11) ẋ + y = t^2 + 1 ẏ = x + t + 1 12) ẋ = e^t

Transcript

00:01 So this is asking us to explain the steps we would use to solve the system of equations using linear combinations and then solve the system.

00:08 So the first step is i have to get everything lined up.

00:12 So i would rewrite the top equation into standard form.

00:22 So that way it's a x plus b y equals c because you can see right now i've got a y on top of an x and equal on top of a plus.

00:30 It's all kind of not really lined up.

00:33 But i do like the way the second equation looks.

00:36 Now, in order to do this, it's going to require subtracting 2x.

00:40 And i would put that in front of the y, and you're going to see this play out in a second.

00:43 But when you subtract 2x, that's going to make it negative 2x.

00:46 And so then negative 2x and 2x would be opposites.

00:49 So we won't need to multiply for linear combinations.

00:52 So then we can add our equations together.

00:56 This would cause the x to cancel, which means we'd find the y value.

01:04 And then once we have the y value, we need to sub back in, substitute back in to find x so that's what we would do we're going to rewrite this into standard form we're going to add our equations together that's going to allow us to find y and then we're going to substitute back in to find x so again like i said for this top equation we would subtract 2x from both sides and so that would become negative 2x plus 2y equals negative 2...

Question

11) \(\dot{x} + y = t^2 + 1\) \(\dot{y} = x + t + 1\) 12) \(\dot{x} = e^t - z\) \(\dot{x} + \dot{y} + \dot{z} = x\) \(\dot{z} = e^t - x - 2y\)

Question

11) \(\dot{x} + y = t^2 + 1\) \(\dot{y} = x + t + 1\) 12) \(\dot{x} = e^t - z\) \(\dot{x} + \dot{y} + \dot{z} = x\) \(\dot{z} = e^t - x - 2y\)

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