Answer:
a) Linear
b) Linear
c) Linear
d) Neither
See explanation below.
Step-by-step explanation:
a) ![\frac{dy}{dx} +e^x y = x^2 y^2](https://tex.z-dn.net/?f=%20%5Cfrac%7Bdy%7D%7Bdx%7D%20%2Be%5Ex%20y%20%3D%20x%5E2%20y%5E2%20)
For this case the differential equation have the following general form:
![y' +p(x) y = q(x) y^n](https://tex.z-dn.net/?f=%20y%27%20%2Bp%28x%29%20y%20%3D%20q%28x%29%20y%5En)
Where
and
and since n>1 we can see that is a linear differential equation.
b) ![y + sin x = x^3 y'](https://tex.z-dn.net/?f=%20y%20%2B%20sin%20x%20%3D%20x%5E3%20y%27)
We can rewrite the following equation on this way:
![y' -\frac{1}{x^3} y= \frac{sin (x)}{x^3}](https://tex.z-dn.net/?f=%20y%27%20-%5Cfrac%7B1%7D%7Bx%5E3%7D%20y%3D%20%5Cfrac%7Bsin%20%28x%29%7D%7Bx%5E3%7D)
For this case the differential equation have the following general form:
![y' +p(x) y = q(x) y^n](https://tex.z-dn.net/?f=%20y%27%20%2Bp%28x%29%20y%20%3D%20q%28x%29%20y%5En)
Where
and
and since n=0 we can see that is a linear differential equation.
c) ![ln x -x^2 y =xy'](https://tex.z-dn.net/?f=ln%20x%20-x%5E2%20y%20%3Dxy%27)
For this case we can write the differential equation on this way:
![y' +xy = \frac{ln(x)}{x}](https://tex.z-dn.net/?f=%20y%27%20%2Bxy%20%3D%20%5Cfrac%7Bln%28x%29%7D%7Bx%7D)
For this case the differential equation have the following general form:
![y' +p(x) y = q(x) y^n](https://tex.z-dn.net/?f=%20y%27%20%2Bp%28x%29%20y%20%3D%20q%28x%29%20y%5En)
Where
and
and since n=0 we can see that is a linear differential equation.
d) ![\frac{dy}{dx} + cos y = tan x](https://tex.z-dn.net/?f=%5Cfrac%7Bdy%7D%7Bdx%7D%20%2B%20cos%20y%20%3D%20tan%20x)
For this case we can't express the differential equation in terms:
![y' +p(x) y = q(x) y^n](https://tex.z-dn.net/?f=%20y%27%20%2Bp%28x%29%20y%20%3D%20q%28x%29%20y%5En)
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.