Answer to Question #325141 in Differential Equations for Ankush

Question

x(dy)/(dx)=x^(2) + 5y

Answer 1

$x\frac{dy}{dx}=x^2+5y$

$y’-5\frac{y}{x}=x$ (1)

Let’s solve the following equation 

$y’-5\frac{y}{x}=0$

$\frac {dy}{y}=5\frac{dx}{x}$

$y=C(x)x^{5}$

To find solution of the equation (1) we should think C is a function of x

$y'=C'x^{5}+5Cx^{4}$

Substitution y and y’ into (1) gives

$C'x^{5}+5Cx^{4} -5\frac{Cx^{5}}{x}=x$

$C'=x^{-4}$

$C=\frac{1}{-3x^3}+C_1$

$y=(\frac{1}{-3x^3}+C_1)x^5=-\frac13x^2+C_1x^5$

Answer: $y=(\frac{1}{-3x^3}+C_1)x^5=-\frac13x^2+C_1x^5$ .