What is the general solution for a DE involving cosh and sinh?

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Homework Help Overview

The discussion revolves around solving a differential equation involving hyperbolic functions, specifically cosh and sinh. The original poster presents their attempt at using variation of parameters to find the general solution for the equation y'' - y = cosh(x).

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the method of variation of parameters and the calculation of particular solutions. The original poster expresses confusion regarding the appearance of the term xsinh(x) in the solution provided by their textbook, questioning the validity of their own derived solution.

Discussion Status

Some participants have pointed out potential errors in the integration process, suggesting alternative approaches to the integral involving cosh(x). There is acknowledgment of a hyperbolic identity that may clarify the situation for the original poster.

Contextual Notes

The original poster mentions a lack of prior exposure to hyperbolic functions, indicating that this is their first encounter with such terms in a differential equation context.

Rijad Hadzic
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Homework Statement


Solve the DE by variation of parameters:

y'' - y = cosh(x)

Homework Equations

The Attempt at a Solution


I got m = 1 and m = -1 so

y = c_1e^x + c_2e^{-x} + y_p

y_p = u_1e^x + u_2e^{-x}

The wonksian gave me -2

so

u_1&#039; = \frac{\begin{vmatrix}<br /> 0 &amp; e^{-x} \\<br /> cosh(x) &amp; -e^{-x} <br /> \end{vmatrix} }{-2}

I get -\frac12 \int{e^{-x}cosh(x) dx} = u_1

I do parts, u = e^-x and dv = coshx and eventually I get

-\frac 14 e^{-x}sinh(x) + \frac 14 e^{-x}cosh(x) = u_1
then

u_2&#039; = \frac{<br /> \begin{vmatrix}<br /> e^x &amp; 0 \\<br /> e^x &amp; cosh(x)<br /> \end{vmatrix} }{-2}

I get -\frac12 \int{e^{x}cosh(x) dx} = u_2

I do parts with u = e^x and dv = coshx and I get

\frac 12 e^xsinh(x) + \frac 12 e^xcosh(x) = u_2

u_1*e^x = -\frac 14 sinh(x) + \frac 14 cosh(x)
u_2*e^{-x} = \frac 12 sinh(x) + \frac 12 cosh(x)

adding I get \frac 14 sinh(x) + \frac 34 cosh(x)

so my general solution is

y_{gen} = c_1e^x +c_2e^{-x} + \frac 14 sinh(x) + \frac 34 cosh(x)

but my book got

y_{gen} = c_1e^x +c_2e^{-x} + \frac 12 xsinh(x)

how could it be possible to get a term called xsinh(x)??

This is the first time I've been given a problem with sinh and cosh. I've never been taught a single thing about them in school so I'm having trouble... I'm guessing there is an identity or something?

Would my answer of

y_{gen} = c_1e^x +c_2e^{-x} + \frac 14 sinh(x) + \frac 34 cosh(x)

satisfy the question?
 
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Rijad Hadzic said:

Homework Statement


Solve the DE by variation of parameters:

y&#039;&#039; - y = cosh(x)

Homework Equations

The Attempt at a Solution


I got m = 1 and m = -1 so

y = c_1e^x + c_2e^{-x} + y_p

y_p = u_1e^x + u_2e^{-x}

The wonksian gave me -2

so

u_1&#039; = \frac{\begin{vmatrix}<br /> 0 &amp; e^{-x} \\<br /> cosh(x) &amp; -e^{-x}<br /> \end{vmatrix} }{-2}

I get -\frac12 \int{e^{-x}cosh(x) dx} = u_1

I do parts, u = e^-x and dv = coshx and eventually I get

-\frac 14 e^{-x}sinh(x) + \frac 14 e^{-x}cosh(x) = u_1
You did something wrong with the integral.
Instead of integrating by parts, write cosh(x)=0.5(ex+e-x)
 
ehild said:
You did something wrong with the integral.
Instead of integrating by parts, write cosh(x)=0.5(ex+e-x)

Thank you I didn't even know this identity existed.
 

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