Question

Solve the initial-value problem y = 24 – 3yy(1) = 1.

Answer 1

`y = 8 -7e^(-3(x-1))`

Explanation:

Assuming that the equation is
`y '= 24 -3y` with initial condition
`y(1) = 1`. We have,

`(dy)/(dx) + 3y = 24`, hence we can say that
`P (x) = 3` and
`Q (x) = 24` in the general form of the first order linear differential equation:

`(dy)/(dx) + P(x)y = Q(x)`

The integrating factor is given by:

`e^{\int{3} \, dx } = e^(3x)`. Thus, multiplying the entire equation by the integrating factor:

`e^(3x)(dy)/(dx) + 3e^(3x)y = 24e^(3x)`. This means that:

`(d[e^(3x)y])/(dx) = 24e^(3x)`

`e^(3x)y = 8e^(3x) + C` then

`y = 8 + Ce^(-3x)`. Applying the initial condition:

`C = -7e^(3)` and therefore,
`y = 8 -7e^(-3(x-1))`

Assuming that the equation is
`y = 24 -3y'` with initial condition
`y(1) = 1`. We have,

`(dy)/(dx) + (1)/(3)y = 8`, hence we can say that
`P (x) = (1)/(3)` and
`Q (x) = 8` in the general form of the first order linear differential equation:

`(dy)/(dx) + P(x)y = Q(x)`

The integrating factor is given by:

`e^{\int{(1)/(3)} \, dx } = e^{(x)/(3)}`. Thus, multiplying the entire equation by the integrating factor:

`e^{(x)/(3)}(dy)/(dx) + (1)/(3)e^{(x)/(3)}y = 8e^{(x)/(3)}`. This means that:

`\frac{d[e^{(x)/(3)}y]}{dx} = 8e^{(x)/(3)}`

`e^{(x)/(3)}y = 24e^{(x)/(3)} + C` then

`y = 24 + Ce^{-(x)/(3)}`. Applying the initial condition:

`C = -23e^{(1)/(3)}` and therefore,
`y = 24 -23e^{-(1)/(3)(x-1)}`