Access Advanced Engineering Mathematics 10th Edition solutions now. Our solutions are written by Chegg experts so you can be assured of the highest quality! Advanced Engineering Mathematics ERWIN KREYSZIG Professor of Mathematics Ohio State University. Advanced engineering mathematics / Erwin Kreyszig.—9th ed. Accompanied by instructor’s manual. Includes bibliographical references and index. ISBN 0-471-48885-2 (cloth: acid-free paper) 1. Mathematical physics. Advanced Engineering Mathematics (10th Ed) by ERWIN KREYSZIG pdf. This market-leading text is known for its comprehensive coverage, careful and correct mathematics, outstanding exercises, and self-contained subject matter parts for maximum flexibility.

  1. Advanced Engineering Mathematics By Erwin Kreyszig Pdf Free Download For Windows 10

In mathematics, a Cauchy-Euler equation (most commonly known as the Euler-Cauchy equation, or simply Euler's equation) is a linearhomogeneousordinary differential equation with variable coefficients. It is sometimes referred to as an equidimensional equation. Because of its particularly simple equidimensional structure the differential equation can be solved explicitly.

  • 1The equation

The equation[edit]

Let y(n)(x) be the nth derivative of the unknown function y(x). Then a Cauchy–Euler equation of order n has the form

anxny(n)(x)+an1xn1y(n1)(x)++a0y(x)=0.{displaystyle a_{n}x^{n}y^{(n)}(x)+a_{n-1}x^{n-1}y^{(n-1)}(x)+cdots +a_{0}y(x)=0.}

The substitution x=eu{displaystyle x=e^{u}} (that is, u=ln(x){displaystyle u=ln(x)}; for x<0{displaystyle x<0}, one might replace all instances of x{displaystyle x} by x{displaystyle x }, which extends the solution's domain to R0{displaystyle R_{0}}) may be used to reduce this equation to a linear differential equation with constant coefficients. Alternatively, the trivial solution y=xm{displaystyle y=x^{m}} may be used to directly solve for the basic solutions.[1]

Second order – solving through trial solution[edit]

Typical solution curves for a second-order Euler–Cauchy equation for the case of two real roots
Typical solution curves for a second-order Euler–Cauchy equation for the case of a double root
Typical solution curves for a second-order Euler–Cauchy equation for the case of complex roots

The most common Cauchy–Euler equation is the second-order equation, appearing in a number of physics and engineering applications, such as when solving Laplace's equation in polar coordinates. The second order Cauchy-Euler equation is[1]

x2d2ydx2+axdydx+by=0.{displaystyle x^{2}{frac {d^{2}y}{dx^{2}}}+ax{frac {dy}{dx}}+by=0.,}

We assume a trial solution[1]

y=xm.{displaystyle y=x^{m}.,}

Differentiating gives

dydx=mxm1{displaystyle {frac {dy}{dx}}=mx^{m-1},}

and

d2ydx2=m(m1)xm2.{displaystyle {frac {d^{2}y}{dx^{2}}}=m(m-1)x^{m-2}.,}

Substituting into the original equation leads to requiring

x2(m(m1)xm2)+ax(mxm1)+b(xm)=0{displaystyle x^{2}(m(m-1)x^{m-2})+ax(mx^{m-1})+b(x^{m})=0,}

Rearranging and factoring gives the indicial equation

m2+(a1)m+b=0.{displaystyle m^{2}+(a-1)m+b=0.,}

We then solve for m. There are three particular cases of interest:

  • Case #1 of two distinct roots, m1 and m2;
  • Case #2 of one real repeated root, m;
  • Case #3 of complex roots, α ± βi.

In case #1, the solution is

y=c1xm1+c2xm2{displaystyle y=c_{1}x^{m_{1}}+c_{2}x^{m_{2}},}

In case #2, the solution is

y=c1xmln(x)+c2xm{displaystyle y=c_{1}x^{m}ln(x)+c_{2}x^{m},}

To get to this solution, the method of reduction of order must be applied after having found one solution y = xm.

In case #3, the solution is

y=c1xαcos(βln(x))+c2xαsin(βln(x)){displaystyle y=c_{1}x^{alpha }cos(beta ln(x))+c_{2}x^{alpha }sin(beta ln(x)),}
α=Re(m){displaystyle alpha =mathop {rm {Re}} (m),}
β=Im(m){displaystyle beta =mathop {rm {Im}} (m),}

For c1,c2{displaystyle c_{1},c_{2}} ∈ ℝ .

