Let c be a positive number. A differential equation of the form dy dt = ky1 + c where k is a positive constant, is called a doomsday equation because the exponent in the expression ky1 + c is larger than the exponent 1 for natural growth. (a) Determine the solution that satisfies the initial condition y(0) = y0.

Respuesta :

Answer:

[tex]y(t)=(\frac{1}{-ckt+y^{-c}_0})^{\frac{1}{c}}[/tex]

Step-by-step explanation:

We are given that  differential equation

[tex]\frac{dy}{dt}=ky^{1+c}[/tex]

Where k=Positive constant

c=Positive number

[tex]\int \frac{dy}{y^{1+c}}=\int kdt[/tex]

[tex]-\frac{1}{c}y^{-c}=kt+C[/tex]

Initial condition

[tex]y(0)=y_0[/tex]

[tex]-\frac{1}{cy^c}=kt+c[/tex]

Substitute the value of initial condition

[tex]-\frac{1}{c(y_0)^{c}}=C[/tex]

[tex]-\frac{1}{c(y)^{c}}=kt-\frac{1}{c(y_0)^{c}}[/tex]

[tex]\frac{1}{y^c}=-ckt+(y_0)^{-c}[/tex]

[tex]y^c=\frac{1}{-ckt+y^{-c}_0}[/tex]

[tex]y(t)=(\frac{1}{-ckt+y^{-c}_0})^{\frac{1}{c}}[/tex]

Differential equation is used to relate equations and their derivatives.

The solution that satisfies the initial condition is: [tex]\mathbf{y(t) = ((y0)^{-c} -c kt)^{-c}}[/tex]

The differential equation is given as:

[tex]\mathbf{\frac{dy}{dt} = ky^{1 + c}}[/tex]

Multiply both sides by dt

[tex]\mathbf{dy = ky^{1 + c} dt}[/tex]

Divide both sides by [tex]\mathbf{y^{1 + c}}[/tex]

[tex]\mathbf{\frac{dy}{y^{1 + c}} = k dt}[/tex]

Integrate both sides

[tex]\mathbf{ \int\limits\frac{dy}{y^{1 + c}} = \int\limits k dt}[/tex]

Remove constants

[tex]\mathbf{ \int\limits\frac{dy}{y^{1 + c}} = k \int\limits dt}[/tex]

Rewrite as:

[tex]\mathbf{ \int\limits dy\ y^{-1 - c} = k \int\limits dt}[/tex]

Now, integrate both sides

[tex]\mathbf{ \frac{y^{1-1 - c}}{1-1 - c} = kt + c}[/tex]

[tex]\mathbf{ \frac{y^{- c}}{- c} = kt + c}[/tex]

This gives

[tex]\mathbf{ -\frac{1}{cy^c} = kt + c}[/tex]

At the initial condition, [tex]\mathbf{y(0) = y0}[/tex]

So, we have:

[tex]\mathbf{ C = -\frac{1}{c(y0)^c} }[/tex]

Substitute [tex]\mathbf{ C = -\frac{1}{c(y0)^c} }[/tex] in [tex]\mathbf{ -\frac{1}{cy^c} = kt + c}[/tex]

[tex]\mathbf{-\frac{1}{cy^c} = kt - \frac{1}{c(y0)^c}}[/tex]

Multiply through by c

[tex]\mathbf{-\frac{1}{y^c} = ckt - \frac{1}{(y0)^c}}[/tex]

Rewrite as:

[tex]\mathbf{-\frac{1}{y^c} = ckt - (y0)^{-c}}[/tex]

Take inverse of both sides

[tex]\mathbf{-y^c = \frac{1}{ckt - (y0)^{-c}}}[/tex]

Multiply both sides by -1

[tex]\mathbf{y^c = \frac{1}{(y0)^{-c} -c kt}}[/tex]

Take c-th root of both sides

[tex]\mathbf{y = \frac{1}{((y0)^{-c} -c kt)^c}}[/tex]

Express as exponents

[tex]\mathbf{y = ((y0)^{-c} -c kt)^{-c}}[/tex]

Express y as a function of t

[tex]\mathbf{y(t) = ((y0)^{-c} -c kt)^{-c}}[/tex]

Hence, the solution that satisfies the initial condition is: [tex]\mathbf{y(t) = ((y0)^{-c} -c kt)^{-c}}[/tex]

Read more about differential equations at:

https://brainly.com/question/14620493

ACCESS MORE
EDU ACCESS