Question

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.

Answer 1

Answer:

$y(t)=(\frac{1}{-ckt+y^{-c}_0})^{\frac{1}{c}}$

Step-by-step explanation:

We are given that  differential equation

$\frac{dy}{dt}=ky^{1+c}$

Where k=Positive constant

c=Positive number

$\int \frac{dy}{y^{1+c}}=\int kdt$

$-\frac{1}{c}y^{-c}=kt+C$

Initial condition

$y(0)=y_0$

$-\frac{1}{cy^c}=kt+c$

Substitute the value of initial condition

$-\frac{1}{c(y_0)^{c}}=C$

$-\frac{1}{c(y)^{c}}=kt-\frac{1}{c(y_0)^{c}}$

$\frac{1}{y^c}=-ckt+(y_0)^{-c}$

$y^c=\frac{1}{-ckt+y^{-c}_0}$

$y(t)=(\frac{1}{-ckt+y^{-c}_0})^{\frac{1}{c}}$

Answer 2

Differential equation is used to relate equations and their derivatives.

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

The differential equation is given as:

$\mathbf{\frac{dy}{dt} = ky^{1 + c}}$

Multiply both sides by dt

$\mathbf{dy = ky^{1 + c} dt}$

Divide both sides by $\mathbf{y^{1 + c}}$

$\mathbf{\frac{dy}{y^{1 + c}} = k dt}$

Integrate both sides

$\mathbf{ \int\limits\frac{dy}{y^{1 + c}} = \int\limits k dt}$

Remove constants

$\mathbf{ \int\limits\frac{dy}{y^{1 + c}} = k \int\limits dt}$

Rewrite as:

$\mathbf{ \int\limits dy\ y^{-1 - c} = k \int\limits dt}$

Now, integrate both sides

$\mathbf{ \frac{y^{1-1 - c}}{1-1 - c} = kt + c}$

$\mathbf{ \frac{y^{- c}}{- c} = kt + c}$

This gives

$\mathbf{ -\frac{1}{cy^c} = kt + c}$

At the initial condition, $\mathbf{y(0) = y0}$

So, we have:

$\mathbf{ C = -\frac{1}{c(y0)^c} }$

Substitute $\mathbf{ C = -\frac{1}{c(y0)^c} }$ in $\mathbf{ -\frac{1}{cy^c} = kt + c}$

$\mathbf{-\frac{1}{cy^c} = kt - \frac{1}{c(y0)^c}}$

Multiply through by c

$\mathbf{-\frac{1}{y^c} = ckt - \frac{1}{(y0)^c}}$

Rewrite as:

$\mathbf{-\frac{1}{y^c} = ckt - (y0)^{-c}}$

Take inverse of both sides

$\mathbf{-y^c = \frac{1}{ckt - (y0)^{-c}}}$

Multiply both sides by -1

$\mathbf{y^c = \frac{1}{(y0)^{-c} -c kt}}$

Take c-th root of both sides

$\mathbf{y = \frac{1}{((y0)^{-c} -c kt)^c}}$

Express as exponents

$\mathbf{y = ((y0)^{-c} -c kt)^{-c}}$

Express y as a function of t

$\mathbf{y(t) = ((y0)^{-c} -c kt)^{-c}}$

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

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