This form of the solution is derived by setting x = et and using Euler's formula

Second order – solution through change of variables[edit]

x2d2ydx2+axdydx+by=0{displaystyle x^{2}{frac {d^{2}y}{dx^{2}}}+ax{frac {dy}{dx}}+by=0,}

We operate the variable substitution defined by

t=ln(x).{displaystyle t=ln(x).,}
y(x)=ϕ(ln(x))=ϕ(t).{displaystyle y(x)=phi (ln(x))=phi (t).,}

Differentiating gives

dydx=1xdϕdt{displaystyle {frac {dy}{dx}}={frac {1}{x}}{frac {dphi }{dt}}}
d2ydx2=1x2(d2ϕdt2dϕdt).{displaystyle {frac {d^{2}y}{dx^{2}}}={frac {1}{x^{2}}}{bigg (}{frac {d^{2}phi }{dt^{2}}}-{frac {dphi }{dt}}{bigg )}.}

Substituting ϕ(t){displaystyle phi (t)} the differential equation becomes

d2ϕdt2+(a1)dϕdt+bϕ=0.{displaystyle {frac {d^{2}phi }{dt^{2}}}+(a-1){frac {dphi }{dt}}+bphi =0.,}

This equation in ϕ(t){displaystyle phi (t)} is solved via its characteristic polynomial

λ2+(a1)λ+b=0.{displaystyle lambda ^{2}+(a-1)lambda +b=0.}

Now let λ1{displaystyle lambda _{1}} and λ2{displaystyle lambda _{2}} denote the two roots of this polynomial. We analyze the two main cases: distinct roots and double roots:

If the roots are distinct, the general solution is

ϕ(t)=c1eλ1t+c2eλ2t{displaystyle phi (t)=c_{1}e^{lambda _{1}t}+c_{2}e^{lambda _{2}t}}, where the exponentials may be complex.

If the roots are equal, the general solution is

ϕ(t)=c1eλ1t+c2teλ1t.{displaystyle phi (t)=c_{1}e^{lambda _{1}t}+c_{2}te^{lambda _{1}t}.}

In both cases, the solution y(x){displaystyle y(x)} may be found by setting t=ln(x){displaystyle t=ln(x)}.

Hence, in the first case,

y(x)=c1xλ1+c2xλ2{displaystyle y(x)=c_{1}x^{lambda _{1}}+c_{2}x^{lambda _{2}}},

and in the second case,

y(x)=c1xλ1+c2ln(x)xλ1.{displaystyle y(x)=c_{1}x^{lambda _{1}}+c_{2}ln(x)x^{lambda _{1}}.}

Example[edit]

Given

x2u3xu+3u=0,{displaystyle x^{2}u'-3xu'+3u=0,,}

we substitute the simple solution xm:

x2(m(m1)xm2)3x(mxm1)+3xm=m(m1)xm3mxm+3xm=(m24m+3)xm=0.{displaystyle x^{2}(m(m-1)x^{m-2})-3x(mx^{m-1})+3x^{m}=m(m-1)x^{m}-3mx^{m}+3x^{m}=(m^{2}-4m+3)x^{m}=0,.}

For xm to be a solution, either x = 0, which gives the trivial solution, or the coefficient of xm is zero. Solving the quadratic equation, we get m = 1, 3. The general solution is therefore

u=c1x+c2x3.{displaystyle u=c_{1}x+c_{2}x^{3},.}

Difference equation analogue[edit]

There is a difference equation analogue to the Cauchy–Euler equation. For a fixed m > 0, define the sequence ƒm(n) as

fm(n):=n(n+1)(n+m1).{displaystyle f_{m}(n):=n(n+1)cdots (n+m-1).}

Applying the difference operator to fm{displaystyle f_{m}}, we find that

Dfm(n)=fm(n+1)fm(n)=m(n+1)(n+2)(n+m1)=mnfm(n).{displaystyle {begin{aligned}Df_{m}(n)&=f_{m}(n+1)-f_{m}(n)&=m(n+1)(n+2)cdots (n+m-1)={frac {m}{n}}f_{m}(n).end{aligned}}}

