Question

Put the differential equation 9ty+ety′=yt2+81 into the form y′+p(t)y=g(t) and find p(t) and g(t). p(t)= help (formulas) g(t)= help (formulas) Is the differential equation 9ty+ety′=yt2+81 linear and homogeneous, linear and nonhomogeneous, or nonlinear?

Answer 1

Answer:

$p(t) = \frac{9t^{3} + 729t - 1}{e^{t}(t^{2} + 81) }$

g(t) = 0

And

The differential equation $9ty + e^{t}y' = \frac{y}{t^{2} + 81 }$ is  linear and homogeneous

Step-by-step explanation:

Given that,

The differential equation is -

$9ty + e^{t}y' = \frac{y}{t^{2} + 81 }$

$e^{t}y' + (9t - \frac{1}{t^{2} + 81 } )y = 0\\e^{t}y' + (\frac{9t(t^{2} + 81 ) - 1}{t^{2} + 81 } )y = 0\\e^{t}y' + (\frac{9t^{3} + 729t - 1}{t^{2} + 81 } )y = 0\\y' + [\frac{9t^{3} + 729t - 1}{e^{t}(t^{2} + 81) } ]y = 0$

By comparing with y′+p(t)y=g(t), we get

$p(t) = \frac{9t^{3} + 729t - 1}{e^{t}(t^{2} + 81) }$

g(t) = 0

And

The differential equation $9ty + e^{t}y' = \frac{y}{t^{2} + 81 }$ is  linear and homogeneous.