Solve differential equation y3y′+1xy4=sin⁡xx4

Let $y^4=z$ then, $4y^3{y}^{\prime }=\frac{dz}{dx}$
$y^3{y}^{\prime }+\frac{1}{x}{y}^4=\frac{\sin x}{x^4}\Rightarrow \frac{1}{4}\frac{dz}{dx}+\frac{1}{x}z=\frac{\sin x}{x^4}$
$\frac{dz}{dx}+\frac{4}{x}z=\frac{4\sin x}{x^4}$
The given ordinary equation is of the form
$\frac{dz}{dx}+P(x)z=Q(x)$
Integrating factor: $IF=e^{\int Pdx}=e^{\int \frac{4}{x}}dx=x^4$
$\frac{d}{dx}(IFz)=IFQ(x)$
$IF\ast z=\int IF\ast 1(x)dx$
$x^4\ast z=\int (x^4)(\frac{4\sin x}{x})dx$
Multiply the original differential equation by the integral factor
$\frac{d}{dx}(IF\ast z)=IF\ast Q(x)$
$IF\ast z=\int IF\ast Q(x)dx$
$x^4\ast z=\int (x^4)(\frac{4\sin x}{x^4})dx$
$x^4\ast z=\int (4\sin x)dx$
$x^{4}z=4\int \sin xdx$
$x^{4}z=-4\cos x+c$
$z=\frac{-4\cos x}{x^4}+\frac{c}{x^4}$
Put the value of z in the obtained equation
$z=\frac{-4\cos x}{x^4}+\frac{c}{x^4}$
$z=y^4$
$y^4=\frac{-4\cos x}{x^4}+\frac{c}{x^4}$
${\displaystyle y=\sqrt[\frac{1}{4}]{(\frac{-4\cos x}{x^4}+\frac{c}{x^4})}}$