If we do this k times, we find that

fm(k)(n)=m(m1)(mk+1)n(n+1)(n+k1)fm(n)=m(m1)(mk+1)fm(n)fk(n),{displaystyle {begin{aligned}f_{m}^{(k)}(n)&={frac {m(m-1)cdots (m-k+1)}{n(n+1)cdots (n+k-1)}}f_{m}(n)&=m(m-1)cdots (m-k+1){frac {f_{m}(n)}{f_{k}(n)}},end{aligned}}}

where the superscript (k) denotes applying the difference operator k times. Comparing this to the fact that the k-th derivative of xm equals

m(m1)(mk+1)xmxk{displaystyle m(m-1)cdots (m-k+1){frac {x^{m}}{x^{k}}}}

suggests that we can solve the N-th order difference equation

fN(n)y(N)(n)+aN1fN1(n)y(N1)(n)++a0y(n)=0,{displaystyle f_{N}(n)y^{(N)}(n)+a_{N-1}f_{N-1}(n)y^{(N-1)}(n)+cdots +a_{0}y(n)=0,}

in a similar manner to the differential equation case. Indeed, substituting the trial solution

y(n)=fm(n){displaystyle y(n)=f_{m}(n),}

brings us to the same situation as the differential equation case,

m(m1)(mN+1)+aN1m(m1)(mN+2)++a1m+a0=0.{displaystyle m(m-1)cdots (m-N+1)+a_{N-1}m(m-1)cdots (m-N+2)+cdots +a_{1}m+a_{0}=0.}

One may now proceed as in the differential equation case, since the general solution of an N-th order linear difference equation is also the linear combination of N linearly independent solutions. Applying reduction of order in case of a multiple root m1 will yield expressions involving a discrete version of ln,

φ(n)=k=1n1km1.{displaystyle varphi (n)=sum _{k=1}^{n}{frac {1}{k-m_{1}}}.}

(Compare with: ln(xm1)=1+m1x1tm1dt.{displaystyle ln(x-m_{1})=int _{1+m_{1}}^{x}{frac {1}{t-m_{1}}},dt.})

Advanced Engineering Mathematics By Erwin Kreyszig Pdf Free Download

In cases where fractions become involved, one may use

fm(n):=Γ(n+m)Γ(n){displaystyle f_{m}(n):={frac {Gamma (n+m)}{Gamma (n)}}}

instead (or simply use it in all cases), which coincides with the definition before for integer m.

See also[edit]

References[edit]

  1. ^ abcKreyszig, Erwin (May 10, 2006). Advanced Engineering Mathematics. Wiley. ISBN978-0-470-08484-7.

Bibliography[edit]

  • Weisstein, Eric W.'Cauchy–Euler equation'. MathWorld.
Retrieved from 'https://en.wikipedia.org/w/index.php?title=Cauchy–Euler_equation&oldid=894113422'
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Advanced Engineering Mathematics By Erwin Kreyszig – The tenth edition of this bestselling text includes examples in more detail and more applied exercises; both changes are aimed at making the material more relevant and accessible to readers. Kreyszig introduces engineers and computer scientists to advanced math topics as they relate to practical problems. It goes into the following topics at great depth differential equations, partial differential equations, Fourier analysis, vector analysis, complex analysis, and linear algebra/differential equations.

Advanced Engineering Mathematics By Erwin Kreyszig Pdf Free Download For Windows 10

Advanced Engineering Mathematics By Erwin Kreyszig – PDF Free Download


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Part A: Ordinary Differential Equations (Ode’s).

Part B: Linear Algebra, Vector Calculus.

Part C: Fourier Analysis, Partial Differential Equations.

Part E: Numerical Analysis Software.

Part F: Optimization, Graphs.

Part G: Probability; Statistics.

Appendix 1: References.

Appendix 2: Answers to Odd-Numbered Problems.

Appendix 3: Auxiliary Material.

Appendix 4: Additional Proofs.

Appendix 5: Tables.

Index.

About Author

Erwin Kreyszig, Ohio State University

Book Details

  • Publisher:[email protected]; 10th Edition edition
  • Language: English
  • ISBN-10: 9780470458365
  • ISBN-13: 978-0470458365
  • ASIN: 0470458364

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