Question

Diff EQ.
Looked at attached picture.

Answer 1

did you try to take the integral of sin(t^2) + t

Answer 2

you mean like a separable ode

Answer 3

yes

Answer 4

also it will help to keep in mind
$$\large x(t) - x(a) = \int_{a}^{t} x' (t) ~dt $$

Answer 5

I just got outta class, thank you. I get what you did there. : )

Answer 6

Woah why did you delete the explanation.

Answer 7

\[\large {x(t) - x(a) = \int_{a}^{t} x' (t) ~dt
\\ \text{plugging in } \\
x(t)-x(0)= \int_{0}^{t} (\sin(t^2) + t ) dt \\=
\int_{0}^{t} \sin(t^2)dt + \int_{0}^{t} t \ dt = \int_{0}^{t} \sin(t^2)dt + \frac{t^2}{2}
\\ \ \\ \text{therefore}
\\ \ \\ x(t)-x(0) = \int_{0}^{t} \sin(t^2)dt + \frac{t^2}{2}
\\ \ \\ \text{solving for x(t)}
\\ \ \\ x(t) = \int_{0}^{t} \sin(t^2)dt + \frac{t^2}{2} + x(0)
\\ \ \\ \text{replace x(0)= 20}
\\ \ \\ x(t) = \int_{0}^{t} \sin(t^2)dt + \frac{t^2}{2} + 20

}\]

Answer 8

We can't simplify that integral sin(t^2) any further