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

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The discussion revolves around solving the differential equation y'' - y = cosh(x) using variation of parameters. The user derived a particular solution but ended up with a general solution that differed from the textbook answer, which included a term xsinh(x). Confusion arose due to unfamiliarity with hyperbolic functions like sinh and cosh, prompting questions about potential identities. A suggestion was made to use the identity cosh(x) = 0.5(e^x + e^{-x}) to simplify the integration process. The user expressed gratitude for the new identity, indicating a learning moment in understanding hyperbolic functions.
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.
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

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