Differential equations

Ed Aboud

1. Homework Statement

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

2. Homework Equations

3. 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.
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

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Mark44

Mentor
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

Homework Helper
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

$$u = sinh^-^1 (y)$$ ?

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