Answer:
[tex]p(t) = \frac{9t^{3} + 729t - 1}{e^{t}(t^{2} + 81) }[/tex]
g(t) = 0
And
The differential equation [tex]9ty + e^{t}y' = \frac{y}{t^{2} + 81 }[/tex] is linear and homogeneous
Step-by-step explanation:
Given that,
The differential equation is -
[tex]9ty + e^{t}y' = \frac{y}{t^{2} + 81 }[/tex]
[tex]e^{t}y' + (9t - \frac{1}{t^{2} + 81 } )y = 0\\e^{t}y' + (\frac{9t(t^{2} + 81 ) - 1}{t^{2} + 81 } )y = 0\\e^{t}y' + (\frac{9t^{3} + 729t - 1}{t^{2} + 81 } )y = 0\\y' + [\frac{9t^{3} + 729t - 1}{e^{t}(t^{2} + 81) } ]y = 0[/tex]
By comparing with y′+p(t)y=g(t), we get
[tex]p(t) = \frac{9t^{3} + 729t - 1}{e^{t}(t^{2} + 81) }[/tex]
g(t) = 0
And
The differential equation [tex]9ty + e^{t}y' = \frac{y}{t^{2} + 81 }[/tex] is linear and homogeneous.
