Problem

Find the solution of the differential equation th…

01:58

Problem 16 Medium Difficulty

Find the solution of the differential equation that satisfies the given initial condition.
$ \frac {dP}{dt} = \sqrt {Pt}, P(1) = 2 $

Name: find the solution of the differential equation that satisfies the given initial condition frac dpdt
Uploaded: 2020-02-27
Duration: 1 min 57 s
Channel: Numerade
Description: Find the solution of the differential equation that satisfies the given initial condition. $ \frac {dP}{dt} = \sqrt {Pt}, P(1) = 2 $

Answer

$P=\left(\frac{1}{3} t^{3 / 2}-\frac{1}{3}+\sqrt{2}\right)^{2}$

Topics

Differential Equations

Calculus: Early Transcendentals

Chapter 9

Differential Equations

Section 3

Separable Equations

Discussion

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Video Transcript

this question is asking us to find a solution of the differential equation that satisfies the given initial condition. We know that we have dp over. DT is square root of P t, which is Petey to the power of 1/2. Now that we have this, we know we want to put all the pieces stuff on left hand side and all the t's stuff on the right hand side. So d p over Pete. The 1/2 is cheap to the 1/2 DT. As you can see, Tea's on the right piece on the left. Now get rid of the fraction by writing this with negative expert is negative one to the power of negative 1/2 because now this makes a lot easier to integrate. Integrate this and we now have p to the 1/2 equals 1/3. Teach the three over two plus 1/2 times. See Cesar constant of integration. You know what this means is that we can now use the initial value of p of oneness to what this means is that our X or in this context, R t is one and our why or in this context or pee is gonna be, too. Take the squirt of both sides to get rid of the squared. Now, writing this just in terms of sea we get C is negative. 1/3 plus square root of two. You can see how we're just writing this in terms of C. Lastly, plug into your P equation where we previously had see, we should now have our value of C, which we have just figured out minus 1/3 plus square root of two, and this original was squared.

Amrita B.

Topics

Differential Equations

Calculus: Early Transcendentals

Chapter 9

Differential Equations

Section 3

Separable Equations

Problem