Respuesta :
Answer:
A) e^(1/y)
Step-by-step explanation:
Re-write the equation
[tex]\displaystyle (x+2y^3)\frac{dy}{dx}=y^2\\\\\frac{dy}{dx}=\frac{y^2}{x+2y^3}\\ \\\frac{dx}{dy}=\frac{x+2y^3}{y^2}\\ \\ \frac{dx}{dy}=\frac{x}{y^2}+2y\\ \\\frac{dx}{dy}-\frac{x}{y^2}=2y[/tex]
Since we have a first-order linear differential equation in the form of [tex]\displaystyle \frac{dx}{dy}+p(y)x=q(y)[/tex], then the integrating factor is [tex]\displaystyle IF=e^{\int {p(y)} \, dy}[/tex]. Comparing with the above form, we have [tex]\displaystyle p(x)=-\frac{1}{y^2}[/tex] and [tex]q(x)=2y[/tex]:
[tex]\displaystyle IF=e^{\int {p(y)} \, dy}\\\\IF=e^{\int {-\frac{1}{y^2} } \, dy}\\\\IF=e^\frac{1}{y}[/tex]
Thus, A is the correct answer
the answer is A
(x+2y³)dy/dx = y²
dx/dy = x/y² + 2y
Integrating factor=
IF= e^(∫−1/y^2 dy)
IF= e^(-∫1/y^2 dy)
IF= e^(-(-y^-1))
IF= e^1/y
(x+2y³)dy/dx = y²
dx/dy = x/y² + 2y
Integrating factor=
IF= e^(∫−1/y^2 dy)
IF= e^(-∫1/y^2 dy)
IF= e^(-(-y^-1))
IF= e^1/y
