Differential stuff!

y/x+ dy/dx =[f(xy)/f '(xy)]
IF=x

yx=∫x.[f(xy)/f '(xy)] dx

solve it now

y= λ[sin(3x+α)] - μe^(2x)

=> 3

finishing off with qsn 1...

y + x dydx = x f(xy) f'(xy)

yx + dydx = f(xy) f'(xy)

I.F. = e^{∫1x dx} = x

so...

y + x dydx = x f(xy) f '(xy)

d(xy) = x f(xy) f '(xy) dx

∫ f '(xy) f (xy) d(xy) = ∫ x dx

ln (f(xy)) = x²2 + ln k

or f(xy) = k e^x2/2

so answer a...