Solving Second Order Differential Equation

In summary, the conversation discusses how to solve the differential equation d^2y/dt^2 = y and provides two possible methods: using linear equations with constant coefficients or using a hyperbolic substitution. The hyperbolic substitution method is explained in more detail, with the final solution being u = sinh^-1(y).
  • #1
Ed Aboud
201
0

Homework Statement



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

Homework Equations





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.


 
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  • #2
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.
 
  • #3
Oh wow how did I not see that.

Thanks for the help!
 
  • #4
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.
 
  • #5
I see, quite true. I don't know too much about hyperbolic functions but I see what your saying. Thanks for the help!
 
  • #6
But should

[tex] u = sinh^-^1 (y) [/tex] ?
 

What is a differential equation?

A differential equation is a mathematical equation that relates a function with its derivatives. It is used to model and describe various phenomena in science and engineering.

What is the purpose of studying differential equations?

Studying differential equations allows scientists to understand and predict the behavior of systems in nature and engineering. It also helps in finding solutions to problems that involve rates of change.

What are the different types of differential equations?

There are several types of differential equations, including ordinary differential equations, partial differential equations, and stochastic differential equations. These types differ based on the type of function and its variables involved in the equation.

How are differential equations solved?

Differential equations can be solved analytically or numerically. Analytical solutions involve finding a functional expression for the solution, while numerical solutions involve using computer algorithms to approximate the solution.

What are the applications of differential equations?

Differential equations have a wide range of applications in various fields, such as physics, chemistry, biology, economics, and engineering. They are used to model and analyze complex systems and phenomena, including population growth, chemical reactions, and electrical circuits.

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