Answer:
Linear and non-homogeneous.
Step-by-step explanation:
We are given that
[tex]\frac{xy'}{(x^2+7)y}=cosx+\frac{e^{3x}}{y}[/tex]
We have to convert into y'+P(x)y=g(x) and determine P(x) and g(x).
We have also find type of differential equation.
[tex]y'=\frac{(x^2+7)y}{x}(cosx+\frac{e^{3x}}{y}}[/tex]
[tex]y'=\frac{(x^2+7)cosx}{x}y+\frac{(x^2+7)e^{3x}}{x}[/tex]
[tex]y'-\frac{cosx(x^2+7)}{x}y=\frac{e^{3x}(x^2+7)}{x}[/tex]
It is linear differential equation because this equation is of the form
y'+P(x)y=g(x)
Compare it with first order first degree linear differential equation
[tex]y'+P(x)y=g(x)[/tex]
[tex]P(x)=-\frac{cosx (x^2+7)}{x},g(x)=\frac{e^{3x}(x^2+7)}{x}[/tex]
[tex]\frac{dy}{dx}=\frac{(x^2+7)(ycosx+e^{3x})}{x}[/tex]
Homogeneous equation
[tex]\frac{dy}{dx}=\frac{f(x,y)}{g(x,y)}[/tex]
Degree of f and g are same.
[tex]f(x,y)=(x^2+7)(ycosx+e^{3x}),g(x,y)=x[/tex]
Degree of f and g are not same .
Therefore, it is non- homogeneous .
Linear and non-homogeneous.