Answer to Question #139576 in Differential Equations for Moel Tariburu

Question

Solve the differential equations xy' = x^2 + 3y, for x.

Answer 1

xy'=x²+3y

xy'-3y=x²

divide through by x

(dy/dx)-(3/x)y=x

when we compare with the standard form

(dy/dx)+P(x)y=Q(x)

now we have

P(x)=(-3/x)*Q(x)=x

since $\int$ p(x)dx=-3$\int$ (1/x)dx

=-3lnx

=lnx^-3

integrating factor

I=e$\int$ ^q(x)dx

=e^lnx-q

=x^-3

x^-3(dy/dx)-3x^-4y=x^-1

d/dx(x^-3y)=x^-1

=x^-3y=$\int$ (1/x)dx

=x^-3y=ln |x|+C

y=x³ln|x|+Cx³