Answer:
[tex]y = 8 -7e^{-3(x-1)}[/tex]
Step-by-step explanation:
Assuming that the equation is [tex]y '= 24 -3y[/tex] with initial condition [tex]y(1) = 1[/tex]. We have,
[tex]\frac{dy}{dx} + 3y = 24[/tex], hence we can say that [tex]P (x) = 3[/tex] and [tex]Q (x) = 24[/tex] in the general form of the first order linear differential equation:
[tex]\frac{dy}{dx} + P(x)y = Q(x)[/tex]
The integrating factor is given by:
[tex]e^{\int{3} \, dx } = e^{3x}[/tex]. Thus, multiplying the entire equation by the integrating factor:
[tex]e^{3x}\frac{dy}{dx} + 3e^{3x}y = 24e^{3x}[/tex]. This means that:
[tex]\frac{d[e^{3x}y]}{dx} = 24e^{3x}[/tex]
[tex]e^{3x}y = 8e^{3x} + C[/tex] then
[tex]y = 8 + Ce^{-3x}[/tex]. Applying the initial condition:
[tex]C = -7e^{3}[/tex] and therefore, [tex]y = 8 -7e^{-3(x-1)}[/tex]
Assuming that the equation is [tex]y = 24 -3y' [/tex] with initial condition [tex]y(1) = 1[/tex]. We have,
[tex]\frac{dy}{dx} + \frac{1}{3}y = 8[/tex], hence we can say that [tex]P (x) = \frac{1}{3}[/tex] and [tex]Q (x) = 8[/tex] in the general form of the first order linear differential equation:
[tex]\frac{dy}{dx} + P(x)y = Q(x)[/tex]
The integrating factor is given by:
[tex]e^{\int{\frac{1}{3}} \, dx } = e^{\frac{x}{3}}[/tex]. Thus, multiplying the entire equation by the integrating factor:
[tex]e^{\frac{x}{3}}\frac{dy}{dx} + \frac{1}{3}e^{\frac{x}{3}}y = 8e^{\frac{x}{3}}[/tex]. This means that:
[tex]\frac{d[e^{\frac{x}{3}}y]}{dx} = 8e^{\frac{x}{3}}[/tex]
[tex]e^{\frac{x}{3}}y = 24e^{\frac{x}{3}} + C[/tex] then
[tex]y = 24 + Ce^{-\frac{x}{3}}[/tex]. Applying the initial condition:
[tex]C = -23e^{\frac{1}{3}}[/tex] and therefore, [tex]y = 24 -23e^{-\frac{1}{3}(x-1)}[/tex]
