Question

Determine whether each first-order differential equation is separable, linear, both, or neither. 1. ????y????x+????xy=x2y2 2. y+sinx=x3y′ 3. lnx−x2y=xy′ 4. ????y????x+cosy=tanx

Answer 1

Answer:

a) Linear

b) Linear

c) Linear

d) Neither

See explanation below.

Step-by-step explanation:

a) $\frac{dy}{dx} +e^x y = x^2 y^2$

For this case the differential equation have the following general form:

$y' +p(x) y = q(x) y^n$

Where $p(x) =e^x$ and $q(x) = x^2$ and since n>1 we can see that is a linear differential equation.

b) $y + sin x = x^3 y'$

We can rewrite the following equation on this way:

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

For this case the differential equation have the following general form:

$y' +p(x) y = q(x) y^n$

Where $p(x) =-\frac{1}{x^3}$ and $q(x) = \frac{sin(x)}{x^3}$ and since n=0 we can see that is a linear differential equation.

c) $ln x -x^2 y =xy'$

For this case we can write the differential equation on this way:

$y' +xy = \frac{ln(x)}{x}$

For this case the differential equation have the following general form:

$y' +p(x) y = q(x) y^n$

Where $p(x) =x$ and $q(x) = \frac{ln(x)}{x}$ and since n=0 we can see that is a linear differential equation.

d) $\frac{dy}{dx} + cos y = tan x$

For this case we can't express the differential equation in terms:

$y' +p(x) y = q(x) y^n$

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.