Question

solve y+dx(xy)=x(sinx+logx)

Answer 1

y + d/dx(xy) = x(sin x + ln x)
y + y + xy' = x (sin x + ln x)
2y + xy' = x (sin x + ln x)
2xy + x²y' = x² (sin x + ln x)
(d/dx x²)y + x²y' = x² (sin x + ln x)
d/dx (x²y) = x² (sin x + ln x)
∫ d/dx (x²y) dx = ∫ x² (sin x + ln x) dx
x²y = ∫ x² (sin x + ln x) dx
x²y = ⅓x³(sin x + ln x) − ∫ ⅓x³ (cos x + 1/x) dx
x²y = ⅓x³(sin x + ln x) − ∫ ⅓x³ cos x + ⅓x² dx
x²y = ⅓x³(sin x + ln x) − ∫ ⅓x³ cos x dx + x³ / 9
x²y = ⅓x³(sin x + ln x) + x³ / 9 − ∫ ⅓x³ cos x dx
y = ⅓x(sin x + ln x) + x / 9 − [ ∫ ⅓x³ cos x dx ] / x²