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.

\({\left( {\frac{{{d^4}y}}{{d{x^4}}}} \right)^{\frac{1}{2}}} = {\left[ {1 + {{\left( {\frac{{{d^2}y}}{{d{x^2}}}} \right)}^2}} \right]^{\frac{1}{3}}}\)

Highest derivative in the given differential equation is 4

Hence order is 4.

To find the degrees, we need to change the differential equation in to form which is free from radicals.

\({\left[ {{{\left( {\frac{{{d^4}y}}{{d{x^4}}}} \right)}^{\frac{1}{2}}}} \right]^6} = {\left[ {{{\left[ {1 + {{\left( {\frac{{{d^2}y}}{{d{x^2}}}} \right)}^2}} \right]}^{\frac{1}{3}}}} \right]^6}\)

\(\Rightarrow {\left( {\frac{{{d^4}y}}{{d{x^4}}}} \right)^3} = {\left[ {1 + {{\left( {\frac{{{d^2}y}}{{d{x^2}}}} \right)}^2}} \right]^2}\)

Now the differential equation is free from radicals

Degree of highest derivative = 3

Order and degree of the given differential equation = 4 and 3

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

\(D\left( {D - 1} \right)y - Dy + y = 4t\)

\( \Rightarrow \left( {{D^2} - 2D + 1} \right)y = 4t\)

\( \Rightarrow {\left( {D - 1} \right)^2}y = 4t\)

Particular integral is,

\(P.I. = \frac{1}{{{{\left( {D - 1} \right)}^2}}}4t = 4{\left( {D - 1} \right)^{ - 2}}t\)

\( = 4\left( {1 + 2D + 3{D^2} + \ldots } \right)t\)

A linear second-order partial differential is written as:

\(a\frac{{{\partial ^2}u}}{{\partial {x^2}}} + b\frac{{{\partial ^2}u}}{{\partial x\partial y}} + c\frac{{{\partial ^2}u}}{{\partial {y^2}}} + d\frac{{\partial u}}{{\partial X}} + e\frac{{\partial u}}{{\partial y}} + fu = 0\)

\(a{u_{xx}} + b{u_{xy}} + c{u_{yy}} + d{u_x} + e{u_y} + fu = 0\)

1) If, b² – 4ac > 0: Hyperbolic PDE

2) If, b² – 4ac = 0: Parabolic PDE

3) If, b² – 4ac < 0: Elliptic PDE

The wave equation: u_tt - u_xx = 0 is hyperbolic

The Laplace equation: u_xx + u_yy = 0 is elliptic

The heat equation: u_t - u_xx = 0 is parabolic

Calculation:

\(\frac{{\partial u}}{{\partial t}} = \alpha \frac{{{\partial ^2}u}}{{\partial {x^2}}}\)

Here, a = α, b = 0, c = 0 and α > 0

So, b² – 4ac = 0

So, given PDE is parabolic.

Given:

x dy/dx + y = 0, (x,y) = (1,1)

x dy/dx = -y

On rearranging,

dy/y = -dx/x

\(\smallint \frac{1}{y}dy = \smallint - \frac{1}{x}dx\)

On integrating,

lny  = -lnx  +  c

Since equation passing through the point (1,1), therefore when x=1, y=1

0 = 0 + c

⇒ c = 0

\({\rm{lny}} = - {\rm{lnx}} = \frac{1}{{{\rm{lnx}}}}\)  or y = 1/x

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

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

\(\Rightarrow \frac{{dx}}{{dy}} + x = y\)

Now it is in the form of first order linear equation.

Integrating factor \(= {e^{\smallint 1dy}} = {e^y}\)

Now, the solution is: \(y{e^y} = \smallint y{e^y}dy + C\)

\(\Rightarrow x{e^y} = \left( {y{e^y} - {e^y}} \right) + C\)

\(\Rightarrow \left( {x + 1} \right){e^y} = y{e^y} + C\)

The above curve passes through (0, 0)

⇒ C = 1

Now, the equation becomes \(\left( {x + 1} \right){e^y} = y{e^y} + 1\)

(α, 1) is also passes through the above equation

\(\Rightarrow \left( {\alpha + 1} \right)e = e + 1\)

(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.

\(\frac{{{d^2}y}}{{d{x^2}}} + \frac{{dy}}{{dx}} - 6y = 0\)

With, \(y\left( 0 \right) = 0,\;and\frac{{dy}}{{dx}}\left( 0 \right) = 1\)

Y (1) = -?

∵ D²y + Dy - 6y = 0

Characteristic equation will be -

m² + m - 6 = 0

m² + 3m - 2m - 6 = 0

m = -3, 2

Since roots are real & different thus:

\(y\left( x \right) = {C_1}{e^{ - 3x}} + {C_2}{e^{2x}}\)

Given y (0) = 0

0 = C₁ + C₂      ----(1)

\(\frac{{dy}}{{dx}} = - 3{C_1}{e^{ - 3x}} + 2{C_2}{e^{2x}}\)

\(\frac{{dy}}{{dx}}\left( 0 \right) = - 3{C_1} + 2{C_2}\)

1 = -3C₁ + 2C₂      ----(2)

By equation (1) & (2)

5C₁ = -1

\({C_1} = - \frac{1}{5}\)

\({C_2} = - \frac{1}{5}\) 

Thus, \(y\left( x \right) = - \frac{1}{5}{e^{ - 3x}} + \frac{1}{5}{e^{2x}}\)

\(y\left( 1 \right) = - \frac{1}{5}{e^{ - 3 \times 1}} + \frac{1}{5}{e^{2 \times 1}}\) 

= 1.467

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