Question

x^4dy/dx+x^3y=-sec(xy)​

Answer 1

$x^4\dfrac{\mathrm dy}{\mathrm dx}+x^3y=-\sec(xy)$

Divide both sides by $x^3$:

$x\dfrac{\mathrm dy}{\mathrm dx}+y=\dfrac{\mathrm d}{\mathrm dx}[xy]=-\dfrac{\sec(xy)}{x^3}$

Integrate both sides to get

$xy=y_0=\displaystyle\int_{x_0}^x\frac{\sec(ty)}{t^3}\,\mathrm dt$

where $x_0$ and $y_0=y(x_0)$ are constants, and divide by $x$ to solve for $y(x)$:

$\boxed{y(x)=\dfrac{y_0}x+\displaystyle\frac1x\int_{x_0}^x\frac{\sec(ty)}{t^3}\,\mathrm dt}$

An exact solution (not in terms of a definite integral, anyway) doesn't seem likely...