[tex]y'-3xy=y^5x^3[/tex]
Divide both sides by [tex]y^5[/tex]:
[tex]y^{-5}y'-3xy^{-4}=x^3[/tex]
Substitute [tex]u(x)=y(x)^{-4}[/tex], so that [tex]u'=-4y^{-5}y'[/tex]. Then we get a new ODE that is linear in [tex]u[/tex],
[tex]-\dfrac{u'}4-3xu=x^3\implies u'+12xu=4x^3[/tex]
Multiply both sides by [tex]e^{6x^2}[/tex]:
[tex]e^{6x^2}u'+12xe^{6x^2}u=4x^3e^{6x^2}[/tex]
Notice that [tex]\left(e^{6x^2}u(x)\right)'=e^{6x^2}u'(x)+12xe^{6x^2}u(x)[/tex]. Integrating both sides with respect to [tex]x[/tex] gives
[tex]\displaystyle\int\left(e^{6x^2}u\right)'\,\mathrm dx=e^{6x^2}u=\int 4x^3e^{6x^2}\,\mathrm dx[/tex]
[tex]\implies u=\dfrac{e^{6x^2}(6x^2-1)}{18}+C[/tex]
[tex]\implies y^4=\dfrac1{\frac{e^{6x^2}(6x^2-1)}{18}+C}[/tex]
Given that [tex]y(1)=1[/tex], we find
[tex]1=\dfrac1{\frac{5e^6}{18}+C}\implies C=\dfrac{18-5e^6}{18}[/tex]
so the particular solution is
[tex]y^4=\dfrac1{\frac{e^{6x^2}(6x^2-1)}{18}+\frac{18-5e^6}{18}}[/tex]
[tex]\implies\boxed{y=\sqrt[4]{\dfrac1{\frac{e^{6x^2}(6x^2-1)}{18}+\frac{18-5e^6}{18}}}}[/tex]
where we take the positive root because the initial condition tells us to expect [tex]y>0[/tex] when [tex]x=1[/tex].
