# 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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