Find the solution of the differential equation that satisfies the given initial condition. `(dP)/(dt) = sqrt(Pt)` P(1) = 2

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Chapter 9, 9.3 - Problem 16 - Calculus: Early Transcendentals (7th Edition, James Stewart).
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lemjay | High School Teacher | (Level 3) Senior Educator

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`(dP)/(dt) = sqrt(Pt)`

`P(1) = 2`

To solve, separate the variables.

`(dP)/(dt)=sqrtPsqrt t`

`(dP)/sqrtP=sqrt t dt`

`P^(-1/2)dP=t^(1/2) dt`

Then, take the integral of each side. Apply the formula `int x^n dx = x^(n+1)/(n+1)+C` .

`2P^(1/2) + C = 2/3t^(3/2) + C`

`2sqrtP+C=(2sqrt t^3)/3+C`

`2sqrtP+C=(2tsqrtt)/3+C`

Since C is a constant, then, we may express the C's in our equation as a single C.

`2sqrtP=(2tsqrtt)/3+C`

Then, apply the condition P(1)=2 to get the value of C. So, plug-in t=1 and P=2.

`2sqrt2 =(2*1sqrt1)/3+C`

`2sqrt2 =2/3+C`

`2sqrt2-2/3=C`

`(6sqrt2-2)/3=C`

Then, plug-in the value of C to `2sqrtP=(2tsqrtt)/3+C` .

Hence, the implicit solution of the differential equation is

`2sqrtP=(2tsqrtt)/3 + (6sqrt2-2)/3`

After that, determine the explicit solution. To do so, isolate the P.

`2sqrtP=(2tsqrt t + 6sqrt2 - 2)/3`

`2sqrtP=(2(tsqrtt + 3sqrt2-1))/3`

`sqrtP=(tsqrtt + 3sqrt2-1)/3`

`(sqrtP)^2=((tsqrtt + 3sqrt2-1)/3)^2`

`P=(tsqrtt+3sqrt2-1)^2/9`

Therefore, the explicit solution of the given differential equation is `P(t)=(tsqrtt+3sqrt2-1)^2/9` .

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