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