Question

solve the differential equations ysin2xdx= (1+y^2+ cos^2x)dy​

Answer 1

Answer:

Write in standard form Pdx + Qdy = 0 :

ysin(2x)dx + (−1−y²−cos²(x))dy = 0

∂P/∂y = sin(2x) and ∂Q/∂x

= − 2cos(x)(−sin(x))

= 2sin(x)cos(x)

But sin(2x) = 2sin(x)cos(x)

so ∂P/∂y = ∂Q/∂x and so equation is exact differential

∫ Pdx = −½ycos(2x) + g(y) and

∫ Qdy = −y − ⅓y³ − ycos²(x) + h(x)

Noting that cos²(x) = ½(1+cos(2x)) : g(y) = −y − ⅓y³ − ½y and h(x) = 0

∴ integrated differential is −½ycos(2x) −3y/2 − ⅓y³ = c