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Differential equations

  • Thread starter Ed Aboud
  • Start date
197
0
1. Homework Statement

Solve [tex] \frac{d^2 y}{dt^2} = y [/tex]

2. Homework Equations



3. The Attempt at a Solution

[tex] \frac{dy}{dt} = v [/tex]

[tex] \frac{d^2 y}{dt^2} = v \frac{dv}{dy} [/tex]


[tex] v \frac{dv}{dy} = y [/tex]

[tex] v dv = y dy [/tex]

[tex] \int v dv = \int y dy [/tex]

[tex] v^2 = y^2 + C [/tex]

[tex] ( \frac{dy}{dt} )^2 = y^2 + C [/tex]

[tex] \frac{dy}{dt} = \sqrt{ y^2 + C } [/tex]

[tex] \int \frac{dy}{ \sqrt{ y^2 + C }} = \int dt [/tex]

[tex] \int \frac{dy}{ \sqrt{ y^2 + C }} = t [/tex]

I have no idea how to integrate this. Have I gone wrong somewhere?

Thanks in advance for any help.
1. Homework Statement



2. Homework Equations



3. The Attempt at a Solution
1. Homework Statement



2. Homework Equations



3. The Attempt at a Solution
1. Homework Statement



2. Homework Equations



3. The Attempt at a Solution
 
Last edited by a moderator:
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There is one function whose derivative is exactly equal to the function itself. It follows that the second derivative would also be equal to the original function.
 
197
0
Oh wow how did I not see that.

Thanks for the help!!
 

HallsofIvy

Science Advisor
Homework Helper
41,712
876
The is a "linear equation with constant coefficients" so there is an easy solution as Mark44 pointed out.

But your method is interesting- and right! To integrate
[tex]\int \frac{dy}{\sqrt{y^2+1}}[/tex]
use a "hyperbolic" substitution. Let y= sinh(u). Since sinh2(u)+ 1= cosh2(u), [itex]\sqrt{y^2+ 1}= \sqrt{cosh^2(u)}= cosh(u)[/itex]. Also du= cosh(u) du so this becomes simply
[tex]\int du= u+ C2= sinh(y)+ C[/tex]

Of course, since
[tex]sinh(y)= \frac{e^y- e^{-y}}{2}[/tex]
You can see how that fits the "usual" solution.
 
197
0
I see, quite true. I don't know too much about hyperbolic functions but I see what your saying. Thanks for the help!
 
197
0
But should

[tex] u = sinh^-^1 (y) [/tex] ?
 

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