Question

Determine whether each first-order differential equation is separable, linear, both, or neither.

Answer 1

Answer:

1. $\frac{dy}{dx}+e^xy=x^2y^2$. It is not a first-order linear differential equation. And it's not separable either.

2. $y+\sin \left(x\right)=x^3y'\:$. It is a first-order linear differential equation.

3. $\ln \left(x\right)-x^2y=xy'\:$. It is a first-order linear differential equation.

4. $\frac{dy}{dx} +\cos \left(y\right)=\tan \left(x\right)$. 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

$\frac{dy}{dx} +P(x) y=Q(x)$

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,

$\frac{dy}{dx}=F(x,y)$,

the right-hand side can then be factored as “a formula of just x” times “a formula of just y”,

$F(x,y)=f(x)g(y)$

If this factoring is not possible, the equation is not separable.

Applying the above definitions, we get that

1. For $\frac{dy}{dx}+e^xy=x^2y^2$

$\frac{dy}{dx}+e^xy=x^2y^2\\\\\frac{dy}{dx}=x^2y^2-e^xy\\\\\frac{dy}{dx}=y(x^2y-e^x)$

It is not a first-order linear differential equation. And it's not separable either.

2. For $y+\sin \left(x\right)=x^3y'\:$

$x^3y'=y+\sin \left(x\right)\\\\x^3y'-y=\sin \left(x\right)\\\\y'\:-\frac{1}{x^3}y=\frac{\sin \left(x\right)}{x^3}$

It is a first-order linear differential equation.

3. For $\ln \left(x\right)-x^2y=xy'\:$

$xy'=\ln \left(x\right)-x^2y\\\\xy'+x^2y=\ln \left(x\right)\\\\y'\:+xy=\frac{\ln \left(x\right)}{x}$

It is a first-order linear differential equation.

4. For $\frac{dy}{dx} +\cos \left(y\right)=\tan \left(x\right)$

It is not a first-order linear differential equation. And it's not separable either.