The general solution of xy' = x^3 cos x + y is

Rearrange to standard form. xy' - y = (x^3)*(cos x)

y' - (1/x)*y = (x^2)*(cos x), The integrating factor works out nicely to 1/x

y'/x - y/(x^2) = x*(cos x)

(y/x)' = x*(cos x)

Integrate both sides with respect to x. Hint: use integration by parts for the integral of x*(cos x)

y/x = x*(sin x) + cos x + C

Therefore, the solution is: y = (x^2)(sin x) + x*(cos x) + C*x ; Where C is an arbitrary constant

xy' = x^3*cos(x) + y

y' = x^2*cos(x) + y/x

y' - y/x = x^2*cos(x)

Integrating factor = e^(∫-1/x dx) = 1/x

Multiplying both sides by integrating factor:

y'/x - y/x² = x*cos(x)

d/dx[y/x] = x*cos(x)

integrating both sides:

y/x = x*sin(x) + cos(x) + C

y = x²*sin(x) + x*cos(x) + Cx

y=(-x^3)cos(x)