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Section 16 .2, problem number 32, we're solving a bernoulli equation.
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So what we're accustomed to standard form, y prime plus p of x, y equal q of x.
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The bernouille equation is a little bit different.
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It's y prime plus p of x y equal q of x times y to the nth power.
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The substitution is to let u equal y to the 1 minus n that transformed this equation to u prime plus 1 minus n p of x y is equal to 1 minus n times q of x.
00:44
So it gets us back to a standard form differential equation.
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So in this case, we've got n equal 3.
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So u is going to be y to the 1 minus 3, which is a number.
00:58
Is y to the negative 2 power.
01:03
So to write down this equation, this equation is going to be d -u -d -x plus 1 minus 3...