Solving Second Order Differential Equation

[Ed Aboud]

Homework Statement

Solve \frac{d^2 y}{dt^2} = y

Homework Equations

The Attempt at a Solution

\frac{dy}{dt} = v

\frac{d^2 y}{dt^2} = v \frac{dv}{dy}


v \frac{dv}{dy} = y

v dv = y dy

\int v dv = \int y dy

v^2 = y^2 + C

( \frac{dy}{dt} )^2 = y^2 + C

\frac{dy}{dt} = \sqrt{ y^2 + C }

\int \frac{dy}{ \sqrt{ y^2 + C }} = \int dt

\int \frac{dy}{ \sqrt{ y^2 + C }} = t

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

Thanks in advance for any help.

 

[Mark44]

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.

 

[Ed Aboud]

Oh wow how did I not see that.

Thanks for the help!

 

[HallsofIvy]

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
\int \frac{dy}{\sqrt{y^2+1}}
use a "hyperbolic" substitution. Let y= sinh(u). Since sinh²(u)+ 1= cosh²(u), \sqrt{y^2+ 1}= \sqrt{cosh^2(u)}= cosh(u). Also du= cosh(u) du so this becomes simply
\int du= u+ C2= sinh(y)+ C

Of course, since
sinh(y)= \frac{e^y- e^{-y}}{2}
You can see how that fits the "usual" solution.

 

[Ed Aboud]

I see, quite true. I don't know too much about hyperbolic functions but I see what your saying. Thanks for the help!

 

[Ed Aboud]

But should

u = sinh^-^1 (y) ?