General solution of double integrals

[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?

 

[DonAntonio]

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?

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

DonAntonio

 

[d.arbitman]

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

 

[haruspex]

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

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