}}{{\partial x}} = 6{y^2}\) 

\(\frac{{\frac{{\partial M}}{{\partial y}} - \frac{{\partial N}}{{\partial x}}}}{N} = \frac{{3{y^2} - 6{y^2}}}{{6x{y^2}}} = \frac{{ - 1}}{2}\left( {\frac{1}{x}} \right) = f\left( x \right)\) 

Integrating factor \({e^{\smallint \frac{{ - 1}}{2}\left( {\frac{1}{x}} \right)dx}} = \frac{1}{{\sqrt x }} = {x^{ - 0.5}}\) 

Given that x^r is an integrating factor.

Give equation is (D² – 6D + 9)y = e^3x/x²

Auxiliary equation is, (D² – 6D + 9) = 0

⇒ (D – 3)² = 0

C.F. = (C₁ + C₂x) e^3x

\(y = a\sin \left( {x + b} \right)\)     ----(1)

There are two constants (a, b) so order of differential equations will be of order two.

\(\frac{{dy}}{{dx}} = a\cos \left( {x + b} \right)\)       ----(2)

\(\frac{{{d^2}y}}{{d{x^2}}} = - a\sin \left( {x + b} \right) = - y\)       ----(3)