Answer to Question #281492 in Mechanics | Relativity for Tanaya

Question

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

Answer 1

$y'-5y/x=x$

for

$y'-5y/x=0$ :

$dy/y=5dx/x$

$lny_1=5lnx+lnc_1$

$y_1=c_1x^5$

then:

$y=x^5c(x)$

$5x^4c(x)+x^5c'(x)-5x^4c(x)=x$

$c'(x)=1/x^4$

$c(x)=C-\frac{1}{3x^3}$

$y=(C-\frac{1}{3x^3})x^5$