Basic Differential Equations Question?

Question

I need some help with one problem, thanks in advance.

Please show the verification that the function is a solution to the differential equation.
y'-2ty=1; y=e^(t^2) ∫e^(-s^2)ds+e^(t^2)

The upper bound for the integral is t and the lower bound is 0.

Answer 1

To verify whether y is a solution you need to figure out y'.
y' = (e^t²)'∫e^(-s²)ds + (e^t²)[∫e^(-s²)ds]' + (e^t²)'
y' = 2te^t²∫e^(-s²)ds + (e^t²)[∫e^(-s²)ds]' + 2te^t²
The derivative of an integral is just the function of the integral, so to finish off y'

y' = 2te^t²∫e^(-s²)ds + (e^t²)[e^(-t²)] + 2te^t²
y' = 2te^t²∫e^(-s²)ds + 1 + 2te^t²

Now plug y' into the differential equation

2te^t²∫e^(-s²)ds + 1 + 2te^t² -2t[e^(t²) ∫e^(-s²)ds+e^(t²) ?= 1

2te^t²∫e^(-s²)ds + 1 + 2te^t² -2te^(t²)∫e^(-s²)ds+-2te^(t²) ?=1
{2te^t²∫e^(-s²)ds -2te^(t²)∫e^(-s²)ds} + {2te^t²-2te^(t²)} +1 ?=1
Then 1 = 1

Answer 2

It is sufficient to show that y' - 1 = 2ty.
Assume y=e^(t^2) ∫e^(-s^2)ds+e^(t^2),
y' = [e^(t^2)]' ∫e^(-s^2)ds + e^(t^2) [∫e^(-s^2)ds]' + [e^(t^2)]'
= 2t*e^(t^2) ∫e^(-s^2)ds + e^(t^2)*e^(-t^2) + 2te^(t^2)
= 2t*[e^(t^2) ∫e^(-s^2)ds+e^(t^2)] + 1
= 2t*y + 1
Therefore indeed y' - 1 = 2ty.