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The function y satisfies a differential equation of the form y' = ky for some number k. If you are told that when t = 3 that y is 2 and the rate of change of y is 4 then what is k?

Respuesta :

LammettHash

[tex]y'=ky[/tex] is a separable ODE:

[tex]\dfrac{\mathrm dy}{\mathrm dt}=ky\implies\dfrac{\mathrm dy}y=k\,\mathrm dt[/tex]

Integrating both sides and solving for [tex]y[/tex], we get

[tex]\ln|y|=kt+C\implies y=e^{kt+C}\implies\boxed{y=Ce^{kt}}[/tex]

Given that [tex]y(3)=2[/tex], we know

[tex]2=Ce^{3k}[/tex]

We also have

[tex]y'=Cke^{kt}[/tex]

and given that [tex]y'(3)=4[/tex], we know

[tex]4=Cke^{3k}[/tex]

Then

[tex]Cke^{3k}=2Ce^{3k}\implies Ck=2C\implies k=2\text{ or }C=0[/tex]

  • If [tex]k=2[/tex], then [tex]2=Ce^6\implies C=2e^{-6}[/tex].
  • If [tex]C=0[/tex], then we get a contradiction because we need to have [tex]Ce^{3k}=2[/tex].

So it must the case that [tex]k=2[/tex].

xero099

The value of [tex]k[/tex] is 2.

In this case we have an ordinary linear differential equation with separable variables:

[tex]y' = k\cdot y[/tex] (1)

If we know that [tex]y = 2[/tex] and [tex]y' = 4[/tex], then the value of [tex]k[/tex] is:

[tex]k = \frac{y'}{y}[/tex]

[tex]k = \frac{4}{2}[/tex]

[tex]k = 2[/tex]

The value of [tex]k[/tex] is 2.

We kindly invite to check to this question on differential equations: https://brainly.com/question/14620493

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