The equation of the curve passing through the origin, in the form y = f ( x ) , which satisfies the differential equation dy dx = sin 1 0 x + 6 y   , is

Step-by-step solution

Put 10x + 6y = z; then 1 0 + 6 dy dx = dz dx ∴ dy dx = 1 6 dz dx - 1 0 ∴ the equation becomes 1 6 dz dx - 1 0 = sin z or dz dx = 6 sin z + 1 0 ; ∴ dz 6 sin z + 1 0 = dx or 1 2 ∫ dz 3 sin z + 5 = ∫ dx ∴ 1 2 ∫ dz 3 · 2 tan z 2 1 + tan 2 z 2 + 5 = x + c or 1 2 ∫ sec 2 z 2 dz 5 tan 2 z 2 + 6 tan z 2 + 5 = x + c or 1 2 ∫ 2 dt 5 t 2 + 6 t + 5 = x + c Putting tan z 2 = t or 1 5 ∫ dt t 2 + 6 5 t + 1 = x + c or 1 5 ∫ dt t + 3 5 2 + 1 - 9 2 5 = x + c or 1 5 ∫ dt t + 3 5 2 + 4 5 2 = x + c ∴ 1 5 · 1 4 5 tan -1 t + 3 5 4 5 = x