Respuesta :
Answer:
a) Linear
b) Linear
c) Linear
d) Neither
See explanation below.
Step-by-step explanation:
a) [tex] \frac{dy}{dx} +e^x y = x^2 y^2 [/tex]
For this case the differential equation have the following general form:
[tex] y' +p(x) y = q(x) y^n[/tex]
Where [tex] p(x) =e^x [/tex] and [tex]q(x) = x^2[/tex] and since n>1 we can see that is a linear differential equation.
b) [tex] y + sin x = x^3 y'[/tex]
We can rewrite the following equation on this way:
[tex] y' -\frac{1}{x^3} y= \frac{sin (x)}{x^3}[/tex]
For this case the differential equation have the following general form:
[tex] y' +p(x) y = q(x) y^n[/tex]
Where [tex] p(x) =-\frac{1}{x^3} [/tex] and [tex]q(x) = \frac{sin(x)}{x^3}[/tex] and since n=0 we can see that is a linear differential equation.
c) [tex]ln x -x^2 y =xy'[/tex]
For this case we can write the differential equation on this way:
[tex] y' +xy = \frac{ln(x)}{x}[/tex]
For this case the differential equation have the following general form:
[tex] y' +p(x) y = q(x) y^n[/tex]
Where [tex] p(x) =x [/tex] and [tex]q(x) = \frac{ln(x)}{x}[/tex] and since n=0 we can see that is a linear differential equation.
d) [tex]\frac{dy}{dx} + cos y = tan x[/tex]
For this case we can't express the differential equation in terms:
[tex] y' +p(x) y = q(x) y^n[/tex]
So the is not linear, and since we can separate the variables in order to integrate is not separable. So then the answer for this one is neither.
