Respuesta :
Answer:
1. [tex]\frac{dy}{dx}+e^xy=x^2y^2[/tex]. It is not a first-order linear differential equation. And it's not separable either.
2. [tex]y+\sin \left(x\right)=x^3y'\:[/tex]. It is a first-order linear differential equation.
3. [tex]\ln \left(x\right)-x^2y=xy'\:[/tex]. It is a first-order linear differential equation.
4. [tex]\frac{dy}{dx} +\cos \left(y\right)=\tan \left(x\right)[/tex]. It is not a first-order linear differential equation. And it's not separable either.
Step-by-step explanation:
Definition 1: A first-order linear differential equation is one that can be put into the form
[tex]\frac{dy}{dx} +P(x) y=Q(x)[/tex]
where P and Q are continuous functions on a given interval.
Definition 2: A first-order differential equation is said to be separable if, after solving it for the derivative,
[tex]\frac{dy}{dx}=F(x,y)[/tex],
the right-hand side can then be factored as “a formula of just x” times “a formula of just y”,
[tex]F(x,y)=f(x)g(y)[/tex]
If this factoring is not possible, the equation is not separable.
Applying the above definitions, we get that
1. For [tex]\frac{dy}{dx}+e^xy=x^2y^2[/tex]
[tex]\frac{dy}{dx}+e^xy=x^2y^2\\\\\frac{dy}{dx}=x^2y^2-e^xy\\\\\frac{dy}{dx}=y(x^2y-e^x)[/tex]
It is not a first-order linear differential equation. And it's not separable either.
2. For [tex]y+\sin \left(x\right)=x^3y'\:[/tex]
[tex]x^3y'=y+\sin \left(x\right)\\\\x^3y'-y=\sin \left(x\right)\\\\y'\:-\frac{1}{x^3}y=\frac{\sin \left(x\right)}{x^3}[/tex]
It is a first-order linear differential equation.
3. For [tex]\ln \left(x\right)-x^2y=xy'\:[/tex]
[tex]xy'=\ln \left(x\right)-x^2y\\\\xy'+x^2y=\ln \left(x\right)\\\\y'\:+xy=\frac{\ln \left(x\right)}{x}[/tex]
It is a first-order linear differential equation.
4. For [tex]\frac{dy}{dx} +\cos \left(y\right)=\tan \left(x\right)[/tex]
It is not a first-order linear differential equation. And it's not separable either.
