# General solution of double integrals

1. Nov 19, 2012

### d.arbitman

I have a differential equation of the form and I want to solve it using calculus, as opposed to using a differential equation method.
$\frac{d^2v}{dt}$ = $\alpha$

where $v$ is a function of $t$ i.e., $v(t)$
and $\alpha$ is some constant.

How do I solve for $v(t)$ if the time ranges from $t_0$ to $t$ such that $v'(t_0)$ and $v(t_0)$ are the lower bounds of my double integrals?

Do I setup a double integral, such as the following?

$\int_{v(t_0)}^{v(t)}$ $\int_{v'(t_0)}^{v(t)} dvdv$ = $\alpha$$\int_{t_0}^{t}$ $\int_{t_0}^{t} dτ dτ$

I get the following solution from evaluating that double integral:
$v(t) = \frac{\alpha}{2}{(t-t_0})^2 +v'(t_0)(t-t_0) + v(t_0)$
and I was wondering if the solution is correct?

2. Nov 19, 2012

### DonAntonio

Why wonder? Derive twice what you got and see if you get back your differential equation!

DonAntonio

3. Nov 19, 2012

### d.arbitman

I did, but I want to check if the procedure is correct.

4. Nov 21, 2012

### haruspex

The LHS makes no sense to me.
$v''(u) = f(u)$
$\left[v'(u)\right]_{u=t_0}^w = \int_{u=t_0}^w f(u) du$
$v'(w) - v'(t_0) = \int_{u=t_0}^w f(u) du$
$\left[v(w)\right]_{w=t_0}^t - (t-t_0)v'(t_0) = \int_{w=t_0}^t \int_{u=t_0}^w f(u) du dw$
$v(t) = v(t_0) + (t-t_0)v'(t_0) + \int_{w=t_0}^t \int_{u=t_0}^w f(u) du dw$