So we could say that 1 plus v squared is equal to cx.

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Plus 1.

And actually, I'm going to do that right now.

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Let me put the Laplace transform of and I'm also going to the sides.

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That's s times the Laplace transform of f of t, minus f of 0, right?

And now we got that as being equal to this thing, and we made a substitution.

So let's make v prime is f prime, and let's make u e to the minus st.

And then the right hand side of the equation is this.

This is equal to 3 times 5, 3 times 5.

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The chain rule, derivative of cosine is minus sign.

Repeat this procedure for each particle.

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So I like to stay away from those crazy Latin alphabets, so we'll just use a regular x.

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We figured out what g of 3 was first.

And now we can use this, which we proved several videos ago, to resubstitute it and get it in terms of the Laplace

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So let's use this one, it seems a little bit, marginally simpler.

All these medications, nothing seems to be working.

They're all basically words you'll see that refer to the same concept, which is we turn spaces to pluses.

That means the heap property isn't satisfied but all we have to do to fix it is we just swap the value in the left child and i.

Because we were able to write it as a function of y divided by x.

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